Print all n-digit numbers whose sum of digits equals to given sum - GeeksforGeeks

Print all n-digit numbers whose sum of digits equals to given sum - GeeksforGeeks
Given number of digits n, print all n-digit numbers whose sum of digits adds upto given sum. Solution should not consider leading 0's as digits.

A simple solution would be to generate all N-digit numbers and print numbers that have sum of their digits equal to given sum. The complexity of this solution would be exponential.

A better solution is to generate only those N-digit numbers that satisfy the given constraints. The idea is to use recursion. We basically fill all digits from 0 to 9 into current position and maintain sum of digits so far. We then recurse for remaining sum and number of digits left. We handle leading 0’s separately as they are not counted as digits.

// n, sum --> value of inputs

// out --> output array

// index --> index of next digit to be filled in

//           output array

void findNDigitNumsUtil(int n, int sum, char* out,

                        int index)

{

    // Base case

    if (index > n || sum < 0)

        return;

 

    // If number becomes N-digit

    if (index == n)

    {

        // if sum of its digits is equal to given sum,

        // print it

        if(sum == 0)

        {

            out[index] = '\0';

            cout << out << " ";

        }

        return;

    }

 

    // Traverse through every digit. Note that

    // here we're considering leading 0's as digits

    for (int i = 0; i <= 9; i++)

    {

        // append current digit to number

        out[index] = i + '0';

 

        // recurse for next digit with reduced sum

        findNDigitNumsUtil(n, sum - i, out, index + 1);

    }

}

 

// This is mainly a wrapper over findNDigitNumsUtil.

// It explicitly handles leading digit

void findNDigitNums(int n, int sum)

{

    // output array to store N-digit numbers

    char out[n + 1];

 

    // fill 1st position by every digit from 1 to 9 and

    // calls findNDigitNumsUtil() for remaining positions

    for (int i = 1; i <= 9; i++)

    {

        out[0] = i + '0';

        findNDigitNumsUtil(n, sum - i, out, 1);

    }

}

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