This is an easy Leetcode problem 53. See below image for description:

ALGORITHM

1.Have a global variable currentSum;

2.Have a global variable max that will also be the return value;

3.Loop through the array, check if the current element (nums[i]) is greater than the current sum and current element put together (i.e currentSum + nums[i]);

4.if it is greater, reassign the currentSum variable to be the current element, nums[i];

5.if it is not, reassign the currentSum variable to be the summation of both the existing current sum and the current array element (currentSum = currentSum + nums[i])

6.finally, check if the currentSum is greater than the current maximum. if it is, reassign the current max to be equal to currentSum.

7.Then return max.

CODE

public static int maxSubArray(int[] nums) { int currentSum = nums[0]; int max = nums[0]; for (int i = 1; i < nums.length; i++) { if (nums[i] > (currentSum + nums[i])) currentSum = nums[i]; else currentSum = currentSum + nums[i]; if (currentSum > max) { max = currentSum; } } return max; } 

OUTCOME

IMPROVEMENT
Math.max checks the maximum/bigger value between two numbers. So we can replace those if statements with Math.max(). Cleaner code is shown below:

IMPROVED CODE

 public static int maxSubArray(int[] nums) { int currentSum = nums[0]; int max = nums[0]; for (int i = 1; i < nums.length; i++) { //cleaner and simple currentSum = Math.max(nums[i], currentSum + nums[i]); max = Math.max(currentSum, max); } return max; } 

LESSONS
1.Don’t always be in the box of a particular algorithm you thought would work, just the way i thought Sliding window would work.

2.Using library functions where and when needed.

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