# Longest subarray such that difference of max and min is at-most K

Given an **array arr[]** of length **N**, tark is to find the length of longest subarray such that difference of its maximum element and minumum element is not more than an integer **K**

**Examples:**

Input:K = 3, arr[]= {1, 4, 5, 6, 9}

Output:3

Explanation:

Largest Subarray is {4, 5, 6}

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

Output:5

Explanation:

Largest Subarray is {1, 2, 3, 4, 5}

**Naive approach:**

- Generate all subarray and find the minimum and maximum in the subarray.
- Calculate difference between minimum and maximum and if it is smaller or equal to K, then update answer

**Time Complexity: ** O(N^{3})

**Efficient approach:** The idea is to first sort the array, and then use binary search to optimize the approach.

- Sort the given array
- For every distinct element
**A[i]**in the array find the first element**A[j]**such that**(A[j]-A[i]) > K**. - For its implementation we use
**Binary search**or**lower_bound**and update ans each time as maximum of previous ans and difference of indixes.

Below is the implementation of the above approach:

`// C++ program to find longest ` `// subarray such that difference ` `// of max and min is at-most K ` ` ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to calculate longest ` `// subarray with above condition ` `int` `findLargestSubarray( ` ` ` `vector<` `int` `>& arr, ` ` ` `int` `N, ` `int` `K) ` `{ ` ` ` ` ` `// Sort the array ` ` ` `sort(arr.begin(), arr.end()); ` ` ` ` ` `int` `value1 = arr[0], value2 = 0; ` ` ` `int` `index1, index2, i, MAX; ` ` ` `index1 = index2 = 0; ` ` ` `i = 0, MAX = 0; ` ` ` ` ` `// Loop which will terminate ` ` ` `// when no further elements ` ` ` `// can be included in the subarray ` ` ` ` ` `while` `(index2 != N) { ` ` ` ` ` `// first value such that ` ` ` `// arr[index2] - arr[index1] > K ` ` ` `value2 = value1 + (K + 1); ` ` ` ` ` `// calculate its index using lower_bound ` ` ` `index2 = lower_bound(arr.begin(), ` ` ` `arr.end(), value2) ` ` ` `- arr.begin(); ` ` ` ` ` `// index2- index1 will give ` ` ` `// the accurate length ` ` ` `// of suarray then compare ` ` ` `// for MAX length and store ` ` ` `// in MAX variable ` ` ` `MAX = max(MAX, (index2 - index1)); ` ` ` ` ` `// change the index1 ` ` ` `// to next greater element ` ` ` `// than previous one ` ` ` `// and recalculate the value1 ` ` ` `index1 ` ` ` `= lower_bound( ` ` ` `arr.begin(), ` ` ` `arr.end(), arr[index1] + 1) ` ` ` `- arr.begin(); ` ` ` `value1 = arr[index1]; ` ` ` `} ` ` ` ` ` `// finally return answer MAX ` ` ` `return` `MAX; ` `} ` `// Driver Code ` `int` `main() ` `{ ` ` ` `int` `N, K; ` ` ` `N = 18; ` ` ` `K = 5; ` ` ` `vector<` `int` `> arr{ 1, 1, 1, 2, 2, ` ` ` `2, 2, 2, 3, ` ` ` `3, 3, 6, 6, 7, ` ` ` `7, 7, 7, 7 }; ` ` ` `cout << findLargestSubarray(arr, N, K); ` ` ` `return` `0; ` `} ` |

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**Output:**

15

**Time Complexity: ** O(N*log(N))

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