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 time based 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.
[crayon-5ba91edb2f574386611794/] After that try to create algorithm.
[crayon-5ba91edb2f57e098547834/] OK, what algorithm is doing:
– firstly run throughout all array
[crayon-5ba91edb2f583732139996/] – skipping negative values and move start index
[crayon-5ba91edb2f587233845600/] – just add next value to current sum
[crayon-5ba91edb2f58b002805608/] – check if current sum > result value, if yes, assign sum to result and move end to current index
[crayon-5ba91edb2f58e870261555/] – and last one if sum < 0 assign 0 to sum and move start to next index
[crayon-5ba91edb2f592225968393/] So, that is all 🙂