**Objective: ** The maximum subarray problem is the task of finding the contiguous subarray within a one-dimensional array of numbers which has the largest sum.

Example: int [] A = {−2, 1, −3, 4, −1, 2, 1, −5, 4}; Output: contiguous subarray with the largest sum is 4, −1, 2, 1, with sum 6.

**Approach:**

Naive solution would be use two for loops and check every subarray and return the subarray which has the maximum sum.

**Time Complexity**: O(N^{2})

Can we reduce it??? Yes, there are multiple solutions which solves this problem in **O(N)**. In this article we will see very known algorithm called kadane’s Algorithm.

Click here to read about how to solve this problem using Dynamic Programming.

**Kadane’s Algorithm:**

start: max_so_far = 0 max_ending_here = 0 loop i= 0 to n (i) max_ending_here = max_ending_here + a[i] (ii) if(max_ending_here < 0) max_ending_here = 0 (iii) if(max_so_far < max_ending_here) max_so_far = max_ending_here return max_so_far

Let’s walk through this algorithm with one example

int [] A = {−2, 1, −3, 4, −1, 2, 1};

max_so_far = 0 max_ending_here = 0 i=0 and a[0] = -2 max_ending_here = max_ending_here + (-2) = -2 max_ending_here < 0 => max_ending_here = 0 i=1 and a[1] = 1 max_ending_here = max_ending_here + (1) => 0 + 1 => 1 max_ending_here > 0 max_so_far < max_ending_here is true => max_so_far = max_ending_here max_so_far = 1 i=2 and a[2] = -3 max_ending_here = max_ending_here + (-3) => 1 -3 = -2 max_ending_here < 0 => max_ending_here = 0 max_so_far < max_ending_here is false so do not update max_so_far. max_so_far = 1 i=3 and a[3] = 4 max_ending_here = max_ending_here + 4 => 0 + 4 => 4 max_ending_here > 0 max_so_far < max_ending_here is true => max_so_far = max_ending_here max_so_far = 4 i=4 and a[4] = -1 max_ending_here = max_ending_here + (-1) => 4 - 1 => 3 max_ending_here > 0 max_so_far < max_ending_here is false max_so_far = 4 i=5 and a[5] = 2 max_ending_here = max_ending_here + 2 => 3 + 2 => 5 max_ending_here > 0 max_so_far < max_ending_here is true => max_so_far = max_ending_here max_so_far = 5 i=6 and a[6] = 1 max_ending_here = max_ending_here + 1 => 5 + 1 => 6 max_ending_here > 0 max_so_far < max_ending_here is true => max_so_far = max_ending_here max_so_far = 6 return max_so_far = 6

**NOTE**: The above algorithm works only when there is at least one positive element in the array.

We need to modify the algorithm to work it if all the elements are negative.

start: max_so_far = a[0] max_ending_here = a[0] loop i= 1 to n (i) max_end_here = Max(arrA[i], max_end_here+a[i]); (ii) max_so_far = Max(max_so_far,max_end_here); return max_so_far

**Complete Code for both algorithms:**

**Output:**

Maximum subarray is 10 Maximum Subarray when all elements are negative : -2

**References:**

https://en.wikipedia.org/wiki/Maximum_subarray_problem