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/*
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;
}
}