# Largest sum contiguous increasing subarray

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

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

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++ 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[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 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# 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.

 \$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. ?>

Output :
Largest sum = 12

Time Complexity : O(n)

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