Given an array arr[] of length N, the task is to find the length of the longest subarray which consists of consecutive numbers in increasing order, from the array.
Examples:
Input: arr[] = {2, 3, 4, 6, 7, 8, 9, 10}
Output: 5
Explanation: Subarray {6, 7, 8, 9, 10} is the longest subarray satisfying the given conditions. Therefore, the required output is 5.Input: arr[] = {4, 5, 1, 2, 3, 4, 9, 10, 11, 12}
Output: 4
Naive Approach: The simplest approach to solve the problem is to traverse the array and for every index i, traverse from over-index and find the length of the longest subarray satisfying the given condition starting from i. Shift i to the index which does not satisfy the condition and check from that index. Finally, print the maximum length of such subarray obtained.
Below is the implementation of the above approach:
C++
// C++ implementation for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the longest subarray// with increasing contiguous elementsint maxiConsecutiveSubarray(int arr[], int N){ // Stores the length of // required longest subarray int maxi = 0; for (int i = 0; i < N - 1; i++) { // Stores the length of length of longest // such subarray from ith index int cnt = 1, j; for (j = i; j < N; j++) { // If consecutive elements are // increasing and differ by 1 if (arr[j + 1] == arr[j] + 1) { cnt++; } // Otherwise else { break; } } // Update the longest subarray // obtained so far maxi = max(maxi, cnt); i = j; } // Return the length obtained return maxi;}// Driver Codeint main(){ int N = 11; int arr[] = { 1, 3, 4, 2, 3, 4, 2, 3, 5, 6, 7 }; cout << maxiConsecutiveSubarray(arr, N); return 0;} |
Java
// Java implementation for the above approachimport java.util.*;class GFG{ // Function to find the longest subarray// with increasing contiguous elementspublic static int maxiConsecutiveSubarray(int arr[], int N){ // Stores the length of // required longest subarray int maxi = 0; for(int i = 0; i < N - 1; i++) { // Stores the length of length of // longest such subarray from ith // index int cnt = 1, j; for(j = i; j < N - 1; j++) { // If consecutive elements are // increasing and differ by 1 if (arr[j + 1] == arr[j] + 1) { cnt++; } // Otherwise else { break; } } // Update the longest subarray // obtained so far maxi = Math.max(maxi, cnt); i = j; } // Return the length obtained return maxi;}// Driver Codepublic static void main(String args[]){ int N = 11; int arr[] = { 1, 3, 4, 2, 3, 4, 2, 3, 5, 6, 7 }; System.out.println(maxiConsecutiveSubarray(arr, N));}}// This code is contributed by hemanth gadarla |
Python3
# Python3 implementation for # the above approach# Function to find the longest # subarray with increasing # contiguous elementsdef maxiConsecutiveSubarray(arr, N): # Stores the length of # required longest subarray maxi = 0; for i in range(N - 1): # Stores the length of # length of longest such # subarray from ith index cnt = 1; for j in range(i, N - 1): # If consecutive elements are # increasing and differ by 1 if (arr[j + 1] == arr[j] + 1): cnt += 1; # Otherwise else: break; # Update the longest subarray # obtained so far maxi = max(maxi, cnt); i = j; # Return the length obtained return maxi;# Driver Codeif __name__ == '__main__': N = 11; arr = [1, 3, 4, 2, 3, 4, 2, 3, 5, 6, 7]; print(maxiConsecutiveSubarray(arr, N));# This code is contributed by Rajput-Ji |
C#
// C# implementation for the// above approachusing System;class GFG{ // Function to find the longest // subarray with increasing // contiguous elementspublic static int maxiConsecutiveSubarray(int []arr, int N){ // Stores the length of // required longest subarray int maxi = 0; for(int i = 0; i < N - 1; i++) { // Stores the length of // length of longest such // subarray from ith index int cnt = 1, j; for(j = i; j < N - 1; j++) { // If consecutive elements are // increasing and differ by 1 if (arr[j + 1] == arr[j] + 1) { cnt++; } // Otherwise else { break; } } // Update the longest subarray // obtained so far maxi = Math.Max(maxi, cnt); i = j; } // Return the length // obtained return maxi;}// Driver Codepublic static void Main(String []args){ int N = 11; int []arr = {1, 3, 4, 2, 3, 4, 2, 3, 5, 6, 7}; Console.WriteLine( maxiConsecutiveSubarray(arr, N));}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript program to implement// the above approach// Function to find the longest subarray// with increasing contiguous elementsfunction maxiConsecutiveSubarray(arr, N){ // Stores the length of // required longest subarray let maxi = 0; for(let i = 0; i < N - 1; i++) { // Stores the length of length of // longest such subarray from ith // index let cnt = 1, j; for(j = i; j < N - 1; j++) { // If consecutive elements are // increasing and differ by 1 if (arr[j + 1] == arr[j] + 1) { cnt++; } // Otherwise else { break; } } // Update the longest subarray // obtained so far maxi = Math.max(maxi, cnt); i = j; } // Return the length obtained return maxi;} // Driver Code let N = 11; let arr = [ 1, 3, 4, 2, 3, 4, 2, 3, 5, 6, 7 ]; document.write(maxiConsecutiveSubarray(arr, N)); </script> |
Output
3
Time Complexity: O(N2)
Auxiliary Space: O(1)
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