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.
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 –
12 21 30
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