Hello Everyone,
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.
Examples:
Input: N = 2, Sum = 3
Output: 12 21 30
Input: N = 3, Sum = 6
Output: 105 114 123 132 141 150 204
213 222 231 240 303 312 321
330 402 411 420 501 510 600
Input: N = 4, Sum = 3
Output: 1002 1011 1020 1101 1110 1200
2001 2010 2100 3000
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.
Below is a simple recursive implementation of above idea –
// A C++ recursive program to print all n-digit
// numbers whose sum of digits equals to given sum
#include <bits/stdc++.h>
using namespace std;
// Recursive function to print all n-digit numbers
// whose sum of digits equals to given sum
// 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);
}
}
// Driver program
int main()
{
int n = 2, sum = 3;
findNDigitNums(n, sum);
return 0;
}
Output:
12 21 30