Minimum numbers (smaller than or equal to N) with sum S Last Updated : 27 Aug, 2022 Summarize Comments Improve Suggest changes Share Like Article Like Report Given N numbers(1, 2, 3, ....N) and a number S. The task is to print the minimum number of numbers that sum up to give S. Examples: Input: N = 5, S = 11 Output: 3 Three numbers (smaller than or equal to N) can be any of the given combinations. (3, 4, 4) (2, 4, 5) (1, 5, 5) (3, 3, 5)Input: N = 1, S = 10 Output: 10 Approach: Greedily we choose N as many times we can, and then if anything less than N is left we will choose that number which adds up to give S, hence the total number of numbers are (S/N) + 1(if S%N>0). Below is the implementation of the above approach. C++ // C++ program to find the minimum numbers // required to get to S #include <bits/stdc++.h> using namespace std; // Function to find the minimum // numbers required to get to S int minimumNumbers(int n, int s) { if (s % n) return s / n + 1; else return s / n; } // Driver Code int main() { int n = 5; int s = 11; cout << minimumNumbers(n, s); return 0; } Java // Java program to find the minimum numbers // required to get to S import java.io.*; class GFG { // Function to find the minimum // numbers required to get to S static int minimumNumbers(int n, int s) { if ((s % n)>0) return s / n + 1; else return s / n; } // Driver Code public static void main (String[] args) { int n = 5; int s = 11; System.out.println(minimumNumbers(n, s)); } } // This code is contributed by shs.. Python 3 # Python 3 program to find the # minimum numbers required to get to S # Function to find the minimum # numbers required to get to S def minimumNumbers(n, s): if (s % n): return s / n + 1; else: return s / n; # Driver Code n = 5; s = 11; print(int(minimumNumbers(n, s))); # This code is contributed # by Shivi_Aggarwal C# // C# program to find the minimum numbers // required to get to S using System; class GFG { // Function to find the minimum // numbers required to get to S static int minimumNumbers(int n, int s) { if ((s % n)>0) return s / n + 1; else return s / n; } // Driver Code public static void Main () { int n = 5; int s = 11; Console.WriteLine(minimumNumbers(n, s)); } } // This code is contributed by shs.. PHP <?php // PHP program to find the minimum numbers // required to get to S // Function to find the minimum // numbers required to get to S function minimumNumbers($n, $s) { if ($s % $n) return round($s / $n + 1); else return round($s /$n); } // Driver Code $n = 5; $s = 11; echo minimumNumbers($n, $s); // This code is contributed by shs.. ?> JavaScript <script> // JavaScript program to find the minimum numbers // required to get to S // Function to find the minimum // numbers required to get to S function minimumNumbers(n, s) { if (s % n) return parseInt(s / n) + 1; else return parseInt(s / n); } // Driver Code let n = 5; let s = 11; document.write(minimumNumbers(n, s)); </script> Output: 3 Time Complexity: O(1) Auxiliary Space: O(1), since no extra space has been taken. 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