Find the contiguous subarray within an array (containing at least one number) which has the largest sum.

For example, given the array [−2,1,−3,4,−1,2,1,−5,4],

the contiguous subarray [4,−1,2,1] has the largest sum = 6.

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More practice:

If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.

The idea is like this: at some index A[i], if A[i] + previous max < A[i], then we can start new counting new max at A[i], because if you include the previous elements, the result will be smaller for all the elements after A[i], so better choose A[i] itself for all the elements after A[i]. You will also need another variable max to record the max value you have scanned. Because the max value before A[i] could be greater than the max of subarray including elements after A[i]. Scan the whole array then you will be able to find the max sum of the subarray.

public class Solution {
public int maxSubArray(int[] A) {
if (A.length == 0)
return 0;
int currMax = A[0];
int max = A[0];
for (int i = 1; i < A.length; i++) {
if (A[i] + currMax < A[i]) {
currMax = A[i];
}
else {
currMax = currMax + A[i];
}
max = Math.max(max, currMax);
}
return max;
}
}

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