# Longest subarray whose elements can be made equal by maximum K increments

Given an array arr[] of positive integers of size N and a positive integer K, the task is to find the maximum possible length of subarray which can be made equal by adding some integer value to each element of the sub-array such that the sum of the added elements does not exceed K.

Examples:

Input: N = 5, arr[] = {1, 4, 9, 3, 6}, K = 9
Output: 3
Explanation:
{1, 4} : {1+3, 4} = {4, 4}
{4, 9} : {4+5, 9} = {9, 9}
{3, 6} : {3+3, 6} = {6, 6}
{9, 3, 6} : {9, 3+6, 6+3} = {9, 9, 9}
Hence, the maximum length of such a subarray is 3.

Input: N = 6, arr[] = {2, 4, 7, 3, 8, 5}, K = 10
Output: 4

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

Approach: This problem can be solved by using dynamic programming.

• Initialize:
• dp[]: Stores the sum of elements that are added to the subarray.
• deque: Stores the indices of the maximum element for each subarray.
• pos: Index of the current position of the subarray.
• ans: Length of maximum sub array.
• mx: Maximum element of a sub array
• pre: Previous index of the current subarray.
• Traverse the array and check if deque is empty or not. If yes, then update the maximum element and the index of maximum element along with the indices of pre and pos.
• Check if the currently added element is greater then K. If yes, then remove it from dp[] array and update the indices of pos and pre.
• Finally update the maximum length of the valid sub array.

Below is the implementation of above approach:

## C++

 `// C++ code for the above approach ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to find maximum ` `// possible length of subarray ` `int` `validSubArrLength(``int` `arr[], ` `                      ``int` `N, ``int` `K) ` `{ ` `    ``// Stores the sum of elements ` `    ``// that needs to be added to ` `    ``// the sub array ` `    ``int` `dp[N + 1]; ` ` `  `    ``// Stores the index of the ` `    ``// current position of subarray ` `    ``int` `pos = 0; ` ` `  `    ``// Stores the maximum ` `    ``// length of subarray. ` `    ``int` `ans = 0; ` ` `  `    ``// Maximum element from ` `    ``// each subarray length ` `    ``int` `mx = 0; ` ` `  `    ``// Previous index of the ` `    ``// current subarray of ` `    ``// maximum length ` `    ``int` `pre = 0; ` ` `  `    ``// Deque to store the indices ` `    ``// of maximum element of ` `    ``// each sub array ` `    ``deque<``int``> q; ` ` `  `    ``// For each array element, ` `    ``// find the maximum length of ` `    ``// required subarray ` `    ``for` `(``int` `i = 0; i < N; i++) { ` `        ``// Traverse the deque and ` `        ``// update the index of ` `        ``// maximum element. ` `        ``while` `(!q.empty() ` `               ``&& arr[q.back()] < arr[i]) ` ` `  `            ``q.pop_back(); ` ` `  `        ``q.push_back(i); ` ` `  `        ``// If it is first element ` `        ``// then update maximum ` `        ``// and dp[] ` `        ``if` `(i == 0) { ` `            ``mx = arr[i]; ` `            ``dp[i] = arr[i]; ` `        ``} ` ` `  `        ``// Else check if current ` `        ``// element exceeds max ` `        ``else` `if` `(mx <= arr[i]) { ` `            ``// Update max and dp[] ` `            ``dp[i] = dp[i - 1] + arr[i]; ` `            ``mx = arr[i]; ` `        ``} ` ` `  `        ``else` `{ ` `            ``dp[i] = dp[i - 1] + arr[i]; ` `        ``} ` ` `  `        ``// Update the index of the ` `        ``// current maximum length ` `        ``// subarray ` `        ``if` `(pre == 0) ` `            ``pos = 0; ` `        ``else` `            ``pos = pre - 1; ` ` `  `        ``// While current element ` `        ``// being added to dp[] array ` `        ``// exceeds K ` `        ``while` `((i - pre + 1) * mx ` `                       ``- (dp[i] - dp[pos]) ` `                   ``> K ` `               ``&& pre < i) { ` ` `  `            ``// Update index of ` `            ``// current position and ` `            ``// the previous position ` `            ``pos = pre; ` `            ``pre++; ` ` `  `            ``// Remove elements ` `            ``// from deque and ` `            ``// update the ` `            ``// maximum element ` `            ``while` `(!q.empty() ` `                   ``&& q.front() < pre ` `                   ``&& pre < i) { ` ` `  `                ``q.pop_front(); ` `                ``mx = arr[q.front()]; ` `            ``} ` `        ``} ` ` `  `        ``// Update the maximum length ` `        ``// of the required subarray. ` `        ``ans = max(ans, i - pre + 1); ` `    ``} ` `    ``return` `ans; ` `} ` ` `  `// Driver Program ` `int` `main() ` `{ ` `    ``int` `N = 6; ` `    ``int` `K = 8; ` `    ``int` `arr[] = { 2, 7, 1, 3, 4, 5 }; ` `    ``cout << validSubArrLength(arr, N, K); ` `    ``return` `0; ` `} `

## Java

 `// Java code for the above approach ` `import` `java.util.*; ` ` `  `class` `GFG{ ` ` `  `// Function to find maximum ` `// possible length of subarray ` `static` `int` `validSubArrLength(``int` `arr[], ` `                             ``int` `N, ``int` `K) ` `{ ` `    ``// Stores the sum of elements ` `    ``// that needs to be added to ` `    ``// the sub array ` `    ``int` `[]dp = ``new` `int``[N + ``1``]; ` ` `  `    ``// Stores the index of the ` `    ``// current position of subarray ` `    ``int` `pos = ``0``; ` ` `  `    ``// Stores the maximum ` `    ``// length of subarray. ` `    ``int` `ans = ``0``; ` ` `  `    ``// Maximum element from ` `    ``// each subarray length ` `    ``int` `mx = ``0``; ` ` `  `    ``// Previous index of the ` `    ``// current subarray of ` `    ``// maximum length ` `    ``int` `pre = ``0``; ` ` `  `    ``// Deque to store the indices ` `    ``// of maximum element of ` `    ``// each sub array ` `    ``Deque q = ``new` `LinkedList<>(); ` ` `  `    ``// For each array element, ` `    ``// find the maximum length of ` `    ``// required subarray ` `    ``for``(``int` `i = ``0``; i < N; i++)  ` `    ``{ ` `         `  `       ``// Traverse the deque and ` `       ``// update the index of ` `       ``// maximum element. ` `       ``while` `(!q.isEmpty() &&  ` `               ``arr[q.getLast()] < arr[i]) ` `           ``q.removeLast(); ` `       ``q.add(i); ` `        `  `       ``// If it is first element ` `       ``// then update maximum ` `       ``// and dp[] ` `       ``if` `(i == ``0``) ` `       ``{ ` `           ``mx = arr[i]; ` `           ``dp[i] = arr[i]; ` `       ``} ` `        `  `       ``// Else check if current ` `       ``// element exceeds max ` `       ``else` `if` `(mx <= arr[i]) ` `       ``{ ` `            `  `           ``// Update max and dp[] ` `           ``dp[i] = dp[i - ``1``] + arr[i]; ` `           ``mx = arr[i]; ` `       ``} ` `       ``else` `       ``{ ` `           ``dp[i] = dp[i - ``1``] + arr[i]; ` `       ``} ` `        `  `       ``// Update the index of the ` `       ``// current maximum length ` `       ``// subarray ` `       ``if` `(pre == ``0``) ` `           ``pos = ``0``; ` `       ``else` `           ``pos = pre - ``1``; ` `        `  `       ``// While current element ` `       ``// being added to dp[] array ` `       ``// exceeds K ` `       ``while` `((i - pre + ``1``) * mx -  ` `          ``(dp[i] - dp[pos]) > K && pre < i) ` `       ``{ ` `            `  `           ``// Update index of ` `           ``// current position and ` `           ``// the previous position ` `           ``pos = pre; ` `           ``pre++; ` `            `  `           ``// Remove elements from  ` `           ``// deque and update the ` `           ``// maximum element ` `           ``while` `(!q.isEmpty() &&  ` `                   ``q.peek() < pre && pre < i) ` `           ``{ ` `               ``q.removeFirst(); ` `               ``mx = arr[q.peek()]; ` `           ``} ` `       ``} ` `        `  `       ``// Update the maximum length ` `       ``// of the required subarray. ` `       ``ans = Math.max(ans, i - pre + ``1``); ` `    ``} ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `N = ``6``; ` `    ``int` `K = ``8``; ` `    ``int` `arr[] = { ``2``, ``7``, ``1``, ``3``, ``4``, ``5` `}; ` `     `  `    ``System.out.print(validSubArrLength(arr, N, K)); ` `} ` `} ` ` `  `// This code is contributed by amal kumar choubey `

Output:

```4
```

Time Complexity: O(N2)
Auxillary Space Complexity: O(N)

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