Maximum Subarray

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.

Kadane’s Algorithm:

Initialize:
    max_so_far = 0
    max_ending_here = 0

Loop for each element of the array
  (a) max_ending_here = max_ending_here + a[i]
  (b) if(max_ending_here < 0)
            max_ending_here = 0
  (c) if(max_so_far < max_ending_here)
            max_so_far = max_ending_here
return max_so_far

public int maxSubArray(int[] A) {
        if (A==null || A.length==0) {
            return 0;
        }

        int largestNegative = checkIfAllNegatives(A);
        if (largestNegative<0) {
            return largestNegative;
        }

        //sum is to track the max sum
        int sum = 0;
        //sumEndHere is used to track positive sums
        int sumEndHere = 0;
        for (int i=0; i<A.length; i++) {
            sumEndHere += A[i];
            if (sumEndHere<0) {
                sumEndHere = 0;
            } else if (sum<sumEndHere) {
                sum = sumEndHere;
            }
        }
        return sum;      
    }

    public int checkIfAllNegatives(int[] A) {
        int largestNegative = Integer.MIN_VALUE;
        for (int i=0; i<A.length; i++) {
            if (A[i]>=0) {
                return 0;
            } else {
                if (largestNegative < A[i]) {
                    largestNegative = A[i];
                }                    
            }
        }

        return largestNegative;
    }

DP: We should ignore the sum of the previous n-1 elements if nth element is greater than the sum.

public int maxSubArrayDP(int[] A) {
        if (A==null || A.length==0) {
            return 0;
        }
        int maxSoFar = A[0];
        int sumSoFar = A[0];

        for (int i=1; i<A.length; i++) {
            sumSoFar = Math.max(A[i], sumSoFar+A[i]);
            maxSoFar = Math.max(maxSoFar, sumSoFar);
        }
        return maxSoFar;
    }