Find the maximum sum of lengths of non-overlapping subarrays (contiguous elements) with k as the maximum element.

Example 1:

Input : arr[] = {2, 1, 4, 9, 2, 3, 8, 3, 4} k = 4 Output : 5 {2, 1, 4} => Length = 3 {3, 4} => Length = 2 So, 3 + 2 = 5 is the answer

Example 2:

Input : arr[] = {1, 2, 3, 2, 3, 4, 1} k = 4 Output : 7 {1, 2, 3, 2, 3, 4, 1} => Length = 7

Example 3:

Input : arr = {4, 5, 7, 1, 2, 9, 8, 4, 3, 1} k = 4 Ans = 4 {4} => Length = 1 {4, 3, 1} => Length = 3 So, 1 + 3 = 4 is the answer

**question source :** http://www.geeksforgeeks.org/amazon-interview-experience-set-376-campus-internship/

**Algorithm :**

Traverse the array starting from first element Take a loop and keep on incrementing count If element is less than equal to k if array element is equal to k, then mark a flag If flag is marked, add this count to answer Take another loop and traverse the array till element is greater than k return ans

## C++

// CPP program to calculate max sum lengths of // non overlapping contiguous subarrays with k as // max element #include <bits/stdc++.h> using namespace std; // Returns max sum of lengths with maximum element // as k int calculateMaxSumLength(int arr[], int n, int k) { int ans = 0; // final sum of lengths // number of elements in current subarray int count = 0; // variable for checking if k appeared in subarray int flag = 0; for (int i = 0; i < n;) { count = 0; flag = 0; // count the number of elements which are // less than equal to k while (arr[i] <= k && i < n) { count++; if (arr[i] == k) flag = 1; i++; } // if current element appeared in current // subarray add count to sumLength if (flag == 1) ans += count; // skip the array elements which are // greater than k while (arr[i] > k && i < n) i++; } return ans; } // driver program int main() { int arr[] = { 4, 5, 7, 1, 2, 9, 8, 4, 3, 1 }; int size = sizeof(arr) / sizeof(arr[0]); int k = 4; int ans = calculateMaxSumLength(arr, size, k); cout << "Max Length :: " << ans << endl; return 0; }

## Java

// A Java program to calculate max sum lengths of // non overlapping contiguous subarrays with k as // max element public class GFG { // Returns max sum of lengths with maximum element // as k static int calculateMaxSumLength(int arr[], int n, int k) { int ans = 0; // final sum of lengths // number of elements in current subarray int count = 0; // variable for checking if k appeared in subarray int flag = 0; for (int i = 0; i < n;) { count = 0; flag = 0; // count the number of elements which are // less than equal to k while (i < n && arr[i] <= k) { count++; if (arr[i] == k) flag = 1; i++; } // if current element appeared in current // subarray add count to sumLength if (flag == 1) ans += count; // skip the array elements which are // greater than k while (i < n && arr[i] > k) i++; } return ans; } // driver program to test above method public static void main(String[] args) { int arr[] = { 4, 5, 7, 1, 2, 9, 8, 4, 3, 1 }; int size = arr.length; int k = 4; int ans = calculateMaxSumLength(arr, size, k); System.out.println("Max Length :: " + ans); } } // This code is contributed by Sumit Ghosh

## Python

# Python program to calculate max sum lengths of non # overlapping contiguous subarrays with k as max element # Returns max sum of lengths with max elements as k def calculateMaxSumLength(arr, n, k): ans = 0 # final sum of lengths for i in range(n): # number of elements in current sub array count = 0 # Variable for checking if k appeared in the sub array flag = 0 # Count the number of elements which are # less than or equal to k while i < n and arr[i] <= k : count = count + 1 if arr[i] == k: flag = 1 i = i + 1 # if current element appeared in current # subarray and count to sumLength if flag == 1: ans = ans + count # skip the array elements which are greater than k while i < n and arr[i] > k : i = i + 1 return ans # Driver Program arr = [4, 5, 7, 1, 2, 9, 8, 4, 3, 1] size = len(arr) k = 4 ans = calculateMaxSumLength(arr, size, k) print "Max Length ::",ans # Contributed by Harshit Agrawal

Output:

Max Length :: 4

**Time Complexity :** O(n)

It may look like O(n2), but if you take a closer look, array is traversed only once

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