Dynamic Programming - Maximum Subarray Problem

Objective: The maximum subarray problem is the task of finding the contiguous subarray within a one-dimensional array of numbers which has the largest sum.

Example:

int [] A = {−2, 1, −3, 4, −1, 2, 1, −5, 4};

Output: contiguous subarray with the largest sum is 4, −1, 2, 1, with sum 6.

Approach:

Earlier we have seen how to solve this problem using Kadane’s Algorithm. In this post we will how to solve it using Dynamic programming.

We will solve this problem in bottom-up manner.

let’s say

“MS(i) is maximum sum ending at index i”

To calculate the solution for any element at index “i” has two options

EITHER added to the solution found till “i-1“^th index
OR start a new sum from the index “i“.

Recursive Solution:

MS(i) = Max[MS(i-1) + A[i] , A[i]]

Complete Code:

	import java.util.Arrays;

	public class MaximumSubArrayDP {

	//bottom up approach
	public int solve(int [] a){
	int [] solution = new int[a.length+1];
	solution[0] = 0;

	for (int j = 1; j <solution.length ; j++) {
	solution[j] = Math.max(solution[j–1]+a[j–1],a[j–1]);
	}
	//this will print the solution matrix
	//System.out.println(Arrays.toString(solution));
	//now return the maximum in the solution arrayyy
	int result = solution[0];
	for (int j = 1; j <solution.length ; j++) {
	if(result<solution[j])
	result = solution[j];
	}

	return result;
	}

	public static void main(String[] args) {
	int arrA[] = { 1, 2, –3, –4, 2, 7, –2, 3 };
	MaximumSubArrayDP i = new MaximumSubArrayDP();
	System.out.println("Maximum subarray is " + i.solve(arrA));
	}
	}

view raw

MaximumSubArrayDP.java

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Output:

[0, 1, 3, 0, 0, 2, 9, 7, 10]
Maximum subarray is  10

Dynamic Programming – Maximum Subarray Problem

MS(i) = Max[MS(i-1) + A[i] , A[i]]

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