Home >Java >javaTutorial >Kadanes Algorithm: Leetcode Maximum subarray
We can build the intuition based on the two-point approach.
We will start with two variables maxSum and maxTillNow.
The first variable stores the max sum we have attained overall in the array.
The second variable stores the value of the maximum sum attained till the current index. Since the array has a negative value, this value will fluctuate, but whenever we get maxSum < maxTillNow, we update the maxSum.
The final case we have to handle will be if the maximum sum till any index reaches zero, i.e., maxTillNow < 0, reset the maxTillNow = 0 at the end of loop.
Time complexity: O(N)
Space complexity: O(1)
class Solution { public int maxSubArray(int[] nums) { int maxSum = Integer.MIN_VALUE; int maxTillNow = 0; for(int i =0;i<nums.length;i++){ maxTillNow+=nums[i]; maxSum = Math.max(maxTillNow,maxSum); if(maxTillNow<0) maxTillNow = 0; } return maxSum; } }
GitHub repo for more solutions: Git
Leetcode profile: Leetcode: devn007
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