Largest sum contiguous increasing subarray

11 Jul, 2022
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++
Java
Python3
C#
PHP
Javascript
// C++ implementation of largest sum// contiguous increasing subarray#include <bits/stdc++.h>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 Codeint main(){    int arr[] = {1, 1, 4, 7, 3, 6};    int n = sizeof(arr) / sizeof(arr[0]);    cout << "Largest sum = "         << largestSum(arr, n);    return 0;}
// 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 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 Codearr = [1, 1, 4, 7, 3, 6]n = len(arr)print("Largest sum = ", largestSum(arr, n)) # This code is contributed by Anant Agarwal.
// C# implementation of largest sum// contiguous increasing subarrayusing 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// PHP implementation of largest sum// contiguous increasing subarray // Returns sum of longest// increasing subarray.function largestSum($arr, $n){    $INT_MIN = 0;         // Initialize result    $result = $INT_MIN;     // Note that i is incremented    // by inner loop also, so overall    // time complexity is O(n)    for ($i = 0; $i < $n; $i++)    {        // Find sum of longest        // increasing subarray        // starting from arr[i]        $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{    $arr = array(1, 1, 4, 7, 3, 6);    $n = sizeof($arr) / sizeof($arr[0]);    echo "Largest sum = " ,          largestSum($arr, $n);    return 0;} // This code is contributed by nitin mittal.?>
<script> // Javascript implementation of largest sum// contiguous increasing subarray // Returns sum of longest// increasing subarray.function largestSum(arr, n){    // Initialize result    var result = -1000000000;     // Note that i is incremented    // by inner loop also, so overall    // time complexity is O(n)    for (var i = 0; i < n; i++)    {        // Find sum of longest        // increasing subarray        // starting from arr[i]        var 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 Codevar arr = [1, 1, 4, 7, 3, 6];var n = arr.length;document.write( "Largest sum = "      + largestSum(arr, n)); // This code is contributed by itsok.</script>
Largest sum = 12
Time Complexity : O(n)
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