# Longest subarray with absolute difference between elements less than or equal to K using Heaps

Given an array arr[] of N integers and an integer K, our task is to find the length of the longest subarray such that for all possible pairs in the subarray absolute difference between elements is less than or equal to K.

Examples:

Input: arr[] = {2, 4, 5, 5, 5, 3, 1}, K = 0
Output: 3
Explanation:
The possible subarray with difference in elements as 0 is {5, 5, 5} whose length is 3. Hence the output is 3.

Input: arr[] = {1, 2, 3, 6, 7}, K = 2
Output: 3
Explanation:
The possible subarray with difference in elements at most 2 is {1, 2, 3} whose length is 3. Hence the output is 3.

Naive Approach:

To solve the problem mentioned above the naive method is to use The Brute Force approach that is to generate all the possible subarray of the given array and check if the difference between the maximum and minimum element of the subarray is at most K or not. If it is, then update the length of the current subarray with the maximum length. Print the maximum length of the subarray after all the operations.

Below is the implementation of the above approach:

 `// Java implementation to find the Longest subarray ` `// of the given array with absolute difference between ` `// elements less than or equal to integer K ` ` `  `class` `GFG { ` `    ``public` `static` `int` `computeLongestSubarray(``int` `arr[],  ` `                                                ``int` `k) ` `    ``{ ` `        ``// maxLength is 1 because k >= 0,  ` `        ``// a single element, subarray will always ` `        ``// have absolute difference zero ` `        ``int` `maxLength = ``1``; ` ` `  `        ``// Check for all possible subarrays ` `        ``for` `(``int` `i = ``0``; i < arr.length; i++) { ` `            ``// Initalization of minimum & ` `            ``// maximum of current subarray ` `            ``int` `minOfSub = arr[i]; ` `            ``int` `maxOfSub = arr[i]; ` ` `  `            ``for` `(``int` `j = i + ``1``; j < arr.length; j++) { ` ` `  `                ``// Update the values for minimum & maximum ` `                ``if` `(arr[j] > maxOfSub) ` `                    ``maxOfSub = arr[j]; ` ` `  `                ``if` `(arr[j] < minOfSub) ` `                    ``minOfSub = arr[j]; ` ` `  `                ``// Check if current subarray satisfies  ` `                ``// the given condition ` `                ``if` `((maxOfSub - minOfSub) <= k) { ` `                    ``int` `currLength = j - i + ``1``; ` ` `  `                    ``// Update the value for maxLength ` `                    ``if` `(maxLength < currLength) ` `                        ``maxLength = currLength; ` `                ``} ` `            ``} ` `        ``} ` ` `  `        ``// Return the final result ` `        ``return` `maxLength; ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `arr[] = { ``1``, ``2``, ``3``, ``6``, ``7` `}; ` ` `  `        ``int` `k = ``2``; ` ` `  `        ``int` `maxLength = computeLongestSubarray(arr, k); ` `        ``System.out.println(maxLength); ` `    ``} ` `} `

Output:

```3
```

Time Complexity: O(n2)

Efficient Approach:

To optimize the above approach the idea is to use Heap Data Structure. Initialize a minHeap that will store the indices of the current subarray such that the elements are in ascending order, where the smallest appears at the top and a maxHeap that will store the indices of the current subarray such that the elements are in descending order, where the largest element appears at the top. Then iterate over the entire array and for each iteration check if:

• All the subarray elements satisfy the condition of maxOfSub-minOfSub <= k, then we compare maxLength so far to the length of current subarray and update maxLength to maximum of either maxLength or current subarray length.
• If the condition is not satisfied then increase the beginning pointer for subarray by 1 and remove all indices from minHeap and maxHeap which are not included in the new subarray.
• After each iteration we increase our subarray length by incrementing the end pointer.

Below is the implementation of the above approach:

 `// Java implementation to find the Longest  ` `// subarray of the given array with absolute ` `// difference between elements less than or equal  ` `// to integer K using Heaps ` `import` `java.util.*; ` ` `  `class` `GFG { ` `    ``public` `static` `int` `computeLongestSubarray(``int` `arr[], ` `                                                 ``int` `k) ` `    ``{ ` `        ``// Stores the maximum length subarray so far ` `        ``int` `maxLength = ``0``; ` ` `  `        ``Deque maxHeap = ``new` `LinkedList<>(); ` `        ``Deque minHeap = ``new` `LinkedList<>(); ` ` `  `        ``// Marks to the beginning and end  ` `        ``// pointer for current subarray ` `        ``int` `beg = ``0``, end = ``0``; ` ` `  `        ``while` `(end < arr.length) { ` ` `  `            ``// Stores the current element being  ` `            ``// added to the subarray ` `            ``int` `currEl = arr[end]; ` ` `  `            ``// Remove indices of all elements smaller ` `            ``// than or equal to current from maxHeap ` `            ``while` `(maxHeap.size() > ``0` `&&  ` `                       ``arr[maxHeap.peekLast()] <= currEl) ` `                ``maxHeap.removeLast(); ` ` `  `            ``// Add current element's index to maxHeap ` `            ``maxHeap.addLast(end); ` ` `  `            ``// Remove indices of all elements larger ` `            ``// than or equal to current from minHeap ` `            ``while` `(minHeap.size() > ``0` `&&  ` `                       ``arr[minHeap.peekLast()] >= currEl) ` `                ``minHeap.removeLast(); ` ` `  `            ``// Add current element's index to minHeap ` `            ``minHeap.addLast(end); ` ` `  `            ``// Index of maximum of current subarray ` `            ``int` `maxOfSub = arr[maxHeap.peekFirst()]; ` ` `  `            ``// Index of minimum of current subarray ` `            ``int` `minOfSub = arr[minHeap.peekFirst()]; ` ` `  `            ``// check if the largest possible difference ` `            ``// between a pair of elements <= k ` `            ``if` `(maxOfSub - minOfSub <= k) { ` `                ``// Length of current subarray ` `                ``int` `currLength = end - beg + ``1``; ` ` `  `                ``// Update maxLength ` `                ``if` `(maxLength < currLength) ` `                    ``maxLength = currLength; ` `            ``} ` ` `  `            ``else` `{ ` `                ``// If current subarray doesn't satisy  ` `                ``// the condition then remove the starting  ` `                ``// element from subarray that satisy ` `                ``// increment the beginning pointer ` `                ``beg++; ` ` `  `                ``// Remove elements from heaps that ` `                ``// are not in the subarray anymore ` `                ``while` `(minHeap.size() > ``0` `&&  ` `                               ``minHeap.peekFirst() < beg) ` `                    ``minHeap.removeFirst(); ` ` `  `                ``while` `(maxHeap.size() > ``0` `&&  ` `                               ``maxHeap.peekFirst() < beg) ` `                    ``maxHeap.removeFirst(); ` `            ``} ` ` `  `            ``end++; ` `        ``} ` ` `  `        ``// Return the final answer ` `        ``return` `maxLength; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `arr[] = { ``1``, ``2``, ``3``, ``6``, ``7` `}; ` ` `  `        ``int` `k = ``2``; ` ` `  `        ``int` `maxLength = computeLongestSubarray(arr, k); ` `        ``System.out.println(maxLength); ` `    ``} ` `} `

Output:

```3
```

Time Complexity: O(n) because every element of the array is added and removed from the heaps only once. My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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