# Longest subarray in which absolute difference between any two element is not greater than X

Given an integer array arr[] of size N and an integer X, the task is to find the longest sub-array where the absolute difference between any two elements is not greater than X.

Examples:

Input: arr = { 8, 4, 2, 6, 7 }, X = 4
Output: 4 2 6
Explanation:
The sub-array described by indexs [1, 3], i.e, { 4, 2, 6 } contains no such difference of two elements which is greater than 4.

Input: arr = { 15, 10, 1, 2, 4, 7, 2}, X = 5
Output: 2 4 7 2
Explanation:
The sub-array described by indexs [3, 6], i.e, { 2, 4, 7, 2 } contains no such difference of two elements which is greater than 5.

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

Naive Approach: Simple solution is to consider all subarrays one by one, find the maximum and minimum element of that sub-array and check if their difference is not greater than X. Among all such sub-arrays print the longest sub-array.

Time Complexity: O(N3)

Efficient Approach: The idea is to use the Sliding Window Technique to consider a sub-array and use a Map data structure to find the maximum and minimum element in that sub-array.

• At first the Start and End of the window points to the 0-th index.
• At every iteration, the element at End is inserted in the Map if not already present or otherwise its count is incremented.
• If the difference between the maximum and minimum element is not greater than X, then update the maximum length of the required sub-array and store the beginning of that sub-array in a variable.
• Otherwise, increment the Start of the window until the difference between the maximum and minimum element is not greater than X.
• When incrementing the Start, the size of the window decreases, remove the element at the Start from the Map if and only if the count of that element becomes zero.

Finally, print the sub-array with the longest length, and the absolute difference between any two elements is not greater than the X.

Below is the implementation of the above approach:

## C++

 `// C++ program to find the longest sub-array ` `// where the absolute difference between any ` `// two elements is not greater than X ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function that prints the longest sub-array ` `// where the absolute difference between any ` `// two element is not greater than X ` `void` `longestSubarray(``int``* A, ``int` `N, ``int` `X) ` `{ ` `    ``// Initialize a variable to store ` `    ``// length of longest sub-array ` `    ``int` `maxLen = 0; ` ` `  `    ``// Initialize a variable to store the ` `    ``// beginning of the longest sub-array ` `    ``int` `beginning = 0; ` ` `  `    ``// Initialize a map to store the maximum ` `    ``// and the minimum elements for a given window ` `    ``map<``int``, ``int``> window; ` ` `  `    ``// Initialize the window ` `    ``int` `start = 0, end = 0; ` ` `  `    ``// Loop througth the array ` `    ``for` `(; end < N; end++) { ` `        ``// Increment the count of that ` `        ``// element in the window ` `        ``window[A[end]]++; ` ` `  `        ``// Find the maximum and minimum element ` `        ``// in the current window ` `        ``auto` `minimum = window.begin()->first; ` `        ``auto` `maximum = window.rbegin()->first; ` ` `  `        ``// If the difference is not ` `        ``// greater than X ` `        ``if` `(maximum - minimum <= X) { ` `            ``// Update the length of the longest ` `            ``// sub-array and store the beginning ` `            ``// of the sub-array ` `            ``if` `(maxLen < end - start + 1) { ` `                ``maxLen = end - start + 1; ` `                ``beginning = start; ` `            ``} ` `        ``} ` `        ``// Decrease the size of the window ` `        ``else` `{ ` `            ``while` `(start < end) { ` `                ``// Remove the element at start ` `                ``window[A[start]]--; ` ` `  `                ``// Remove the element from the window ` `                ``// if its count is zero ` `                ``if` `(window[A[start]] == 0) { ` ` `  `                    ``window.erase(window.find(A[start])); ` `                ``} ` `                ``// Increment the start of the window ` `                ``start++; ` ` `  `                ``// Find the maximum and minimum element ` `                ``// in the current window ` `                ``auto` `minimum = window.begin()->first; ` `                ``auto` `maximum = window.rbegin()->first; ` ` `  `                ``// Stop decreasing the size of window ` `                ``// when difference is not greater ` `                ``if` `(maximum - minimum <= X) ` `                    ``break``; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// Print the longest sub-array ` `    ``for` `(``int` `i = beginning; i < beginning + maxLen; i++) ` `        ``cout << A[i] << ``" "``; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 15, 10, 1, 2, 4, 7, 2 }, X = 5; ` ` `  `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` ` `  `    ``longestSubarray(arr, n, X); ` ` `  `    ``return` `0; ` `} `

## Python3

 `# Python3 program to find the longest sub-array ` `# where the absolute difference between any ` `# two elements is not greater than X ` ` `  `# Function that prints the longest sub-array ` `# where the absolute difference between any ` `# two element is not greater than X ` `def` `longestSubarray(A, N, X): ` `     `  `    ``# Initialize a variable to store ` `    ``# length of longest sub-array ` `    ``maxLen ``=` `0` ` `  `    ``# Initialize a variable to store the ` `    ``# beginning of the longest sub-array ` `    ``beginning ``=` `0` ` `  `    ``# Initialize a map to store the maximum ` `    ``# and the minimum elements for a given window ` `    ``window ``=` `{} ` ` `  `    ``# Initialize the window ` `    ``start ``=` `0` ` `  `    ``# Loop througth the array ` `    ``for` `end ``in` `range``(N): ` ` `  `        ``# Increment the count of that ` `        ``# element in the window ` `        ``if` `A[end] ``in` `window: ` `            ``window[A[end]] ``+``=` `1` `        ``else``: ` `            ``window[A[end]] ``=` `1` ` `  `        ``# Find the maximum and minimum element ` `        ``# in the current window ` `        ``minimum ``=` `min``(``list``(window.keys())) ` `        ``maximum ``=` `max``(``list``(window.keys())) ` ` `  `        ``# If the difference is not ` `        ``# greater than X ` `        ``if` `maximum ``-` `minimum <``=` `X: ` `             `  `            ``# Update the length of the longest ` `            ``# sub-array and store the beginning ` `            ``# of the sub-array ` `            ``if` `maxLen < end ``-` `start ``+` `1``: ` `                ``maxLen ``=` `end ``-` `start ``+` `1` `                ``beginning ``=` `start ` ` `  `        ``# Decrease the size of the window ` `        ``else``: ` `            ``while` `start < end: ` ` `  `                ``# Remove the element at start ` `                ``window[A[start]] ``-``=` `1` ` `  `                ``# Remove the element from the window ` `                ``# if its count is zero ` `                ``if` `window[A[start]] ``=``=` `0``: ` `                    ``window.pop(A[start]) ` ` `  `                ``# Increment the start of the window ` `                ``start ``+``=` `1` ` `  `                ``# Find the maximum and minimum element ` `                ``# in the current window ` `                ``minimum ``=` `min``(``list``(window.keys())) ` `                ``maximum ``=` `max``(``list``(window.keys())) ` ` `  `                ``# Stop decreasing the size of window ` `                ``# when difference is not greater ` `                ``if` `maximum ``-` `minimum <``=` `X: ` `                    ``break` `                     `  `    ``# Print the longest sub-array ` `    ``for` `i ``in` `range``(beginning, beginning ``+` `maxLen): ` `        ``print``(A[i], end ``=` `' '``) ` ` `  `# Driver Code ` `arr ``=` `[``15``, ``10``, ``1``, ``2``, ``4``, ``7``, ``2``] ` `X ``=` `5` `n ``=` `len``(arr) ` `longestSubarray(arr, n, X) ` ` `  `# This code is contributed by Shivam Singh `

Output:

```2 4 7 2
```

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

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Improved By : SHIVAMSINGH67