C Exercises: Count the numbers without digit 7, from 1 to a given number

C Programming Mathematics: Exercise-22 with Solution

Write a C program to count the numbers without the digit 7, from 1 to a given number.

Example 1:
Input: n = 10
Output: 9
Example 2:
Input: n = 687
Output: 555

Visual Presentation:

Sample Solution:

C Code:

#include <stdio.h>
#include <stdlib.h>

// Function to count the numbers without digit 7
int count_nums_not_7(int num) { 
    if (num < 7) 
        return num; // If the number is less than 7, return the number itself
    if (num >= 7 && num < 10) 
        return num - 1; // If the number is 7 or 8 or 9, return num - 1

    int r = 1; 
    while (num / r > 9) 
        r = r * 10; // Find the divisor 'r' to split the number

    int m = num / r; // Extract the most significant digit of the number

    if (m != 7) 
        return count_nums_not_7(m) * count_nums_not_7(r - 1) + count_nums_not_7(m) + count_nums_not_7(num % r); 
    else
        return count_nums_not_7(m * r - 1); 
}      

// Main function
int main(void) { 
    int n = 10;  
    if (n > 0)
        printf("Count the numbers without digit 7, from 1 to %d : %d", n, count_nums_not_7(n));

    n = 687;  
    if (n > 0)
        printf("\nCount the numbers without digit 7, from 1 to %d : %d", n, count_nums_not_7(n));

    return 0; // End of the main function
}

Sample Output:

Count the numbers without digit 7, from 1 to 10 : 9
Count the numbers without digit 7, from 1 to 687 : 555

Flowchart:

C Programming Code Editor:

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