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

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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 index [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 indexes [3, 6], i.e, { 2, 4, 7, 2 } contains no such difference of two elements which is greater than 5. 

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)

Longest subarray in which absolute difference between any two element is not greater than X using Sliding Window and Map:

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 <bits/stdc++.h>
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 through 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;
}


Java




// JAVA program to find the longest sub-array
// where the absolute difference between any
// two elements is not greater than X
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG
{
 
  // Function that prints the longest sub-array
  // where the absolute difference between any
  // two element is not greater than X
  static 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
    TreeMap<Integer, Integer> window = new TreeMap<>();
 
    // Initialize the window
    int start = 0, end = 0;
 
    // Loop through the array
    for (; end < N; end++)
    {
 
      // Increment the count of that
      // element in the window
      window.put(A[end],
                 window.getOrDefault(A[end], 0) + 1);
 
      // Find the maximum and minimum element
      // in the current window
      int minimum = window.firstKey();
      int maximum = window.lastKey();
 
      // 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.put(A[start],
                     window.get(A[start]) - 1);
 
          // Remove the element from the window
          // if its count is zero
          if (window.get(A[start]) == 0) {
 
            window.remove(A[start]);
          }
 
          // Increment the start of the window
          start++;
 
          // Find the maximum and minimum element
          // in the current window
          minimum = window.firstKey();
          maximum = window.lastKey();
 
          // 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++)
      System.out.print(A[i] + " ");
  }
 
  // Driver Code
  public static void main(String[] args)
  {
 
    // Given array
    int arr[] = { 15, 10, 1, 2, 4, 7, 2 }, X = 5;
 
    // store the size of the array
    int n = arr.length;
 
    // Function call
    longestSubarray(arr, n, X);
  }
}
 
// This code is contributed by Kingash.


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 through 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


C#




// C# program to find the longest sub-array
// where the absolute difference between any
// two elements is not greater than X
using System;
using System.Linq;
using System.Collections.Generic;
 
class GFG
{
 
  // Function that prints the longest sub-array
  // where the absolute difference between any
  // two element is not greater than X
  static 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
    Dictionary<int, int> window = new Dictionary<int, int>();
 
    // Initialize the window
    int start = 0, end = 0;
 
    // Loop through the array
    for (; end < N; end++)
    {
 
      // Increment the count of that
      // element in the window
      if (!window.ContainsKey(A[end]))
        window[A[end]] = 0;
      window[A[end]]++;
       
 
      // Find the maximum and minimum element
      // in the current window
      int minimum = window.Keys.Min();
      int maximum = window.Keys.Max();
 
      // 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.Remove(A[start]);
          }
 
          // Increment the start of the window
          start++;
 
          // Find the maximum and minimum element
          // in the current window
          minimum =window.Keys.Min();
          maximum = window.Keys.Max();
 
          // 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++)
        Console.Write(A[i] + " ");
  }
 
  // Driver Code
  public static void Main(string[] args)
  {
 
    // Given array
    int[] arr = { 15, 10, 1, 2, 4, 7, 2 };
    int X = 5;
 
    // store the size of the array
    int n = arr.Length;
 
    // Function call
    longestSubarray(arr, n, X);
  }
}
 
// This code is contributed by phasing17.


Javascript




<script>
 
// JavaScript 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
function longestSubarray(A, N, X){
     
    // Initialize a variable to store
    // length of longest sub-array
    let maxLen = 0
 
    // Initialize a variable to store the
    // beginning of the longest sub-array
    let beginning = 0
 
    // Initialize a map to store the maximum
    // and the minimum elements for a given window
    let window = new Map()
 
    // Initialize the window
    let start = 0
 
    // Loop through the array
    for(let end=0;end<N;end++){
 
        // Increment the count of that
        // element in the window
        if(window.has(A[end]))
            window.set(A[end],window.get(A[end]) + 1)
        else
            window.set(A[end] , 1)
 
        // Find the maximum and minimum element
        // in the current window
        let minimum = Math.min(...window.keys())
        let maximum = Math.max(...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.set(A[start],window.get(A[start]) - 1)
 
                // Remove the element from the window
                // if its count is zero
                if(window.get(A[start]) == 0)
                    window.delete(A[start])
 
                // Increment the start of the window
                start += 1
 
                // Find the maximum and minimum element
                // in the current window
                minimum = Math.min(...window.keys())
                maximum = Math.max(...window.keys())
 
                // Stop decreasing the size of window
                // when difference is not greater
                if(maximum - minimum <= X)
                    break
            }
        }
    }
                     
    // Print the longest sub-array
    for(let i=beginning;i<beginning+maxLen;i++){
        document.write(A[i],' ')
    }
}
 
// Driver Code
let arr = [15, 10, 1, 2, 4, 7, 2]
let X = 5
let n = arr.length
longestSubarray(arr, n, X)
 
// This code is contributed by Shinjanpatra
 
</script>


Output

2 4 7 2 








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

Longest subarray in which absolute difference between any two element is not greater than X using Monotonic Queues:

Maintain a sliding window and keep track of the minimum and maximum element in the window. Keep adding the elements to the window by moving the end pointer till the difference between the maximum and minimum element does not exceed X. If the difference becomes greater than X, then keep decreasing the window by moving the start pointer until the difference becomes less than X. Also pop the elements as the size of the window decreases. Find the longest sliding window with difference less than X and print it.
Maintain a monotonically increasing min queue to keep track of minimum elements and a monotonically decreasing max queue to keep track of maximum elements.

Steps to solve the problem:

  • Initialize two empty dequeues, minQueue and maxQueue.
  • Maintain a variable ansStart and ansEnd to mark the start and end of largest subarray.
  • Maintain two pointers start = 0 and end = 0 to mark the range of sliding window.
  • Iterate over all the elements of the array. For each element,
    • Pop the elements from the back of minQueue till they are greater than the current element.
    • Pop the elements from the back of maxQueue till they are smaller than the current element.
    • Push the current index to both minQueue and maxQueue.
    • Compare the minimum element from minQueue and Maximum element from maxQueue. If the difference is greater than X, move start to start + 1 and pop from the queues accordingly.
    • Update ansStart and ansEnd if the current window has length greater than ans.
  • Return the value of ans.

Below is the implementation of the above approach:

C++




#include <bits/stdc++.h>
using namespace std;
 
int longestSubarray(vector<int>& arr, int X)
{
    // Monotonic Queue to store maximum and minimum elements
    deque<int> minQueue, maxQueue;
    // Pointers to mark the range of current subarray
    int n = arr.size(), start = 0, end = 0;
    // Pointers to mark the range of maximum subarray
    int ansStart = 0, ansEnd = 0;
    while (end < n) {
        // Pop the elements greater than current element
        // from min Queue
        while (!minQueue.empty()
               && arr[minQueue.back()] > arr[end])
            minQueue.pop_back();
        // Pop the elements smaller than current element
        // from max Queue
        while (!maxQueue.empty()
               && arr[maxQueue.back()] < arr[end])
            maxQueue.pop_back();
        // Push the current index to both the queues
        minQueue.push_back(end);
        maxQueue.push_back(end);
        // Check if the subarray has maximum difference less
        // than X
        while (arr[maxQueue.front()] - arr[minQueue.front()]
               > X) {
            // Reduce the length of sliding window by moving
            // the start pointer
            if (start == minQueue.front())
                minQueue.pop_front();
            if (start == maxQueue.front())
                maxQueue.pop_front();
            start += 1;
        }
        // Maximize the subarray length
        if (end - start > ansEnd - ansStart) {
            ansStart = start;
            ansEnd = end;
        }
        end += 1;
    }
    for (int i = ansStart; i <= ansEnd; i++)
        cout << arr[i] << " ";
}
 
int main()
{
    vector<int> arr{ 15, 10, 1, 2, 4, 7, 2 };
    int X = 5;
    longestSubarray(arr, X);
    return 0;
}


Java




import java.util.LinkedList;
import java.util.Queue;
 
class GFG {
    public static void main(String[] args) {
        LinkedList<Integer> arr = new LinkedList<Integer>();
        arr.add(15);
        arr.add(10);
        arr.add(1);
        arr.add(2);
        arr.add(4);
        arr.add(7);
        arr.add(2);
 
        int X = 5;
        LongestSubarray(arr, X);
    }
 
    static void LongestSubarray(LinkedList<Integer> arr, int X) {
        // Monotonic Queue to store maximum and minimum elements
        Queue<Integer> minQueue = new LinkedList<Integer>();
        Queue<Integer> maxQueue = new LinkedList<Integer>();
 
        // Pointers to mark the range of the current subarray
        int n = arr.size();
        int start = 0, end = 0;
 
        // Pointers to mark the range of the maximum subarray
        int ansStart = 0, ansEnd = 0;
 
        while (end < n) {
            // Pop the elements greater than the current element from min Queue
            while (!minQueue.isEmpty() && arr.get(minQueue.peek()) > arr.get(end))
                minQueue.poll();
 
            // Pop the elements smaller than the current element from max Queue
            while (!maxQueue.isEmpty() && arr.get(maxQueue.peek()) < arr.get(end))
                maxQueue.poll();
 
            // Push the current index to both queues
            minQueue.offer(end);
            maxQueue.offer(end);
 
            // Check if the subarray has a maximum difference less than X
            while (arr.get(maxQueue.peek()) - arr.get(minQueue.peek()) > X) {
                // Reduce the length of the sliding window by moving the start pointer
                if (start == minQueue.peek())
                    minQueue.poll();
                if (start == maxQueue.peek())
                    maxQueue.poll();
                start += 1;
            }
 
            // Maximize the subarray length
            if (end - start > ansEnd - ansStart) {
                ansStart = start;
                ansEnd = end;
            }
 
            end += 1;
        }
 
        // Print the elements of the longest subarray
        for (int i = ansStart; i <= ansEnd; i++) {
            System.out.print(arr.get(i) + " ");
        }
    }
}


Python3




from collections import deque
 
def longest_subarray(arr, X):
    # Monotonic Queue to store maximum and minimum elements
    min_queue, max_queue = deque(), deque()
    # Pointers to mark the range of the current subarray
    n, start, end = len(arr), 0, 0
    # Pointers to mark the range of the maximum subarray
    ans_start, ans_end = 0, 0
 
    while end < n:
        # Pop the elements greater than the current element from the min Queue
        while min_queue and arr[min_queue[-1]] > arr[end]:
            min_queue.pop()
 
        # Pop the elements smaller than the current element from the max Queue
        while max_queue and arr[max_queue[-1]] < arr[end]:
            max_queue.pop()
 
        # Push the current index to both queues
        min_queue.append(end)
        max_queue.append(end)
 
        # Check if the subarray has a maximum difference less than X
        while arr[max_queue[0]] - arr[min_queue[0]] > X:
            # Reduce the length of the sliding window by moving the start pointer
            if start == min_queue[0]:
                min_queue.popleft()
            if start == max_queue[0]:
                max_queue.popleft()
            start += 1
 
        # Maximize the subarray length
        if end - start > ans_end - ans_start:
            ans_start, ans_end = start, end
 
        end += 1
 
    # Print the elements of the longest subarray
    for i in range(ans_start, ans_end + 1):
        print(arr[i], end=" ")
 
# Driver code
arr = [15, 10, 1, 2, 4, 7, 2]
X = 5
longest_subarray(arr, X)


C#




using System;
using System.Collections.Generic;
 
class GFG
{
    static void Main()
    {
        List<int> arr = new List<int> { 15, 10, 1, 2, 4, 7, 2 };
        int X = 5;
        LongestSubarray(arr, X);
    }
 
    static void LongestSubarray(List<int> arr, int X)
    {
        // Monotonic Queue to store maximum and minimum elements
        Queue<int> minQueue = new Queue<int>();
        Queue<int> maxQueue = new Queue<int>();
         
        // Pointers to mark the range of the current subarray
        int n = arr.Count;
        int start = 0, end = 0;
 
        // Pointers to mark the range of the maximum subarray
        int ansStart = 0, ansEnd = 0;
 
        while (end < n)
        {
            // Pop the elements greater than the current element from min Queue
            while (minQueue.Count > 0 && arr[minQueue.Peek()] > arr[end])
                minQueue.Dequeue();
 
            // Pop the elements smaller than the current element from max Queue
            while (maxQueue.Count > 0 && arr[maxQueue.Peek()] < arr[end])
                maxQueue.Dequeue();
 
            // Push the current index to both queues
            minQueue.Enqueue(end);
            maxQueue.Enqueue(end);
 
            // Check if the subarray has a maximum difference less than X
            while (arr[maxQueue.Peek()] - arr[minQueue.Peek()] > X)
            {
                // Reduce the length of the sliding window by moving the start pointer
                if (start == minQueue.Peek())
                    minQueue.Dequeue();
                if (start == maxQueue.Peek())
                    maxQueue.Dequeue();
                start += 1;
            }
 
            // Maximize the subarray length
            if (end - start > ansEnd - ansStart)
            {
                ansStart = start;
                ansEnd = end;
            }
 
            end += 1;
        }
 
        // Print the elements of the longest subarray
        for (int i = ansStart; i <= ansEnd; i++)
        {
            Console.Write(arr[i] + " ");
        }
    }
}


Javascript




function longestSubarray(arr, X) {
    // Monotonic Queue to store maximum and minimum elements
    const minQueue = [];
    const maxQueue = [];
    // Pointers to mark the range of current subarray
    let n = arr.length;
    let start = 0;
    let end = 0;
    // Pointers to mark the range of maximum subarray
    let ansStart = 0;
    let ansEnd = 0;
 
    while (end < n) {
        // Pop the elements greater than the current element from the min Queue
        while (minQueue.length > 0 && arr[minQueue[minQueue.length - 1]] > arr[end]) {
            minQueue.pop();
        }
        // Pop the elements smaller than the current element from the max Queue
        while (maxQueue.length > 0 && arr[maxQueue[maxQueue.length - 1]] < arr[end]) {
            maxQueue.pop();
        }
        // Push the current index to both queues
        minQueue.push(end);
        maxQueue.push(end);
        // Check if the subarray has a maximum difference less than X
        while (arr[maxQueue[0]] - arr[minQueue[0]] > X) {
            // Reduce the length of the sliding window by moving the start pointer
            if (start === minQueue[0]) {
                minQueue.shift();
            }
            if (start === maxQueue[0]) {
                maxQueue.shift();
            }
            start += 1;
        }
        // Maximize the subarray length
        if (end - start > ansEnd - ansStart) {
            ansStart = start;
            ansEnd = end;
        }
        end += 1;
    }
 
    for (let i = ansStart; i <= ansEnd; i++) {
        console.log(arr[i] + " ");
    }
}
 
// Test
const arr = [15, 10, 1, 2, 4, 7, 2];
const X = 5;
longestSubarray(arr, X);


Output

2 4 7 2 








Time Complexity: O(N), where N is the size of the array
Auxiliary Space: O(N)



Last Updated : 07 Jan, 2024
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