### Largest value sub-array [part 1]

Wednesday, November 11th, 2009I was teaching a bunch of freshman the other day. One of the questions that came up was to determine the largest value sub-array given an array of integers. Obviously, the value is the array can be negative. There are two variants of the questions, one to find the actual maximum value when each of the members of the sub-array are added together, and the second variant is to actually find the start and ending location of the sub-array.

We will first consider the first variant to arrive at optimum answer.

A simple answer would be to perform brute force search in the array, considering each position as possible starting and ending point. Time complexity: O(n^2). Not very good eh? Can we improve on this? Yes, we can if we use dynamic programming. For many of us, the algorithm came up as obvious, but I realize that it is really not that simple for students new to programming to come up with a correct algorithm. To answer this, I tried to explain the concept behind the algorithm: for each element of the array, we find the maximum value sub-array that ends at that element. This value can either be the value of the element itself (meaning starting at the element and ending at the particular element) or the maximum value sub-array ending at the previous element added to the value of the element. (The latter only occurs if the maximum value sub-array at previous location is not negative.) There you go, we have a recurrence relation: given an array A, let v(k) be the maximum value of any sub-array ending at position k, where k ranging from 0 inclusive to N exclusive, N is the length of the array. Hence we have:

v(0) = A[0]

v(k) = max{v(k-1) + A[k], A[k]} if k > 0

The maximum value will be given by max{v_i} for i from 0 to N-1. To implement this in Java, we make several optimization. Firstly, we can opt for recursive method or dynamic programming. Both will take O(n) time but since we’re using Java, we would rather use dynamic programming since there is no tail call optimization in Java. Then, rather than finding the maximum value using max{v_i} at the end, we keep a running maximum throughout the running of the algorithm and return it. We also note that v(k-1) + A[k] will only be greater than A[k] iff v(k-1) is greater than 0. Here is how the code goes (this is a rather faithful implementation, it uses the same variable names as the recurrence relation above to make it easier to see):

public static int calculateMaxValue(int[] A) { int[] v = new int[A.length]; int max = v[0] = A[0]; for (int i = 1; i < A.length; ++i) { v[i] = A[i]; if (v[i-1] > 0) { v[i] += v[i-1]; } if (v[i] > max) { max = v[i]; } } return max; }

This algorithm has O(n) time complexity.

What else can we do to optimize this code? Well, we don’t really need to keep an array of maximum values. We notice that at any point of time, to calculate the value of v(k), we only need the value of A[k] and v(k-1). We just turn an O(n) space requirement to O(1).

public static int calculateMaxValue(int[] A) { int max = A[0], current = A[0]; for (int i = 1; i < A.length; ++i) { if (current < 0) current = 0; current += A[i]; if (current > max) max = current; } return max; }

Voila! We shall continue with the second variant next time. (: