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

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.