# Largest sum contiguous increasing subarray

• Difficulty Level : Easy
• Last Updated : 08 Jun, 2021

Given an array of n positive distinct integers. The problem is to find the largest sum of contiguous increasing subarray in O(n) time complexity.
Examples :

```Input : arr[] = {2, 1, 4, 7, 3, 6}
Output : 12
Contiguous Increasing subarray {1, 4, 7} = 12

Input : arr[] = {38, 7, 8, 10, 12}
Output : 38```

A simple solution is to generate all subarrays and compute their sums. Finally return the subarray with maximum sum. Time complexity of this solution is O(n2).
An efficient solution is based on the fact that all elements are positive. So we consider longest increasing subarrays and compare their sums. To increasing subarrays cannot overlap, so our time complexity becomes O(n).
Algorithm:

```Let arr be the array of size n
Let result be the required sum

int largestSum(arr, n)
result = INT_MIN  // Initialize result

i = 0
while i < n

// Find sum of longest increasing subarray
// starting with i
curr_sum = arr[i];
while i+1 < n && arr[i] < arr[i+1]
curr_sum += arr[i+1];
i++;

// If current sum is greater than current
// result.
if result < curr_sum
result = curr_sum;

i++;
return result```

Below is the implementation of above algorithm.

## C++

 `// C++ implementation of largest sum``// contiguous increasing subarray``#include ``using` `namespace` `std;` `// Returns sum of longest``// increasing subarray.``int` `largestSum(``int` `arr[], ``int` `n)``{``    ``// Initialize result``    ``int` `result = INT_MIN;` `    ``// Note that i is incremented``    ``// by inner loop also, so overall``    ``// time complexity is O(n)``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``// Find sum of longest``        ``// increasing subarray``        ``// starting from arr[i]``        ``int` `curr_sum = arr[i];``        ``while` `(i + 1 < n &&``               ``arr[i + 1] > arr[i])``        ``{``            ``curr_sum += arr[i + 1];``            ``i++;``        ``}` `        ``// Update result if required``        ``if` `(curr_sum > result)``            ``result = curr_sum;``    ``}` `    ``// required largest sum``    ``return` `result;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = {1, 1, 4, 7, 3, 6};``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``    ``cout << ``"Largest sum = "``         ``<< largestSum(arr, n);``    ``return` `0;``}`

## Java

 `// Java implementation of largest sum``// contiguous increasing subarray` `class` `GFG``{``    ``// Returns sum of longest``    ``// increasing subarray.``    ``static` `int` `largestSum(``int` `arr[], ``int` `n)``    ``{``        ``// Initialize result``        ``int` `result = -``9999999``;``    ` `        ``// Note that i is incremented``        ``// by inner loop also, so overall``        ``// time complexity is O(n)``        ``for` `(``int` `i = ``0``; i < n; i++)``        ``{``            ``// Find sum of longest``            ``// increasing subarray``            ``// starting from arr[i]``            ``int` `curr_sum = arr[i];``            ``while` `(i + ``1` `< n &&``                   ``arr[i + ``1``] > arr[i])``            ``{``                ``curr_sum += arr[i + ``1``];``                ``i++;``            ``}``    ` `            ``// Update result if required``            ``if` `(curr_sum > result)``                ``result = curr_sum;``        ``}``    ` `        ``// required largest sum``        ``return` `result;``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `arr[] = {``1``, ``1``, ``4``, ``7``, ``3``, ``6``};``        ``int` `n = arr.length;``        ``System.out.println(``"Largest sum = "` `+``                         ``largestSum(arr, n));``    ``}``}`

## Python3

 `# Python3 implementation of largest``# sum contiguous increasing subarray` `# Returns sum of longest``# increasing subarray.``def` `largestSum(arr, n):``    ` `    ``# Initialize result``    ``result ``=` `-``2147483648` `    ``# Note that i is incremented``    ``# by inner loop also, so overall``    ``# time complexity is O(n)``    ``for` `i ``in` `range``(n):``    ` `        ``# Find sum of longest increasing``        ``# subarray starting from arr[i]``        ``curr_sum ``=` `arr[i]``        ``while` `(i ``+` `1` `< n ``and``               ``arr[i ``+` `1``] > arr[i]):``        ` `            ``curr_sum ``+``=` `arr[i ``+` `1``]``            ``i ``+``=` `1``        ` `        ``# Update result if required``        ``if` `(curr_sum > result):``            ``result ``=` `curr_sum``    ` `    ``# required largest sum``    ``return` `result` `# Driver Code``arr ``=` `[``1``, ``1``, ``4``, ``7``, ``3``, ``6``]``n ``=` `len``(arr)``print``(``"Largest sum = "``, largestSum(arr, n))` `# This code is contributed by Anant Agarwal.`

## C#

 `// C# implementation of largest sum``// contiguous increasing subarray``using` `System;` `class` `GFG``{``    ` `    ``// Returns sum of longest``    ``// increasing subarray.``    ``static` `int` `largestSum(``int` `[]arr,``                          ``int` `n)``    ``{``        ` `        ``// Initialize result``        ``int` `result = -9999999;``        ` `        ``// Note that i is incremented by``        ``// inner loop also, so overall``        ``// time complexity is O(n)``        ``for` `(``int` `i = 0; i < n; i++)``        ``{``            ` `            ``// Find sum of longest increasing``            ``// subarray starting from arr[i]``            ``int` `curr_sum = arr[i];``            ``while` `(i + 1 < n &&``                   ``arr[i + 1] > arr[i])``            ``{``                ``curr_sum += arr[i + 1];``                ``i++;``            ``}``    ` `            ``// Update result if required``            ``if` `(curr_sum > result)``                ``result = curr_sum;``        ``}``    ` `        ``// required largest sum``        ``return` `result;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main ()``    ``{``        ``int` `[]arr = {1, 1, 4, 7, 3, 6};``        ``int` `n = arr.Length;``        ``Console.Write(``"Largest sum = "` `+``                    ``largestSum(arr, n));``    ``}``}` `// This code is contributed``// by Nitin Mittal.`

## PHP

 ` ``\$arr``[``\$i``])``        ``{``            ``\$curr_sum` `+= ``\$arr``[``\$i` `+ 1];``            ``\$i``++;``        ``}` `        ``// Update result if required``        ``if` `(``\$curr_sum` `> ``\$result``)``            ``\$result` `= ``\$curr_sum``;``    ``}` `    ``// required largest sum``    ``return` `\$result``;``}` `// Driver Code``{``    ``\$arr` `= ``array``(1, 1, 4, 7, 3, 6);``    ``\$n` `= sizeof(``\$arr``) / sizeof(``\$arr``);``    ``echo` `"Largest sum = "` `,``          ``largestSum(``\$arr``, ``\$n``);``    ``return` `0;``}` `// This code is contributed by nitin mittal.``?>`

## Javascript

 ``

Output :

`Largest sum = 12`

Time Complexity : O(n)
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