Given an array of distinct integers, find length of the longest subarray which contains numbers that can be arranged in a continuous sequence.

Examples:

Input: arr[] = {10, 12, 11}; Output: Length of the longest contiguous subarray is 3 Input: arr[] = {14, 12, 11, 20}; Output: Length of the longest contiguous subarray is 2 Input: arr[] = {1, 56, 58, 57, 90, 92, 94, 93, 91, 45}; Output: Length of the longest contiguous subarray is 5

**We strongly recommend to minimize the browser and try this yourself first.**

The important thing to note in question is, it is given that all elements are distinct. If all elements are distinct, then a subarray has contiguous elements if and only if the difference between maximum and minimum elements in subarray is equal to the difference between last and first indexes of subarray. So the idea is to keep track of minimum and maximum element in every subarray.

The following is the implementation of above idea.

## C++

#include<iostream> using namespace std; // Utility functions to find minimum and maximum of // two elements int min(int x, int y) { return (x < y)? x : y; } int max(int x, int y) { return (x > y)? x : y; } // Returns length of the longest contiguous subarray int findLength(int arr[], int n) { int max_len = 1; // Initialize result for (int i=0; i<n-1; i++) { // Initialize min and max for all subarrays starting with i int mn = arr[i], mx = arr[i]; // Consider all subarrays starting with i and ending with j for (int j=i+1; j<n; j++) { // Update min and max in this subarray if needed mn = min(mn, arr[j]); mx = max(mx, arr[j]); // If current subarray has all contiguous elements if ((mx - mn) == j-i) max_len = max(max_len, mx-mn+1); } } return max_len; // Return result } // Driver program to test above function int main() { int arr[] = {1, 56, 58, 57, 90, 92, 94, 93, 91, 45}; int n = sizeof(arr)/sizeof(arr[0]); cout << "Length of the longest contiguous subarray is " << findLength(arr, n); return 0; }

## Java

class LargestSubArray2 { // Utility functions to find minimum and maximum of // two elements int min(int x, int y) { return (x < y) ? x : y; } int max(int x, int y) { return (x > y) ? x : y; } // Returns length of the longest contiguous subarray int findLength(int arr[], int n) { int max_len = 1; // Initialize result for (int i = 0; i < n - 1; i++) { // Initialize min and max for all subarrays starting with i int mn = arr[i], mx = arr[i]; // Consider all subarrays starting with i and ending with j for (int j = i + 1; j < n; j++) { // Update min and max in this subarray if needed mn = min(mn, arr[j]); mx = max(mx, arr[j]); // If current subarray has all contiguous elements if ((mx - mn) == j - i) max_len = max(max_len, mx - mn + 1); } } return max_len; // Return result } public static void main(String[] args) { LargestSubArray2 large = new LargestSubArray2(); int arr[] = {1, 56, 58, 57, 90, 92, 94, 93, 91, 45}; int n = arr.length; System.out.println("Length of the longest contiguous subarray is " + large.findLength(arr, n)); } } // This code has been contributed by Mayank Jaiswal

Output:

Length of the longest contiguous subarray is 5

Time Complexity of the above solution is O(n^{2}).

We will soon be covering solution for the problem where duplicate elements are allowed in subarray.

Length of the largest subarray with contiguous elements | Set 2

This article is contributed by **Arjun**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above