# Maximum sum of lengths of non-overlapping subarrays with k as the max element.

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
```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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

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

This article is contributed by Mandeep Singh. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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