Given an array arr[] of N integers, the task is to find the length of the longest increasing subarray such that elements in the subarray are consecutive integers.
Examples:
Input: arr[] = {1, 9, 3, 4, 20, 2}
Output: 2
Explanation: The subarray {3, 4} is the longest subarray of consecutive elementsInput: arr[] = {36, 41, 56, 32, 33, 34, 35, 43, 32, 42}
Output: 4
Explanation: The subarray {32, 33, 34, 35} is the longest subarray of consecutive elements
Approach: The idea is to run a loop and keep a count and max (both initially zero). Follow the steps mentioned below:
- Run a loop from start to end.
- If the current element is not equal to the (previous element+1) then set the count to 1.
- Else increase the count.
- Update max with a maximum of count and max.
Below is the implementation of the above approach.
C++
// C++ program to find longest// increasing consecutive subarray#include <bits/stdc++.h>using namespace std;// Returns length of the longest// consecutive subarrayint findLongestConseqSubarr(vector<int>& v){ int ans = 0, count = 0; // find the maximum length // by traversing the array for (int i = 0; i < v.size(); i++) { // Check if the current element // is equal to previous element + 1 if (i > 0 && v[i] == v[i - 1] + 1) count++; // reset the count else count = 1; // update the maximum ans = max(ans, count); } return ans;}// Driver codeint main(){ vector<int> arr = { 1, 9, 3, 4, 20, 2 }; cout << findLongestConseqSubarr(arr); return 0;} |
Java
// Java program for the above approachimport java.io.*;import java.lang.*;import java.util.*;class GFG { // Returns length of the longest // consecutive subarray static int findLongestConseqSubarr(int arr[ ]) { int ans = 0, count = 0; // find the maximum length // by traversing the array for (int i = 0; i < arr.length; i++) { // Check if the current element // is equal to previous element + 1 if (i > 0 && arr[i] == arr[i - 1] + 1) count++; // reset the count else count = 1; // update the maximum ans = Math.max(ans, count); } return ans; } // Driver code public static void main (String[] args) { int arr[ ] = { 1, 9, 3, 4, 20, 2 }; System.out.print(findLongestConseqSubarr(arr)); }}// This code is contributed by hrithikgarg03188. |
Python3
# python3 program to find longest# increasing consecutive subarray# Returns length of the longest# consecutive subarraydef findLongestConseqSubarr(v): ans, count = 0, 0 # find the maximum length # by traversing the array for i in range(0, len(v)): # Check if the current element # is equal to previous element + 1 if (i > 0 and v[i] == v[i - 1] + 1): count += 1 # reset the count else: count = 1 # update the maximum ans = max(ans, count) return ans# Driver codeif __name__ == "__main__": arr = [1, 9, 3, 4, 20, 2] print(findLongestConseqSubarr(arr)) # This code is contributed by rakeshsahni |
C#
// C# program for the above approachusing System;class GFG{ // Returns length of the longest // consecutive subarray static int findLongestConseqSubarr(int[] arr) { int ans = 0, count = 0; // find the maximum length // by traversing the array for (int i = 0; i < arr.Length; i++) { // Check if the current element // is equal to previous element + 1 if (i > 0 && arr[i] == arr[i - 1] + 1) count++; // reset the count else count = 1; // update the maximum ans = Math.Max(ans, count); } return ans; } // Driver code public static void Main() { int[] arr = { 1, 9, 3, 4, 20, 2 }; Console.Write(findLongestConseqSubarr(arr)); }}// This code is contributed by gfgking |
Javascript
<script> // JavaScript code for the above approach // Returns length of the longest // consecutive subarray function findLongestConseqSubarr(v) { let ans = 0, count = 0; // find the maximum length // by traversing the array for (let i = 0; i < v.length; i++) { // Check if the current element // is equal to previous element + 1 if (i > 0 && v[i] == v[i - 1] + 1) count++; // reset the count else count = 1; // update the maximum ans = Math.max(ans, count); } return ans; } // Driver code let arr = [1, 9, 3, 4, 20, 2]; document.write(findLongestConseqSubarr(arr)); // This code is contributed by Potta Lokesh </script> |
Output
2
Time Complexity: O(N)
Auxiliary Space: O(1)
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