/*

input : array of integer

output : largest sum of subarray

constraints :

1) is the input array not empty?

yes. at least one el

2) range of integers

[-10^4, 10^4]

3) maximum lenght of input

[10^5]

>> maximum sum = 10^5 * 10^4 = 10 ^ 9 < INTEGER

sol1) brute force

nested for loop : O(n^2)

tc : O(n^2), sc : O(1)

sol2) dp?

Subarray elements are continuous in the original array, so we can use dp.

let dp[i] represent the largest sum of a subarray where the ith element is the last element of the subarray.

if dp[i-1] + curval < cur val : take curval

if dp[i-1] + cur val >= curval : take dp[i-1] + curval

tc : O(n) sc : O(n)

*/

class Solution {

public int maxSubArray(int[] nums) {

int n = nums.length;

int[] dp = new int[n];

int maxSum = nums[0];

dp[0] = nums[0];

for(int i = 1; i < n; i++) {

if(dp[i-1] + nums[i] < nums[i]) {

dp[i] = nums[i];

} else {

dp[i] = nums[i] + dp[i-1];

}

maxSum = Math.max(maxSum, dp[i]);

}

return maxSum;

}

}