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Maximum sum of lengths of non-overlapping subarrays with k as the max element.
• Difficulty Level : Medium
• Last Updated : 29 Apr, 2021

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

```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

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

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```

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 ``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);``    ``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``    ``i``=``0``    ``while` `i < 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 Rohit`

## C#

 `// A C# program to calculate max``// sum lengths of non overlapping``// contiguous subarrays with k as``// max element``using` `System;``class` `GFG {``    ` `    ``// Returns max sum of lengths``    ``// with maximum element as k``    ``static` `int` `calculateMaxSumLength(``int` `[]arr,``                                     ``int` `n,``                                     ``int` `k)``    ``{``        ` `        ``// final sum of lengths``        ``int` `ans = 0;` `        ``// 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 Code``    ``public` `static` `void` `Main()``    ``{``        ``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);``        ``Console.WriteLine(``"Max Length :: "` `+ ans);``    ``}``}` `// This code is contributed by anuj_67.`

## PHP

 ` ``\$k` `&& ``\$i` `< ``\$n``)``            ``\$i``++;    ``    ``}``    ``return` `\$ans``;``}` `// Driver Code``\$arr` `= ``array``( 4, 5, 7, 1, 2,   ``              ``9, 8, 4, 3, 1 );``\$size` `= sizeof(``\$arr``);``\$k` `= 4;``\$ans` `= calculateMaxSumLength(``\$arr``, ``\$size``, ``\$k``);``echo` `"Max Length :: "` `. ``\$ans` `. ``"\n"``;` `// This code is contributed by ita_c``?>`

## Javascript

 ``

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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