# Minimum capacity of small arrays needed to contain all element of the given array

Given an array of positive integers and a value K, The task is to empty the array in less than or equal to K small arrays such that each small array can only contain at max P elements from a single slot / index of the given array. Find the minimum value of P.

Examples:

Input: arr[] = {1, 2, 3, 4, 5}, K = 7
Output: 3

Explanation:
We put 1 into the first small array,
2 into the second array,
3 into the third array,
After this, we divide the other elements as 1 + 3 and 2 + 3
These 4 elements can be put into the remaining 4 boxes.
So, the required value of P is 3.

Input: arr[] = {23, 1, 43, 66, 220}, K = 102
Output: 4

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: To solve this problem we need to Binary Search the answer.

1. First, we set the lower limit to 1 and the upper limit to the maximum value of the given array.
2. Now, we can perform a binary search in this range. For a particular value of capacity, we calculate the number of small arrays we need to contain all values from items.
3. If this required number of small arrays is more than K, the answer is definitely bigger hence, we trim the left side of search. If it is less than or equal to K, we keep this value as a potential answer and trim the right side of the search.

Below is the implementation of the above approach.

## C++

 `// C++ program to find the Minimum  ` `// capacity of small arrays needed  ` `// to contain all element of ` `// the given array ` `#include ` `using` `namespace` `std; ` ` `  ` `  `// Function returns the value ` `// of Minimum capacity needed ` `int` `MinimumCapacity(vector<``int``> arr, ` `                    ``int` `K) ` `{ ` `    ``// Initializing maximum ` `    ``// value ` `    ``int` `maxVal = arr;  ` ` `  `    ``// Finding maximum value  ` `    ``// in arr ` `    ``for` `(``auto` `x : arr) ` `        ``maxVal = max(maxVal, x);  ` ` `  `    ``int` `l = 1, r = maxVal, m; ` `    ``int` `ans, req; ` ` `  `    ``// Binary Search the answer ` `    ``while` `(l <= r) ` `    ``{  ` ` `  `        ``// m is the mid-point ` `        ``// of the range ` `        ``m = l + (r - l) / 2;  ` ` `  `        ``req = 0; ` ` `  `        ``// Finding the total number of  ` `        ``// arrays needed to completely  ` `        ``// contain the items array ` `        ``// with P = req ` `        ``for` `(``auto` `x : arr) ` `            ``req += x / m + (x % m > 0);  ` `         `  `        ``// If the required number of  ` `        ``// arrays is more than K, it  ` `        ``// means we need to increase ` `        ``// the value of P ` `        ``if` `(req > K) ` `            ``l = m + 1;  ` ` `  `        ``else` `            ``// If the required number of ` `            ``// arrays is less than or equal  ` `            ``// to K, it means this is a  ` `            ``// possible answer and we go to ` `            ``// check if any smaller possible  ` `            ``// value exists for P ` `            ``ans = m, r = m - 1;  ` `    ``} ` ` `  `    ``return` `ans; ` `} ` `     `  `// Driver Code ` `int` `main() ` `{ ` ` `  `    ``// Given array ` `    ``vector<``int``> arr = { 1, 2, 3, 4, 5 };  ` ` `  `    ``// Number of available small arrays ` `    ``int` `K = 7; ` ` `  `    ``cout << MinimumCapacity(arr, K); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to find the Minimum  ` `// capacity of small arrays needed  ` `// to contain all element of ` `// the given array ` `class` `GFG{ ` ` `  `// Function returns the value ` `// of Minimum capacity needed ` `static` `int` `MinimumCapacity(``int` `[]arr, ` `                           ``int` `K) ` `{ ` `    ``// Initializing maximum ` `    ``// value ` `    ``int` `maxVal = arr[``0``];  ` ` `  `    ``// Finding maximum value  ` `    ``// in arr ` `    ``for` `(``int` `x : arr) ` `        ``maxVal = Math.max(maxVal, x);  ` ` `  `    ``int` `l = ``1``, r = maxVal, m; ` `    ``int` `ans = ``0``, req; ` ` `  `    ``// Binary Search the answer ` `    ``while` `(l <= r) ` `    ``{  ` ` `  `        ``// m is the mid-point ` `        ``// of the range ` `        ``m = l + (r - l) / ``2``;  ` ` `  `        ``req = ``0``; ` ` `  `        ``// Finding the total number of  ` `        ``// arrays needed to completely  ` `        ``// contain the items array ` `        ``// with P = req ` `        ``for` `(``int` `x : arr) ` `            ``req += x / m + (x % m > ``0` `? ``1` `: ``0``);  ` `         `  `        ``// If the required number of  ` `        ``// arrays is more than K, it  ` `        ``// means we need to increase ` `        ``// the value of P ` `        ``if` `(req > K) ` `            ``l = m + ``1``;  ` ` `  `        ``else`  `        ``{ ` `            ``// If the required number of ` `            ``// arrays is less than or equal  ` `            ``// to K, it means this is a  ` `            ``// possible answer and we go to ` `            ``// check if any smaller possible  ` `            ``// value exists for P ` `            ``ans = m; ` `            ``r = m - ``1``;  ` `        ``} ` `    ``} ` ` `  `    ``return` `ans; ` `} ` `     `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` ` `  `    ``// Given array ` `    ``int` `[]arr = { ``1``, ``2``, ``3``, ``4``, ``5` `};  ` ` `  `    ``// Number of available small arrays ` `    ``int` `K = ``7``; ` ` `  `    ``System.out.print(MinimumCapacity(arr, K)); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## Python3

 `# Python3 program to find the Minimum  ` `# capacity of small arrays needed  ` `# to contain all element of  ` `# the given array  ` ` `  `# Function returns the value  ` `# of Minimum capacity needed  ` `def` `MinimumCapacity(arr, K):  ` ` `  `    ``# Initializing maximum  ` `    ``# value  ` `    ``maxVal ``=` `arr[``0``]  ` ` `  `    ``# Finding maximum value  ` `    ``# in arr  ` `    ``for` `x ``in` `arr:  ` `        ``maxVal ``=` `max``(maxVal, x) ` ` `  `    ``l ``=` `1` `    ``r ``=` `maxVal ` `    ``m ``=` `0` `    ``ans ``=` `0` `    ``req ``=` `0` ` `  `    ``# Binary Search the answer  ` `    ``while` `l <``=` `r:  ` ` `  `        ``# m is the mid-point  ` `        ``# of the range  ` `        ``m ``=` `l ``+` `(r ``-` `l) ``/``/` `2` ` `  `        ``req ``=` `0` ` `  `        ``# Finding the total number of  ` `        ``# arrays needed to completely  ` `        ``# contain the items array  ` `        ``# with P = req  ` `        ``for` `x ``in` `arr:  ` `            ``req ``+``=` `x ``/``/` `m ``+` `(x ``%` `m > ``0``)  ` `         `  `        ``# If the required number of  ` `        ``# arrays is more than K, it  ` `        ``# means we need to increase  ` `        ``# the value of P  ` `        ``if` `req > K:  ` `            ``l ``=` `m ``+` `1` ` `  `        ``else``: ` `             `  `            ``# If the required number of  ` `            ``# arrays is less than or equal  ` `            ``# to K, it means this is a  ` `            ``# possible answer and we go to  ` `            ``# check if any smaller possible  ` `            ``# value exists for P  ` `            ``ans ``=` `m ` `            ``r ``=` `m ``-` `1` ` `  `    ``return` `ans ` `     `  `#Driver Code ` `# Given array  ` `arr ``=` `[ ``1``, ``2``, ``3``, ``4``, ``5` `]  ` ` `  `# Number of available small arrays  ` `K ``=` `7` `print``(MinimumCapacity(arr, K)) ` ` `  `# This code is contributed by divyamohan123 `

Output:

```3
```

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Improved By : divyamohan123, 29AjayKumar