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

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

A sequence *a* is a subsequence of a sequence *b* if ????a can be obtained from *b* by deletion of several (possibly, zero) elements. For example, [3,1][3,1] is a subsequence of [3,2,1][3,2,1] and [4,3,1][4,3,1], but not a subsequence of [1,3,3,7][1,3,3,7] and [3,10,4][3,10,4].

**Examples:**

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

Largest Subarray is {4, 5, 6}

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

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

**Naive approach:**

- Generate all subsequences 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(2^{N})

**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++

`// 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 subarray 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;` `}` |

**Output:**

15

**Time Complexity: **O(N*log(N))**Auxiliary Space: **O(1)

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