Output of a char pattern program

Here are 10 pattern type problems statements with solutions. All the programs are written in C language.  This post is a part of the pattern programs series.

1. Write a C program to print the following pattern:

   
   A
   B A
   C B A
   D C B A
   E D C B A

2. Write a C program to print the following pattern:

   A B C D E
   B C D E
   C D E
   D E
   E
   

3. Write a C program to print the following pattern:

            A
   a   a
   B   B   B
   b   b   b   b
   C   C   C   C   C
   c   c   c   c
   D   D   D
   d   d
   E
   

4. Write a C program to print the following pattern:

   
   A
   Z
   B   C
   Y   X
   D   E   F
   W   V   U
   G   H   I   J
   T   S   R   Q
   K   L   M   N   O
   

5. Write a C program to print the following pattern:

    A   B   C   D   E
   Z   Y   X   W
   A   B   C   D
   Z   Y   X
   A   B   C
   Z   Y
   A   B
   Z
   A
   Z
   A   B
   Z   Y
   A   B   C
   Z   Y   X
   A   B   C   D
   Z   Y   X   W
   A   B   C   D   E
   

6. Write a C program to print the following pattern:

   A B C D E D C B A
   A B C D   D C B A
   A B C       C B A
   A B           B A
   A               A
   A B           B A
   A B C       C B A
   A B C D   D C B A
   A B C D E D C B A
   
   

7. Write a C program to print the following pattern:

              A
   B C D
   E F G H I
   J K L M N O P
   Q R S T U V W X Y
   Z
   Y X W V U T S R Q
   P O N M L K J
   I H G F E
   D C B
   A
   

8. Write a C program to print the following pattern:

   N     N
   N N   N
   N   N N
   N     N
   

9. Write a C program to print the following pattern:

    X               X
   X           X
   X       X
   X   X
   X
   X   X
   X       X
   X           X
   X               X
   

10. Write a C program to print the following pattern:

    K             K
    K           K
    K         K
    K       K
    K     K
    K   K
    K K
    K
    K K
    K   K
    K     K
    K       K
    K         K
    K           K
    K             K
    

1. Write a C program to print the following pattern:

A
B A
C B A
D C B A
E D C B A

Program:

#include <stdio.h>

int main() {
	char prnt = 'A';
	int i, j;
	for (i = 1; i <= 5; i++) { // Prints the right angle triangle
		for (j = 1; j <= i; j++) {
			printf("%2c", ((prnt + (i - j))));  // Prints the characters in the required order
		}
		printf("\n");
	}
	return 0;
}

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2. Write C program to print the following pattern:

   A B C D E
     B C D E
       C D E
         D E
           E

Program:

#include<stdio.h>
int main() {
	int i, j;
	char c = 'A';
	for (i = 0; i < 5; i++) {
		for (j = 0; j < 5; j++) {
			if (j < i) {
				printf(" ");
			} else {
				printf("%c", c + j);  // The order of the alphabets are increasing in terms of j
			}
			printf(" ");
		}
		printf("\n");
	}
	return 0;
}

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3. Write C program to print the following pattern:

           A
         a   a
       B   B   B
     b   b   b   b
   C   C   C   C   C
     c   c   c   c
       D   D   D
         d   d
           E

Program:

#include<stdio.h>
int main() {
	char prnt_even = 'a';
	char prnt_odd = 'A';
	int i, j, s, nos = 4,line=0;  //line tracks if the row is odd or even
	for (i = 1; i <= 9; (i = i + 2)) {
		for (s = nos; s >= 1; s--) {  // For the required spacings
			printf("  ");
		}
		for (j = 1; j <= i; j++) {
			if(line==0){
				if ((i % 2) != 0 && (j % 2) != 0) {
					printf("%2c", prnt_odd);   // Prints capital letter if l=0
					} else {
						printf("  ");
					}
				}
				else{
					if ((i % 2) != 0 && (j % 2) != 0) {
						printf("%2c", prnt_even);  //else prints a small letter
						} else {
							printf("  ");
						}
					}
				}
/* After one column the value of line is interchanged and prnt_even or prnt_odd is incremented accordingly */
				if(line==0){
					prnt_odd++;
					line=1;
				}
				else{
					prnt_even++;
					line=0;
				}
			nos--;  //no of space decremented by 1
			printf("\n");
		}
	nos = 1;
	/* Second half of the diamond. This one skips its first row rest goes pretty much the same */
	for (i = 7; i >= 1; (i = i - 2)) {
		for (s = nos; s >= 1; s--) {
			printf("  ");
		}
		for (j = 1; j <= i; j++) {
			if(line==0){
				if ((i % 2) != 0 && (j % 2) != 0) {
					printf("%2c", prnt_odd);
					} else {
						printf("  ");
					}
				}
				else{
					if ((i % 2) != 0 && (j % 2) != 0) {
						printf("%2c", prnt_even);
						} else {
							printf("  ");
						}
					}
				}
				if(line==0){
					prnt_odd++;
					line=1;
				}
				else{
					prnt_even++;
					line=0;
				}
				nos++;  //no of spaces is incremented by one
				printf("\n");
			}
			return 0;
		}

Explanation:
This is a diamond formation where all the even places are hollowed. In case of odd row no a capital letter is printed and a small case alphabet otherwise. Both the capital letters n the small letters increment by one starting from ‘A’ or ‘a’.
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4. Write a C program to print the following pattern:

    A
      Z
    B   C
      Y   X
    D   E   F
      W   V   U
    G   H   I   J
      T   S   R   Q
    K   L   M   N   O

Program:

#include <stdio.h>
int main() {
	char prnt_odd = 'A';
	char prnt_even = 'Z';
	int i, j;
	// The right angle has nine rows
	for (i = 1; i <= 9; i++) {
		for (j = 1; j <= i; j++) {
			//checks for the column printing restrictions..
			if ((i % 2) != 0 && (j % 2) != 0) {
				printf("%2c", prnt_odd);
				prnt_odd++;
			} else if ((i % 2) == 0 && (j % 2) == 0) {
				printf("%2c", prnt_even);
				prnt_even--;
			} else {
				printf("  ");
			}
		}
		printf("\n");
	}
	return 0;
}

Explanation:
This is a right angle triangle of alphabets, having two sets of alphabets. One for the odd rows which starts from ‘A’ and increments by one. And another one for the even rows which starts from ‘Z’ and decrements by 1. The restriction in the columns are: if the row no is odd only the odd places are printed else if the row no is even only the even places are filled, rest are skipped.
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5. Write a C program to print the following pattern:

    A   B   C   D   E
      Z   Y   X   W
    A   B   C   D
      Z   Y   X
    A   B   C
      Z   Y
    A   B
      Z
    A
      Z
    A   B
      Z   Y
    A   B   C
      Z   Y   X
    A   B   C   D
      Z   Y   X   W
    A   B   C   D   E

Program:

#include <stdio.h>
//This prints the columns with guidance of the row number supplied.
int triangle(int i)
{
	char prnt_odd = 'A';
	char prnt_even= 'Z';
	int j;
	for(j=1; j<=i; j++)
	{
		//Checks for the restrictions
		if((i%2)!=0 && (j%2)!=0)
		{
			printf("%2c",prnt_odd);
			prnt_odd++;
		}
		else if((i%2)==0 && (j%2)==0)
		{
			printf("%2c",prnt_even);
			prnt_even--;
		}
		else
		{
			printf("  ");
		}
	}
	return 0;
}

int main()
{
	int i;
	//Prints the first right angle triangle with 9 rows
	for(i=9; i>=1; i--)
	{
		triangle(i);
		printf("\n");
	}
	// This prints the second right angle triangle skipping its tip.
	for(i=2; i<=9; i++)
	{
		triangle(i);
		printf("\n");
	}
	return 0;
}

Explanation:
This pattern can be seen as two right angle triangles of alphabets, having the same tip with two sets of alphabets. One for the odd rows which starts from ‘A’ and increments by one. And another one for the even rows which starts from ‘Z’ and decrements by 1. The restriction in the columns are: if the row number is odd, only the odd places are printed. Else if the row number is even only the even places are filled, rest are skipped.
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6. Write a C program to print the following pattern:

   A B C D E D C B A
   A B C D   D C B A
   A B C       C B A
   A B           B A
   A               A
   A B           B A
   A B C       C B A
   A B C D   D C B A
   A B C D E D C B A

Program:

#include <stdio.h>
/*
 nos   = Number of Spaces required.
 i     = Number of columns
 skip  = The row in which we have to skip the first character
 form  = Whether the alphabets has to printed in Incremental or Decremental order
 */
int triangle(int nos, int i, int skip, int form) {
	char prnt[] = "ABCDE";
	int j, s;
	for (s = nos; s >= 1; s--) {   // Controls the required spacings
		printf("  ");
	}
	for (j = 0; j <= i; j++) {     // prints the characters row wise
		if (skip != 1) {           // If we need to skip a row.
			if (i == skip && j == 0) {
				continue;
			}
		}
		if (form == 0) { /* Decides whether the characters are in incremental or decremental order. */
			printf("%2c", prnt[j]);
		} else {
			printf("%2c", prnt[i - j]);
		}
	}
	return 0;
}

int main() {
	int i, nos = -1;
	for (i = 4; i >= 0; i--) {
		triangle(0, i, 1, 0);    // Prints the first triangle
		triangle(nos, i, 4, 1);  // Prints the second triangle
		nos = nos + 2;    // Spacing factor
		printf("\n");
	}
	nos = 5;
	for (i = 1; i <= 4; i++) {
		triangle(0, i, 0, 0);   // Prints the third triangle
		triangle(nos, i, 4, 1); // Prints the fourth triangle
		nos = nos - 2;     //Spacing factor.
		printf("\n");
	}
	return 0;
}

Explanation:
This pattern can be treated as four individual right angle triangles composed of alphabets in a sequential manner.
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7. Write a C program to print the following pattern:

              A
            B C D
          E F G H I
        J K L M N O P
      Q R S T U V W X Y
                        Z
                          Y X W V U T S R Q
                            P O N M L K J
                              I H G F E
                                D C B
                                  A

Program:

#include <stdio.h>
int main() {
	char prnt = 'A';
	int i, j, s, nos = 5;
	for (i = 1; i <= 9; (i = i + 2)) {
		for (s = nos; s >= 1; s--) {
			printf("  ");
		}
		for (j = 1; j <= i; j++) {
			printf("%2c", prnt);
			++prnt;           //Increments the alphabet
			if (i == 9 && j == 9) //for the extra Z
			{
				printf("\f%2c", prnt);
			}
		}
		nos--;
		//for the continuation
		if (i != 9) {
			printf("\n");
		}
	}
	printf("\f"); //Line Feed
	for (i = 9; i >= 1; (i = i - 2)) {
		//Maintaining the required Spaces
		if (i == 9) {
			nos = 0;
		} else if (i == 7) {
			nos = 12;
		}
		for (s = nos; s >= 1; s--) {
			printf("  ");
		}
		for (j = 1; j <= i; j++) {
			--prnt;   //decrements the alphabet before printing it.
			printf("%2c", prnt);
		}
		nos++;
		printf("\n");
	}
	return 0;
}

Explanation:
This can be seen as two isosceles triangle made up of alphabets incrementing by one starting from ‘A’. The connecting Z can be considered additionally.
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8. Write a C program to print the following pattern:

   N     N
   N N   N
   N   N N
   N     N

Program:

#include <stdio.h>
int main() {
	char prnt = 'N';
	int i, j, k;
	for (i = 1; i <= 4; i++) {
		//First right-angle triangle.
		for (j = 1; j <= i; j++) {
			if (j == 1 || j == i) {    // This Ensures an empty baseless triangle
				printf("%2c", prnt);   //Prints the character after 2 spaces.
			} else {
				printf("  ");
			}
		}
		//Second right-angle triangle
		for (k = 3; k >= i; k--) {
			if (k == i) {    // k=0 has to skipped as it shares the same hypotenuse
				if (i == 4) { //For the joining point
					break;
				}
				printf("%2c", prnt);
			} else {
				printf("  ");
			}
		}
		printf("\n");
	}
	return 0;
}

Explanation:
This pattern can be viewed as two baseless right-angle triangles arranged in such a way that they share a common hypotenuse.
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9. Write a C program to print the following pattern:

    X               X
      X           X
        X       X
          X   X
            X
          X   X
        X       X
      X           X
    X               X

Program:

#include <stdio.h>
int main() {
	char prnt = 'X';
	int i, j, s, nos = 0;  //nos controls the spacing.
	//1st triangle
	for (i = 9; i >= 1; (i = i - 2)) {
		for (s = nos; s >= 1; s--) {  //Spacing control
			printf("  ");
		}
		for (j = 1; j <= i; j++) {
			if (j == 1 || j == i) { //This hollows the triangle
				printf("%2c", prnt);
			} else {
				printf("  ");
			}
		}
		printf("\n");
		nos++;  // In the upper triangle the space increments by 1.
	}
/* Since both the triangles have the same peak point, while printing the second triangle we'll ignore its peak point. Thus i starts from 3 and the nos from 3.*/
	nos = 3;
	for (i = 3; i <= 9; (i = i + 2)) {
		for (s = nos; s >= 1; s--) {
			printf("  ");
		}
		for (j = 1; j <= i; j++) {
			if (j == 1 || j == i) {
				printf("%2c", prnt);
			} else {
				printf("  ");
			}
		}
		printf("\n");
		nos--; //The spaces are in a decrementing order.
	}
	return 0;
}

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10. Write a C program to print the following pattern:

    K             K
    K           K
    K         K
    K       K
    K     K
    K   K
    K K
    K
    K K
    K   K
    K     K
    K       K
    K         K
    K           K
    K             K

Program:

#include <stdio.h>
int main() {
	char prnt = 'K';
	int i, j;
	//First right-angle triangle
	for (i = 7; i >= 1; i--) {
		for (j = 1; j <= i; j++) {
			if (j == 1 || j == i) {  //This hollows the triangle
				printf("%2c", prnt);
			} else {
				printf("  ");
			}
		}
		if (i != 1) {
			printf("\n");
		}
	}
	// Second triangle
	for (i = 1; i <= 7; i++) {
		for (j = 1; j <= i; j++) {
			if (j == 1 || j == i) {
				printf("%2c", prnt);
			} else {
				printf("  ");
			}
		}
		printf("\n");
	}
	return 0;
}

Explanation:
This pattern can be be seen as an inverted right triangle placed over a normal right-angled triangle. In addition to the triangles, there is an extra character in the intersection point.

Since both the triangles share the same tip and in the figure and we have an extra character in the point of intersection, we can start the second right angle triangle from the same line where the first one ends.
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Also read Print Pattern Programs

1. Write a C program to print the following pattern:

   1
   0 1
   1 0 1
   0 1 0 1
   1 0 1 0 1

2. Write a C program to print the following pattern:

   0
   1 1
   2 3 5
   8 13 21

3. Write a C program to print the following pattern:

   1
   121
   12321
   1234321
   12321
   121
   1

4. Write a C program to print the following pattern:

          2
        4 5 6
      6 7 8 9 10
        4 5 6
          2

5. Write a C program to print the following pattern:

    1                         1
    3 3 3                 3 3 3
    5 5 5 5 5         5 5 5 5 5
    7 7 7 7 7 7 7 7 7 7 7 7 7 7
    5 5 5 5 5         5 5 5 5 5
    3 3 3                 3 3 3
    1                         1

6. Write a C program to print the following pattern:

       0
    -2-3 0
   -4-3-2-1 0
    -2-3 0
       0

7. Write a C program to print the following pattern:

   77777777777
            7
           7
          7
         7
        7
       7
      7
     7
    7
   7

8. Write a C program to print the following pattern:

    1                       1
    1 0                   1 0
    1 0 1               1 0 1
    1 0 1 0           1 0 1 0
    1 0 1 0 1       1 0 1 0 1
    1 0 1 0 1 0   1 0 1 0 1 0
    1 0 1 0 1 0 1 0 1 0 1 0 1
    1 0 1 0 1 0   1 0 1 0 1 0
    1 0 1 0 1       1 0 1 0 1
    1 0 1 0           1 0 1 0
    1 0 1               1 0 1
    1 0                   1 0
    1                       1

9. Write a C program to print the following pattern:

   1
   2 4
   3 6 9
   2 4
   1

10. Write a C program to print the following pattern:

     1
     1 0
     1 0 0
     1 0 0 0
     1 0 0 0 0
     1 0 0 0 0 0
     1 0 0 0 0 0 0
     1 0 0 0 0 0
     1 0 0 0 0
     1 0 0 0
     1 0 0
     1 0
     1

1. Write a C program to print the following pattern:

1
0 1
1 0 1
0 1 0 1
1 0 1 0 1

Program:

#include <stdio.h>

int main(void) {
 int i, j;
 for (i = 0; i < 4; i++) {
  for (j = 0; j <= i; j++) {
   if (((i + j) % 2) == 0) {  // Decides on as to which digit to print.
    printf("0");
   } else {
    printf("1");
   }
   printf("\t");
  }
  printf("\n");
 }
 return 0;
}

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Explanation: This is a right angle triangle composed of 0′s and 1′s.
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End of Question1 Start of Question2

2. Write C program to print the following pattern:

   0
   1 1
   2 3 5
   8 13 21

Program:

#include <stdio.h>

int main(void) {
 int i, j, a = 0, b = 1, temp = 1;
 for (i = 1; i <= 4; i++) {
  for (j = 1; j <= i; j++) {
   if (i == 1 && j == 1) { // Prints the '0' individually first
    printf("0");
    continue;
   }
   printf("%d ", temp);  // Prints the next digit in the series
   //Computes the series
   temp = a + b;
   a = b;
   b = temp;
   if (i == 4 && j == 3) { // Skips the 4th character of the base
    break;
   }
  }
  printf("\n");
 }
 return 0;
}

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Explanation: This prints the Fibonacci series in a right angle triangle formation where the base has only three characters.
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End of Question2 Start of Question3

3. Write C program to print the following pattern:

   1
   121
   12321
   1234321
   12321
   121
   1

Program:

#include <stdio.h>

void sequence(int x);
int main() {
 /* c taken for columns */
 int i, x = 0, num = 7;
 for (i = 1; i <= num; i++) {
  if (i <= (num / 2) + 1) {
   x = i;
  } else {
   x = 8 - i;
  }
  sequence(x);
  puts("\n");
 }
 return 0;
}

void sequence(int x) {
 int j;

 for (j = 1; j < x; j++) {
  printf("%d", j);
 }
 for (j = x; j > 0; j--) {
  printf("%d", j);
 }
}

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End of Question3 Start of Question4

4. Write a C program to print the following pattern:

          2
        4 5 6
      6 7 8 9 10
        4 5 6
          2

Program:

#include <stdio.h>

int main(void) {
 int prnt;
 int i, j, k, r, s, sp, nos = 3, nosp = 2; //nos n nosp controls the spacing factor
 // Prints the upper triangle
 for (i = 1; i <= 5; i++) {
  if ((i % 2) != 0) {
   for (s = nos; s >= 1; s--) {
    printf("  ");
   }
   for (j = 1; j <= i; j++) {
    if (i == 5 && j == 5) { //Provides the extra space reqd betn 9 n 10
     printf(" ");        // as 10 is a 2 digit no.
    }
    prnt = i + j;
    printf("%2d", prnt);
   }
  }
  if ((i % 2) != 0) {
   printf("\n");
   nos--;
  }
 }
 // Prints the lower triangle skipin its base..
 for (k = 3; k >= 1; k--) {
  if ((k % 2) != 0) {
   for (sp = nosp; sp >= 1; sp--) {
    printf("  ");
   }
   for (r = 1; r <= k; r++) {
    prnt = k + r;
    printf("%2d", prnt);
   }
  }
  if ((k % 2) != 0) {
   printf("\n");
   nosp++;
  }
 }
 return 0;
}

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Explanation: This is a diamond formation composed of numbers. The numbers are in the following order next_no=i+j where
next_no = The next no to be printed
i = index of the outer for loop
j = index of the inner for loop
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End of Question4 Start of Question5

5. Write a C program to print the following pattern:

    1                           1
    3 3 3                   3 3 3
    5 5 5 5 5           5 5 5 5 5
    7 7 7 7 7 7 7   7 7 7 7 7 7 7
    5 5 5 5 5           5 5 5 5 5
    3 3 3                   3 3 3
    1                           1

Program:

#include <stdio.h>

int main(void) {
 int i, j, k, s, p, q, sp, r, c = 1, nos = 13;
 for (i = 1; c <= 4; i++) {
  if ((i % 2) != 0) {   // Filters out the even line nos.
   for (j = 1; j <= i; j++) { // The upper left triangle
    printf("%2d", i);
   }
   for (s = nos; s >= 1; s--) {  // The spacing factor
    printf("  ");
   }
   for (k = 1; k <= i; k++) { // The upper right triangle
    printf("%2d", i);
   }
   printf("\n");
   nos = nos - 4;  // Space control
   ++c;
  }
 }
 nos = 10;  // Space control re intialized
 c = 1;
 for (p = 5; (c < 4 && p != 0); p--) {
  if ((p % 2) != 0) {  // Filters out the even row nos
   for (q = 1; q <= p; q++) {  // Lower left triangle
    printf("%2d", p);
   }
   for (sp = nos; sp >= 1; sp--) { // Spacing factor
    printf(" ");
   }
   for (r = 1; r <= p; r++) {  // Lower right triangle
    printf("%2d", p);
   }

   printf("\n");
   --c;
   nos = nos + 8;  // Spacing control.
  }
 }

 return 0;
}

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Explanation: Here we are printing only the odd row nos along with thier respective line number. This structure can divided into four identical right angle triangles which are kind of twisted and turned placed in a particular format .
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End of Question5 Start of Question6

6. Write a C program to print the following pattern:

       0
    -2-3 0
   -4-3-2-1 0
    -2-3 0
       0

Program:

#include <stdio.h>

int main(void) {
 int i, j, k, r, s, sp, nos = 2, nosp = 1;
 for (i = 1; i <= 5; i++) {
  if ((i % 2) != 0) {
   for (s = nos; s >= 1; s--) {  //for the spacing factor.
    printf("  ");
   }
   for (j = 1; j <= i; j++) {
    printf("%2d", j-i);
   }
  }
  if ((i % 2) != 0) {
   printf("\n");
   nos--;
  }
 }
 for (k = 3; k >= 1; k--) {
  if ((k % 2) != 0) {
   for (sp = nosp; sp >= 1; sp--) {  // for the spacing factor.
    printf("  ");
   }
   for (r = 1; r <= k; r++) {
    printf("%2d", r-k);
   }
  }
  if ((k % 2) != 0) {
   printf("\n");
   nosp++;
  }
 }
 return 0;
}

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Explanation:This can be seen as a diamond composed of numbers. If we use the conventional nested for loop for its construction the numbers can be seen to flowing the following function f(x) -> j-i
where
j= inner loop index
i= outer loop index
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End of Question6 Start of Question7

7. Write a C program to print the following pattern:

   77777777777
            7
           7
          7
         7
        7
       7
      7
     7
    7
   7

Program:

#include <stdio.h>

int main(void) {
 int i, j;
 for (i = 11; i >= 1; i--) {
  for (j = 1; j <= i; j++) {
   if (i == 11) {
    printf("7");  // Makes sure the base is printed completely
    continue;
   } else if (j == i) { // Hollows the rest
    printf("7");
   } else {
    printf(" ");
   }
  }
  printf("\n");
 }
 return 0;
}

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Explanation: This can be seen as a hollow right-angled triangle composed of 7′s
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End of Question7 Start of Question8

8. Write a C program to print the following pattern:

    1                       1
    1 0                   1 0
    1 0 1               1 0 1
    1 0 1 0           1 0 1 0
    1 0 1 0 1       1 0 1 0 1
    1 0 1 0 1 0   1 0 1 0 1 0
    1 0 1 0 1 0 1 0 1 0 1 0 1
    1 0 1 0 1 0   1 0 1 0 1 0
    1 0 1 0 1       1 0 1 0 1
    1 0 1 0           1 0 1 0
    1 0 1               1 0 1
    1 0                   1 0
    1                       1

Program:

#include <stdio.h>

int main(void) {
    int i,j,k,s,nos=11;
    for (i=1; i<=7; i++) {
        for (j=1; j<=i; j++) {
            if ((j%2)!=0) {   // Applying the condition
                printf(" 1");
            } else {
                printf(" 0");
            }
        }
        for (s=nos; s>=1; s--) {  // Space factor
            printf("  ");
        }
        for (k=1; k<=i; k++) {
            if(i==7 && k==1)  // Skipping the extra 1
            {
                continue;
            }
            if ((k%2)!=0) {  // Applying the condition
                printf(" 1");
            } else {
                printf(" 0");
            }
        }
        printf("\n");
        nos=nos-2;  // Space Control
    }
     nos=1;
     for ( i=6; i>=1; i--) {  // It shares the same base
         for (j=1; j<=i; j++) {
             if (j%2!=0) {
                 printf(" 1");
             } else {
                 printf(" 0");
             }
         }
         for(s=nos; s>=1; s--)  // Spacing factor
         {
             printf("  ");
         }
         for (k=1; k<=i; k++) {
             if (k%2!=0) {
                 printf(" 1");
             } else {
                 printf(" 0");
             }
         }
         printf("\n");
         nos=nos+2;
     }
    return 0;
}

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End of Question8 Start of Question9

9. Write a C program to print the following pattern:

   1
   2 4
   3 6 9
   2 4
   1

Program:

#include <stdio.h>
int main(void) {
    int i,j;
    for (i=1; i<=3 ; i++) {
        for (j=1; j<=i; j++)  {
            printf("%2d", (i*j));
        }
        printf("\n");
    }
    for (i=2; i>=1; i--) { // As they share the same base
        for (j=1; j<=i; j++)  {
            printf("%2d",i*j);
        }
        printf("\n");
    }
    return 0;
}

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Explanation: This can be seen as two right angle triangles sharing th same base
The numbers are following the following function f(x) = i *j
where
i = Index of the Outer loop
j = Index of the inner loop
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End of Question9 Start of Question10

10. Write a C program to print the following pattern:

     1
     1 0
     1 0 0
     1 0 0 0
     1 0 0 0 0
     1 0 0 0 0 0
     1 0 0 0 0 0 0
     1 0 0 0 0 0
     1 0 0 0 0
     1 0 0 0
     1 0 0
     1 0
     1

Program:

#include <stdio.h>

int main(void) {
    int i,j;
    for (i=1; i<=7; i++) {
        for (j=1; j<=i; j++) {
            if (j==1) {           // Applying the condition
                printf(" 1");
            } else {
                printf(" 0");
            }
        }
        printf("\n");
    }
    for (i=6; i>=1; i--) {  //As it shares the same base i=6
        for (j=1; j<=i; j++) {
            if (j==1) {   // Applying the condition
                printf(" 1");
            } else {
                printf(" 0");
            }
        }
        printf("\n");
    }
    return 0;
}

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Explanation: This can be seen as two right angle triangles sharing the same base which is composed of 0′s n 1′s. The first column is filled with 1′s and rest with 0′s
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Here are few challenging C program questions for you. Let’s see how many of them you would be able to write without seeing the program solution given below. These questions are to print patterns using asterisk(star) character. Comment below if you have tougher pattern questions.

Also read Number Pattern Programs

1. Write a C program to print the following pattern:

            *
         *  *
      *  *  *
   *  *  *  *
   

2. Write a C program to print the following pattern:

   *                   *
   * * *           * * *
   * * * * *   * * * * *
   * * * * * * * * * * *
   

3. Write a C program to print the following pattern:

   *                         *
      *                   *
   *     *             *     *
      *     *       *     *
   *     *      *      *     *
      *     *       *     *
   *     *             *     *
      *                   *
   *                         *
   

4. Write a C program to print the following pattern:

   *   *   *   *   *
     *   *   *   *
       *   *   *
         *   *
           *
         *   *
       *   *   *
     *   *   *   *
   *   *   *   *   *
   

5. Write a C program to print the following pattern:

             *
           * * *
         * * * * *
       * * * * * * *
     * * * * * * * * *
       * * * * * * *
         * * * * *
           * * *
             *
           * * *
         * * * * *
   

6. Write a C program to print the following pattern:

        *
          * *
            * * *
              * * * *
            * * *
          * *
        *
   

7. Write a C program to print the following pattern:

   * * * * * * * * *
   * * * *   * * * *
   * * *       * * *
   * *           * *
   *               *
   * *           * *
   * * *       * * *
   * * * *   * * * *
   * * * * * * * * *
   

8. Write a C program to print the following pattern:

   * * * * * * * * * * * * * * * * *
     * * * * * * *   * * * * * * *
       * * * * *       * * * * *
         * * *           * * *
           * * * * * * * * *
             * * * * * * *
               * * * * *
                 * * *
                   *
   

9. Write a C program to print the following pattern:

         *
       * * *
     * * * * *
   * * * * * * *
   *           *
   * *       * *
   * * *   * * *
   * * * * * * *
   * * *   * * *
   * *       * *
   *           *
   * * * * * * *
     * * * * *
       * * *
         *
   

10. Write a C program to print the following pattern:

    * * * * * * * * * * * * * * * * * * * * * * * * *
      *           *   *           *   *           *
        *       *       *       *       *       *
          *   *           *   *           *   *
            *               *               *
          *   *           *   *           *   *
        *       *       *       *       *       *
      *           *   *           *   *           *
    * * * * * * * * * * * * * * * * * * * * * * * * *
    

1. Write a C program to print the following pattern:

         *
      *  *
   *  *  *
*  *  *  *

Program:

/* This is a simple mirror-image of a right angle triangle */

#include <stdio.h>
int main() {
 char prnt = '*';
 int i, j, nos = 4, s;
 for (i = 1; i <= 5; i++) {
for (s = nos; s >= 1; s--) {  // Spacing factor
   printf("  ");
  }
  for (j = 1; j <= i; j++) {
   printf("%2c", prnt);
  }
  printf("\n");
  --nos;   // Controls the spacing factor
 }
 return 0;
}

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2. Write C program to print the following pattern:

   *                   *
   * * *           * * *
   * * * * *   * * * * *
   * * * * * * * * * * *

Program:

#include<stdio.h>
int main() {
 char prnt = '*';
 int i, j, k, s, c = 1, nos = 9;
 for (i = 1; c <= 4; i++) {
     // As we want to print the columns in odd sequence viz. 1,3,5,.etc
  if ((i % 2) != 0) {
   for (j = 1; j <= i; j++) {
 printf("%2c", prnt);
}
for (s = nos; s >= 1; s--) { //The spacing factor
    if (c == 4 && s == 1) {
     break;
    }
    printf("  ");
   }
   for (k = 1; k <= i; k++) {
    if (c == 4 && k == 5) {
     break;
    }
    printf("%2c", prnt);
   }
   printf("\n");
   nos = nos - 4;  // controls the spacing factor
   ++c;
  }
 }
 return 0;
}

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3. Write C program to print the following pattern:

   *                         *
      *                   *
   *     *             *     *
      *     *       *     *
   *     *      *      *     *
      *     *       *     *
   *     *             *     *
      *                   *
   *                         *

Program:

#include<stdio.h>
int main() {
 char prnt = '*';
 int i, j, k, s, p, r, nos = 7;

 for (i = 1; i <= 5; i++) {
  for (j = 1; j <= i; j++) {
 if ((i % 2) != 0 && (j % 2) != 0) {
printf("%3c", prnt);
}
else if ((i % 2) == 0 && (j % 2) == 0) {
printf("%3c", prnt);
}
else {
printf("   ");
}
}
for (s = nos; s >= 1; s--) {  // for the spacing factor
   printf("   ");
  }
  for (k = 1; k <= i; k++) { //Joining seperate figures
if (i == 5 && k == 1) {
 continue;
}
if ((k % 2) != 0) {
printf("%3c", prnt);
}
else {
printf("   ");
}
}
printf("\n");
nos = nos - 2;   // space control
}  nos = 1;  // remaining half..
for (p = 4; p >= 1; p--) {
  for (r = 1; r <= p; r++) {
if ((p % 2) != 0 && (r % 2) != 0) {
printf("%3c", prnt);
}
else if ((p % 2) == 0 && (r % 2) == 0) {
printf("%3c", prnt);
}
else {
printf("   ");
}
}
for (s = nos; s >= 1; s--) {
   printf("   ");
  }
  for (k = 1; k <= p; k++) {
   if ((k % 2) != 0) {
    printf("%3c", prnt);
   }
else {
    printf("   ");
   }
  }
  nos = nos + 2;  // space control
  printf("\n");
 }
 return 0;
}

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Explanation:
This can be seen as an inverted diamond composed of stars. It can be noted that the composition of this figure follows sequential pattern of consecutive stars and spaces.

In case of odd row number, the odd column positions will be filled up with ‘*’, else a space will be spaced and vice-versa in case of even numbered row.

In order to achieve this we will construct four different right angle triangles

aligned as per the requirement.
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4. Write a C program to print the following pattern:

   *   *   *   *   *
     *   *   *   *
       *   *   *
         *   *
           *
         *   *
       *   *   *
     *   *   *   *
   *   *   *   *   *

Program:

#include<stdio.h>
int main() {
 char prnt = '*';
 int i, j, s, nos = 0;
 for (i = 9; i >= 1; (i = i - 2)) {
  for (s = nos; s >= 1; s--) {
   printf("  ");
  }
  for (j = 1; j <= i; j++) {
   if ((i % 2) != 0 && (j % 2) != 0) {
    printf("%2c", prnt);
   } else {
    printf("  ");
   }
  }
  printf("\n");
  nos++;
 }
 nos = 3;
 for (i = 3; i <= 9; (i = i + 2)) {
for (s = nos; s >= 1; s--) {
   printf("  ");
  }
  for (j = 1; j <= i; j++) {

   if ((i % 2) != 0 && (j % 2) != 0) {
    printf("%2c", prnt);
   } else {
    printf("  ");
   }
  }
  nos--;
  printf("\n");
 }
 return 0;
}

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5. Write a C program to print the following pattern:

             *
           * * *
         * * * * *
       * * * * * * *
     * * * * * * * * *
       * * * * * * *
         * * * * *
           * * *
             *
           * * *
         * * * * *

Program:

#include<stdio.h>
int main() {
 char prnt = '*';
 int i, j, k, s, nos = 4;
 for (i = 1; i <= 5; i++) {
for (s = nos; s >= 1; s--) {
   printf("  ");
  }
  for (j = 1; j <= i; j++) {
   printf("%2c", prnt);
  }
  for (k = 1; k <= (i - 1); k++) {
if (i == 1) {     continue;
}
printf("%2c", prnt);
}
 printf("\n");   nos--;
}
 nos = 1;
for (i = 4; i >= 1; i--) {
  for (s = nos; s >= 1; s--) {
   printf("  ");
  }
  for (j = 1; j <= i; j++) {
   printf("%2c", prnt);
  }
  for (k = 1; k <= (i - 1); k++) {
   printf("%2c", prnt);
  }
  nos++;
  printf("\n");
 }
 nos = 3;
 for (i = 2; i <= 5; i++) {
if ((i % 2) != 0) {
for (s = nos; s >= 1; s--) {
    printf("  ");
   }
   for (j = 1; j <= i; j++) {
    printf("%2c", prnt);
   }
  }
  if ((i % 2) != 0) {
   printf("\n");
   nos--;
  }
 }
 return 0;
}

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6. Write a C program to print the following pattern:

        *
          * *
            * * *
              * * * *
            * * *
          * *
        *

Program:

/*
    This can be seen as two right angle triangles sharing the same base
    which is modified by adding few extra shifting spaces
*/
#include <stdio.h>
// This function controls the inner loop and the spacing
// factor guided by the outer loop index and the spacing index.
int triangle(int nos, int i) {
 char prnt = '*';
 int s, j;
 for (s = nos; s >= 1; s--) {    // Spacing factor
  printf("  ");
 }
 for (j = 1; j <= i; j++) {      //The inner loop
  printf("%2c", prnt);
 }
 return 0;
}

int main() {
 int i, nos = 5;
 //draws the upper triangle
 for (i = 1; i <= 4; i++) {
 triangle(nos, i);    //Inner loop construction
 nos++;              // Increments the spacing factor
 printf("\n");  }
nos = 7;  //Draws the lower triangle skipping its base.
for (i = 3; i >= 1; i--) {
  int j = 1;
  triangle(nos, i);  // Inner loop construction
  nos = nos - j;     // Spacing factor
  printf("\n");
 }
 return 0;
}

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7. Write a C program to print the following pattern:

   * * * * * * * * *
   * * * *   * * * *
   * * *       * * *
   * *           * *
   *               *
   * *           * *
   * * *       * * *
   * * * *   * * * *
   * * * * * * * * *

Program:

#include <stdio.h>

int main() {
 char prnt = '*';
 int i, j, k, s, nos = -1;
 for (i = 5; i >= 1; i--) {
  for (j = 1; j <= i; j++) {
 printf("%2c", prnt);
}
for (s = nos; s >= 1; s--) {
   printf("  ");
  }
  for (k = 1; k <= i; k++) {
   if (i == 5 && k == 5) {
    continue;
   }
   printf("%2c", prnt);
  }
  nos = nos + 2;
  printf("\n");
 }
 nos = 5;
 for (i = 2; i <= 5; i++) {
  for (j = 1; j <= i; j++) {
 printf("%2c", prnt);
}
for (s = nos; s >= 1; s--) {
   printf("  ");
  }
  for (k = 1; k <= i; k++) {
   if (i == 5 && k == 5) {
    break;
   }
   printf("%2c", prnt);
  }
  nos = nos - 2;
  printf("\n");
 }
 return 0;
}

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8. Write a C program to print the following pattern:

   * * * * * * * * * * * * * * * * *
     * * * * * * *   * * * * * * *
       * * * * *       * * * * *
         * * *           * * *
           * * * * * * * * *
             * * * * * * *
               * * * * *
                 * * *
                   *

Program:

#include <stdio.h>
int main() {
 char prnt = '*';
 int i, j, k, s, sp, nos = 0, nosp = -1;
 for (i = 9; i >= 3; (i = i - 2)) {
  for (s = nos; s >= 1; s--) {
   printf("  ");
  }
  for (j = 1; j <= i; j++) {
printf("%2c", prnt);
}
for (sp = nosp; sp >= 1; sp--) {
   printf("  ");
  }
  for (k = 1; k <= i; k++) {
 if (i == 9 && k == 1) {
continue;
}
printf("%2c", prnt);
}
nos++;
nosp = nosp + 2;
printf("\n");
}
nos = 4;
for (i = 9; i >= 1; (i = i - 2)) {
  for (s = nos; s >= 1; s--) {
   printf("  ");
  }
  for (j = 1; j <= i; j++) {
   printf("%2c", prnt);
  }
  nos++;
  printf("\n");
 }

 return 0;
}

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9. Write a C program to print the following pattern:

         *
       * * *
     * * * * *
   * * * * * * *
   *           *
   * *       * *
   * * *   * * *
   * * * * * * *
   * * *   * * *
   * *       * *
   *           *
   * * * * * * *
     * * * * *
       * * *
         *

Program:

#include <stdio.h>
/*
 * nos = Num. of spaces required in the triangle.
 * i   = Counter for the num. of charcters to print in each row
 * skip= A flag for checking whether to
 *       skip a character in a row.
 *
 */
int triangle(int nos, int i, int skip) {
 char prnt = '*';
 int s, j;
 for (s = nos; s >= 1; s--) {
  printf("  ");
 }
 for (j = 1; j <= i; j++) {
  if (skip != 0) {
   if (i == 4 && j == 1) {
    continue;
   }
  }
  printf("%2c", prnt);
 }
 return 0;
}

int main() {
 int i, nos = 4;
 for (i = 1; i <= 7; (i = i + 2)) {
  triangle(nos, i, 0);
  nos--;
  printf("\n");
 }
 nos = 5;
 for (i = 1; i <= 4; i++) {
triangle(1, i, 0); //one space needed in each case of the formation
triangle(nos, i, 1); //skip printing one star in the last row.
nos = nos - 2;
printf("\n");
}
nos = 1;
for (i = 3; i >= 1; i--) {
  triangle(1, i, 0);
  triangle(nos, i, 0);
  nos = nos + 2;
  printf("\n");
 }
 nos = 1;
 for (i = 7; i >= 1; (i = i - 2)) {
  triangle(nos, i, 0);
  nos++;
  printf("\n");
 }
 return 0;
}

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10. Write a C program to print the following pattern:

    * * * * * * * * * * * * * * * * * * * * * * * * *
      *           *   *           *   *           *
        *       *       *       *       *       *
          *   *           *   *           *   *
            *               *               *
          *   *           *   *           *   *
        *       *       *       *       *       *
      *           *   *           *   *           *
    * * * * * * * * * * * * * * * * * * * * * * * * *

Program:

#include <stdio.h>

/*
 * nos = Num. of spaces required in the triangle.
 * i   = Counter for the num. of characters to print in each row
 * skip= A flag for check whether to
 *       skip a character in a row.
 *
 */

int triangle(int nos, int i, int skip) {
 char prnt = '*';
 int s, j;
 for (s = nos; s >= 1; s--) {
  printf("  ");
 }
 for (j = 1; j <= i; j++) {
 if (skip != 0) {
 if (i == 9 && j == 1) {
   continue;
 }
 }
if (i == 1 || i == 9) {
 printf("%2c", prnt);
}
else if (j == 1 || j == i) {
 printf("%2c", prnt);
 } else {
 printf("  ");
}  }
return 0; }
int main() {
int i, nos = 0, nosp = -1, nbsp = -1;
for (i = 9; i >= 1; (i = i - 2)) {
  triangle(nos, i, 0);
  triangle(nosp, i, 1);
  triangle(nbsp, i, 1);
  printf("\n");
  nos++;
  nosp = nosp + 2;
  nbsp = nbsp + 2;
 }
 nos = 3, nosp = 5, nbsp = 5;
 for (i = 3; i <= 9; (i = i + 2)) {
  triangle(nos, i, 0);
  triangle(nosp, i, 1);
  triangle(nbsp, i, 1);
  printf("\n");
  nos--;
  nosp = nosp - 2;
  nbsp = nbsp - 2;
 }
 return 0;
}

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Also Read:

• Star Pattern Programs
• Number Pattern Programs

Showing 5 out of 65 C Interview Questions.

1. Write a program to display the multiplication table of a given number.

Program: Multiplication table of a given number

#include <stdio.h>
int main() {
      int num, i = 1;
      printf("\n Enter any Number:");
      scanf("%d", &num);
      printf("Multiplication table of %d: \n", num);
      while (i <= 10) {
            printf("\n %d x %d = %d", num, i, num * i);
            i++;
      }
      return 0;
}

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Output:

Enter any Number:5
 5 x 1 = 5
 5 x 2 = 10
 5 x 3 = 15
 5 x 4 = 20
 5 x 5 = 25
 5 x 6 = 30
 5 x 7 = 35
 5 x 8 = 40
 5 x 9 = 45
 5 x 10 = 50

Explanation: We need to multiply the given number (i.e. the number for which we want the multiplication table) with value of ‘i’ which increments from 1 to 10.

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2. Write C program to print the following pattern:

         1
        2 2
       3 3 3
      4 4 4 4
     5 5 5 5 5

Program:

#include<stdio.h>
int main() {
    int i, j, k, c = 5;
    for (i = 1; i <= 5; i++) {
            /* k is taken for spaces */
            for (k = 1; k <= c; k++) {
                  /* blank space */
                  printf(" ");
            }
            for (j = 1; j <= i; j++) {
                  /* %2d ensures that the number is printed in
two spaces for alignment and the numbers are printed in the order. */
                  printf("%2d", i);
            }
            printf("\n");
            /*c is decremented by 1 */
            c--;
      }
      return 0;
}

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Output:

         1
        2 2
       3 3 3
      4 4 4 4
     5 5 5 5 5

Explanation: Here ‘i’ loop is used for printing the numbers in the respective rows and ‘k’ loop is used for providing spaces. ‘j’ loop prints the numbers. ‘c’ is decremented for numbers to be displayed in alternate columns.

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3. Write C program to print the following pattern:

         1
       1 2 1
     1 2 3 2 1
   1 2 3 4 3 2 1
 1 2 3 4 5 4 3 2 1

Program:

#include<stdio.h>
int main() {
      /* c taken for columns */
      int i, j, c = 9, m, k;
      for (i = 1; i <= 5; i++) {
            /* k is used for spaces */
            for (k = 1; k <= c; k++) {
                  printf(" ");
            }
            for (j = 1; j <= i; j++) {                   printf("%2d", j);             }             for (m = j - 2; m > 0; m--) {
                  /* %2d ensures that the number
                   * is printed in two spaces
                   * for alignment */
                  printf("%2d", m);
            }
            printf("\n");
            /* c is decremented by 2 */
            c = c - 2;
      }
      return 0;
}

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Output:

         1
       1 2 1
     1 2 3 2 1
   1 2 3 4 3 2 1
 1 2 3 4 5 4 3 2 1

Explanation: Here ‘i’ loop is used for printing numbers in rows and ‘k’ loop is used for providing spaces. ‘j’ loop is used for printing numbers in increasing order. ‘m’ loop is used for printing numbers in reverse order.

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4. Write a C program to display the following format:

------
  a b
 ------
  1 5
  2 4
  3 3
  4 2
  5 1
 ------

Program:

#include<stdio.h>
int main() {
      int i = 1, j = 5;
      printf("----------\n");
      printf("a \t b \n");
      printf("----------\n");
      /* logic: while loop repeats
       * 5 times i.e. until
       * the condition i<=5 fails */
      while (i <= 5) {
            /* i and j value printed */
            printf("%d \t %d\n", i, j);
            /* i and j value incremented
             by 1 for every iteration */
            i++;
            j--;
      }
      printf("----------");
      return 0;
}

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Output:

------
  a b
 ------
  1 5
  2 4
  3 3
  4 2
  5 1
 ------

Explanation: Here, ‘i’ is initialized to least value 1 and ‘j’ initialized to highest value 5. We keep incrementing the i’ value and decrementing the ‘j’ value until the condition fails. The value is displayed at each increment and at each decrement. Back to top

5. Write a C program to display the following format:

--------
 no. sum
 --------
  1  1
  2  3
  3  6
  4  10
  5  15
 --------

Program:

#include<stdio.h>
int main() {
      int num = 1, sum = 0;
      printf("-----------\n");
      printf("num \t sum\n");
      printf("-----------\n");
      /* while loop repeats 5 times
       *  i.e. until the condition
       *  num <= 5 fails */
      while (num <= 5) {
            sum = sum + num;
            printf("%d \t %d\n", num, sum);
            /* num incremented by 1
             * for every time
             * the loop is executed */
            num++;
      }
      printf("-----------");
      return 0;
}

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Output:

--------
 no. sum
 --------
  1  1
  2  3
  3  6
  4  10
  5  15
 --------

Explanation: In the above program we have taken two variables ‘num’ and ‘sum’. ‘num’ is used to check the condition and to display the numbers up to 5. ‘sum’ is used to add the numbers which are displayed using variable ‘num’. The ‘sum’ value is initialized to zero. sum is added to the numbers which are incremented by ‘i’ and displayed.

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Also read 10 Challenging Pattern Programs