# Minimise the maximum element of N subarrays of size K

Given an array arr[] and two integers N and K, the task is to choose non-overlapping N subarrays of size K such that maximum element of all subarrays is minimum.

Note: If it is not possible to choose N such subarrays then return -1.

Examples:

Input: arr[] = {1, 10, 3, 10, 2}, N = 3, K = 1
Output: 3
Explanation:
The three non-overlapping subarrays are –
Subarrays => {{1}, {2}, {3}}
Maximum of these subarrays are => 3

Input: arr[] = {7, 7, 7, 7, 12, 7, 7}, N = 2, K = 3
Output: 12
Explanation:
The two non-overlapping subarrays are –
Subarrays => {{7, 7, 7}, {7, 12, 7}}
Maximum element of these subarrays are => 12

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

Approach: The idea is to use binary search. Below is the illustration of the binary search:

• Search space: As we have to find the maximum element from the N subarrays which is one of the element from the array. Therefore, the search space will be the minimum element to the maximum element of the array.
• Function for binary search: The function for binary search is to find the count of K sized array possible in the with all the elements less than the given number which will be the middle of the search space.
• Left Search Space: The condition when the count of K sized subarrays possible is greater than or equal to N, Then the answer possible can lie in the left search space.
• Right Search Space: The condition when the count of K sized subarrays possible is less than N, Then the answer possible ccan lie in the right search space.

Below is the implementation of the above approach:

## C++

 `// C++ implementation to choose  ` `// N subarrays of size K such that  ` `// the maximum element of  ` `// subarrays is minimum ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to choose  ` `// N subarrays of size K such that  ` `// the maximum element of  ` `// subarrays is minimum ` `int` `minDays(vector<``int``>& arr,  ` `                ``int` `n, ``int` `k) ` `{ ` `    ``int` `l = arr.size(),  ` `left = 1, right = 1e9; ` ` `  `    ``// Condition to check if it ` `    ``// is not possible to choose k ` `    ``// sized N subarrays ` `    ``if` `(n * k > l) ` `        ``return` `-1; ` ` `  `    ``// Using binary search ` `    ``while` `(left < right) { ` ` `  `        ``// calculating mid ` `        ``int` `mid = (left + right) / 2,  ` `                 ``cnt = 0, product = 0; ` `                  `  `        ``// Loop to find the count of the  ` `        ``// K sized subarrays possible with  ` `        ``// elements less than  mid ` `        ``for` `(``int` `j = 0; j < l; ++j) { ` `            ``if` `(arr[j] > mid) { ` `                ``cnt = 0; ` `            ``} ` `            ``else` `if` `(++cnt >= k) { ` `                ``product++; ` `                ``cnt = 0; ` `            ``} ` `        ``} ` `     `  `        ``// Condition to check if the ` `        ``// answer is in right subarray ` `        ``if` `(product < n) { ` `            ``left = mid + 1; ` `        ``} ` `        ``else` `{ ` `            ``right = mid; ` `        ``} ` `    ``} ` `    ``return` `left; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``vector<``int``> arr{ 1, 10, 3, 10, 2 }; ` `    ``int` `n = 3, k = 1; ` `     `  `    ``// Function Call ` `    ``cout << minDays(arr, n, k) << endl; ` `    ``return` `0; ` `} `

## Java

 `// Java implementation to choose  ` `// N subarrays of size K such that  ` `// the maximum element of  ` `// subarrays is minimum ` `class` `GFG{ ` ` `  `// Function to choose  ` `// N subarrays of size K such that  ` `// the maximum element of  ` `// subarrays is minimum ` `static` `int` `minDays(``int` `[]arr,  ` `                   ``int` `n, ``int` `k) ` `{ ` `    ``int` `l = arr.length,  ` `        ``left = ``1``, right = (``int``) 1e9; ` ` `  `    ``// Condition to check if it ` `    ``// is not possible to choose k ` `    ``// sized N subarrays ` `    ``if` `(n * k > l) ` `        ``return` `-``1``; ` ` `  `    ``// Using binary search ` `    ``while` `(left < right)  ` `    ``{ ` ` `  `        ``// calculating mid ` `        ``int` `mid = (left + right) / ``2``,  ` `                   ``cnt = ``0``, product = ``0``; ` `                 `  `        ``// Loop to find the count of the  ` `        ``// K sized subarrays possible with  ` `        ``// elements less than mid ` `        ``for` `(``int` `j = ``0``; j < l; ++j) ` `        ``{ ` `            ``if` `(arr[j] > mid)  ` `            ``{ ` `                ``cnt = ``0``; ` `            ``} ` `            ``else` `if` `(++cnt >= k)  ` `            ``{ ` `                ``product++; ` `                ``cnt = ``0``; ` `            ``} ` `        ``} ` `     `  `        ``// Condition to check if the ` `        ``// answer is in right subarray ` `        ``if` `(product < n)  ` `        ``{ ` `            ``left = mid + ``1``; ` `        ``} ` `        ``else`  `        ``{ ` `            ``right = mid; ` `        ``} ` `    ``} ` `    ``return` `left; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `[]arr = { ``1``, ``10``, ``3``, ``10``, ``2` `}; ` `    ``int` `n = ``3``, k = ``1``; ` `     `  `    ``// Function Call ` `    ``System.out.print(minDays(arr, n, k) + ``"\n"``); ` `} ` `} ` ` `  `// This code is contributed by Amit Katiyar `

## C#

 `// C# implementation to choose N   ` `// subarrays of size K such that  ` `// the maximum element of  ` `// subarrays is minimum ` `using` `System; ` `class` `GFG{ ` ` `  `// Function to choose N subarrays   ` `// of size K such that the maximum  ` `// element of subarrays is minimum ` `static` `int` `minDays(``int` `[]arr,  ` `                   ``int` `n, ``int` `k) ` `{ ` `    ``int` `l = arr.Length;  ` `    ``int` `left = 1, right = (``int``)1e9; ` ` `  `    ``// Condition to check if it ` `    ``// is not possible to choose k ` `    ``// sized N subarrays ` `    ``if` `(n * k > l) ` `        ``return` `-1; ` ` `  `    ``// Using binary search ` `    ``while` `(left < right)  ` `    ``{ ` ` `  `        ``// Calculating mid ` `        ``int` `mid = (left + right) / 2,  ` `                   ``cnt = 0, product = 0; ` `                 `  `        ``// Loop to find the count of the  ` `        ``// K sized subarrays possible with  ` `        ``// elements less than mid ` `        ``for``(``int` `j = 0; j < l; ++j) ` `        ``{ ` `           ``if` `(arr[j] > mid)  ` `           ``{ ` `               ``cnt = 0; ` `           ``} ` `           ``else` `if` `(++cnt >= k)  ` `           ``{ ` `               ``product++; ` `               ``cnt = 0; ` `           ``} ` `        ``} ` `     `  `        ``// Condition to check if the ` `        ``// answer is in right subarray ` `        ``if` `(product < n)  ` `        ``{ ` `            ``left = mid + 1; ` `        ``} ` `        ``else` `        ``{ ` `            ``right = mid; ` `        ``} ` `    ``} ` `    ``return` `left; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `[]arr = { 1, 10, 3, 10, 2 }; ` `    ``int` `n = 3, k = 1; ` `     `  `    ``// Function Call ` `    ``Console.Write(minDays(arr, n, k) + ``"\n"``); ` `} ` `} ` ` `  `// This code is contributed by Amit Katiyar `

Output:

```3
```

Time Complexity: O(N*logN)

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Improved By : amit143katiyar