I’ve just opened Introduction to Algorithms on Divide-and-Conquer chapter, and found interesting item – The maximum-subarray problem. In book you can find implementation with complexity O(NlogN) with good explanation, but in Exercises part there is good task:

Use the following ideas to develop a nonrecursive, linear-time algorithm for the maximum-subarray problem. Start at the left end of the array, and progress toward the right, keeping track of the maximum subarray seen so far. Knowing a maximum subarray of A[1..j] , extend the answer to find a maximum subarray ending at index j+1 by using the following observation: a maximum subarray of A[1..j+1] is either a maximum subarray of A[1..j] or a subarray A[i..j+1], for some 1<=i<=j+1. Determine a maximum subarray of the form A[i..j+1] in

constant timebased on knowing a maximum subarray ending at index j .

So, That is easy, because there are some ideas, you just need to stop and think a bit and all will be clear …

As described in the book you need to return 3 params

– max sum

– start index of sub array

– end index of sub array

OK, lets go …

Firstly create return structure.

public class SubArrayResult { final long sum; final int startIndex; final int endIndex; private SubArrayResult(long sum, int startIndex, int endIndex) { this.sum = sum; this.startIndex = startIndex; this.endIndex = endIndex; } public long getSum() { return sum; } public int getStartIndex() { return startIndex; } public int getEndIndex() { return endIndex; } @Override public String toString() { return "SubArrayResult{" + "sum=" + sum + ", startIndex=" + startIndex + ", endIndex=" + endIndex + '}'; } }

After that try to create algorithm.

public SubArrayResult findMaximumSubArray(int[] A) { long sum = 0; long result = 0; int start = 0; int end = 0; for (int i = 0; i < A.length; i++) { if (sum == 0 && A[i] <= 0) { start = i + 1; continue; } sum += A[i]; if (sum > result) { end = i; result = sum; } if (sum < 0) { start = i + 1; sum = 0; } } return new SubArrayResult(result, start, end); }

OK, what algorithm is doing:

– firstly run throughout all array

for (int i = 0; i < A.length; i++)

– skipping negative values and move start index

if (sum == 0 && A[i] <= 0) { start = i + 1; continue; }

– just add next value to current sum

sum += A[i];

– check if current sum > result value, if yes, assign sum to result and move end to current index

if (sum > result) { end = i; result = sum; }

– and last one if sum < 0 assign 0 to sum and move start to next index

if (sum < 0) { start = i + 1; sum = 0; }

So, that is all 🙂