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Minimum numbers (smaller than or equal to N) with sum S

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  • Difficulty Level : Easy
  • Last Updated : 18 Jun, 2021

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

 


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