Related Articles

# Minimum K such that sum of array elements after division by K does not exceed S

• Difficulty Level : Medium
• Last Updated : 26 Aug, 2021

Given an array arr[] of N elements and an integer S. The task is to find the minimum number K such that the sum of the array elements does not exceed S after dividing all the elements by K
Note: Consider integer division.
Examples:

Input: arr[] = {10, 7, 8, 10, 12, 19}, S = 27
Output:
After dividing by 3, the array becomes
{3, 2, 2, 3, 4, 6} and the new sum is 20.
Input: arr[] = {19, 17, 11, 10}, S = 40
Output:

Naive approach: Iterate for all values of K from 1 to the maximum element in the array plus one not maximum element because if we put k as maximum element then the sum will be one and what if the S is given to us as zero. So we iterate from k=1 to max element in the array plus one. and then sum up the array elements by dividing with K, if the sum does not exceed S then the current value will be the answer. The time complexity of this approach will be O(M * N) where M is the maximum element in the array.
Efficient approach: An efficient approach is to find the value of K by performing a binary search on the answer. Initiate a binary search on the value of K and a check is done inside it to see if the sum exceeds K then the binary search is performed on the second half or the first half accordingly.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the minimum value of k``// that satisfies the given condition``int` `findMinimumK(``int` `a[], ``int` `n, ``int` `s)``{``    ``// Find the maximum element``    ``int` `maximum = a;``    ``for` `(``int` `i = 0; i < n; i++) {``        ``maximum = max(maximum, a[i]);``    ``}` `    ``// Lowest answer can be 1 and the``    ``// highest answer can be (maximum + 1)``    ``int` `low = 1, high = maximum + 1;` `    ``int` `ans = high;` `    ``// Binary search``    ``while` `(low <= high) {` `        ``// Get the mid element``        ``int` `mid = (low + high) / 2;``        ``int` `sum = 0;` `        ``// Calculate the sum after dividing``        ``// the array by new K which is mid``        ``for` `(``int` `i = 0; i < n; i++) {``            ``sum += (``int``)(a[i] / mid);``        ``}` `        ``// Search in the second half``        ``if` `(sum > s)``            ``low = mid + 1;` `        ``// First half``        ``else` `{``            ``ans = min(ans, mid);``            ``high = mid - 1;``        ``}``    ``}` `    ``return` `ans;``}` `// Driver code``int` `main()``{``    ``int` `a[] = { 10, 7, 8, 10, 12, 19 };``    ``int` `n = ``sizeof``(a) / ``sizeof``(a);``    ``int` `s = 27;` `    ``cout << findMinimumK(a, n, s);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{``    ` `    ``// Function to return the minimum value of k``    ``// that satisfies the given condition``    ``static` `int` `findMinimumK(``int` `a[],``                            ``int` `n, ``int` `s)``    ``{``        ``// Find the maximum element``        ``int` `maximum = a[``0``];``        ` `        ``for` `(``int` `i = ``0``; i < n; i++)``        ``{``            ``maximum = Math.max(maximum, a[i]);``        ``}``    ` `        ``// Lowest answer can be 1 and the``        ``// highest answer can be (maximum + 1)``        ``int` `low = ``1``, high = maximum + ``1``;``    ` `        ``int` `ans = high;``    ` `        ``// Binary search``        ``while` `(low <= high)``        ``{``    ` `            ``// Get the mid element``            ``int` `mid = (low + high) / ``2``;``            ``int` `sum = ``0``;``    ` `            ``// Calculate the sum after dividing``            ``// the array by new K which is mid``            ``for` `(``int` `i = ``0``; i < n; i++)``            ``{``                ``sum += (``int``)(a[i] / mid);``            ``}``    ` `            ``// Search in the second half``            ``if` `(sum > s)``                ``low = mid + ``1``;``    ` `            ``// First half``            ``else``            ``{``                ``ans = Math.min(ans, mid);``                ``high = mid - ``1``;``            ``}``        ``}``        ``return` `ans;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `a[] = { ``10``, ``7``, ``8``, ``10``, ``12``, ``19` `};``        ``int` `n = a.length;``        ``int` `s = ``27``;``    ` `        ``System.out.println(findMinimumK(a, n, s));``    ``}``}   ` `// This code is contributed by AnkitRai01`

## Python3

 `# Python3 implementation of the approach` `# Function to return the minimum value of k``# that satisfies the given condition``def` `findMinimumK(a, n, s):``    ` `    ``# Find the maximum element``    ``maximum ``=` `a[``0``]``    ``for` `i ``in` `range``(n):``        ``maximum ``=` `max``(maximum, a[i])` `    ``# Lowest answer can be 1 and the``    ``# highest answer can be (maximum + 1)``    ``low ``=` `1``    ``high ``=` `maximum ``+` `1` `    ``ans ``=` `high` `    ``# Binary search``    ``while` `(low <``=` `high):` `        ``# Get the mid element``        ``mid ``=` `(low ``+` `high) ``/``/` `2``        ``sum` `=` `0` `        ``# Calculate the sum after dividing``        ``# the array by new K which is mid``        ``for` `i ``in` `range``(n):``            ``sum` `+``=` `(a[i] ``/``/` `mid)` `        ``# Search in the second half``        ``if` `(``sum` `> s):``            ``low ``=` `mid ``+` `1` `        ``# First half``        ``else``:``            ``ans ``=` `min``(ans, mid)``            ``high ``=` `mid ``-` `1` `    ``return` `ans` `# Driver code``a ``=` `[``10``, ``7``, ``8``, ``10``, ``12``, ``19``]``n ``=` `len``(a)``s ``=` `27` `print``(findMinimumK(a, n, s))` `# This code is contributed by Mohit Kumar`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{``    ` `    ``// Function to return the minimum value of k``    ``// that satisfies the given condition``    ``static` `int` `findMinimumK(``int` `[]a,``                            ``int` `n, ``int` `s)``    ``{``        ``// Find the maximum element``        ``int` `maximum = a;``        ` `        ``for` `(``int` `i = 0; i < n; i++)``        ``{``            ``maximum = Math.Max(maximum, a[i]);``        ``}``    ` `        ``// Lowest answer can be 1 and the``        ``// highest answer can be (maximum + 1)``        ``int` `low = 1, high = maximum + 1;``    ` `        ``int` `ans = high;``    ` `        ``// Binary search``        ``while` `(low <= high)``        ``{``    ` `            ``// Get the mid element``            ``int` `mid = (low + high) / 2;``            ``int` `sum = 0;``    ` `            ``// Calculate the sum after dividing``            ``// the array by new K which is mid``            ``for` `(``int` `i = 0; i < n; i++)``            ``{``                ``sum += (``int``)(a[i] / mid);``            ``}``    ` `            ``// Search in the second half``            ``if` `(sum > s)``                ``low = mid + 1;``    ` `            ``// First half``            ``else``            ``{``                ``ans = Math.Min(ans, mid);``                ``high = mid - 1;``            ``}``        ``}``        ``return` `ans;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main ()``    ``{``        ``int` `[]a = { 10, 7, 8, 10, 12, 19 };``        ``int` `n = a.Length;``        ``int` `s = 27;``    ` `        ``Console.WriteLine(findMinimumK(a, n, s));``    ``}``}` `// This code is contributed by AnkitRai01`

## Javascript

 ``
Output:
`3`

Time Complexity: O(N*(log N)), N=Array length

Auxilliary Space: O(1)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up