Lintcode 45. Maximum Subarray Difference

Given an array with integers.

Find two non-overlapping subarrays A and B, which |SUM(A) - SUM(B)| is the largest.

Return the largest difference.

Example

Example 1:

Input:[1, 2, -3, 1]
Output:6
Explanation:
The subarray are [1,2] and [-3].So the answer is 6.

Example 2:

Input:[0,-1]
Output:1
Explanation:
The subarray are [0] and [-1].So the answer is 1.

Challenge

O(n) time and O(n) space.

Notice

The subarray should contain at least one number

Code (Java):

public class Solution {
    /**
     * @param nums: A list of integers
     * @return: An integer indicate the value of maximum difference between two substrings
     */
    public int maxDiffSubArrays(int[] nums) {
        // write your code here
        if (nums == null || nums.length < 2) {
            return 0;
        }
        
        int n = nums.length;
        int[] leftMax = new int[n];
        int[] leftMin = new int[n];
        int[] rightMax = new int[n];
        int[] rightMin = new int[n];
        
        int curMax = Integer.MIN_VALUE;
        int curMin = Integer.MAX_VALUE;
        int minPreSum = 0;
        int maxPreSum = 0;
        int sum = 0;
        
        for (int i = 0; i < n; i++) {
            sum += nums[i];
            curMax = Math.max(curMax, sum - minPreSum);
            curMin = Math.min(curMin, sum - maxPreSum);
            
            leftMax[i] = curMax;
            leftMin[i] = curMin;
            
            minPreSum = Math.min(minPreSum, sum);
            maxPreSum = Math.max(maxPreSum, sum);
        }
        
        curMax = Integer.MIN_VALUE;
        curMin = Integer.MAX_VALUE;
        minPreSum = 0;
        maxPreSum = 0;
        sum = 0;
        
        for (int i = n - 1; i >= 0; i--) {
            sum += nums[i];
            curMax = Math.max(curMax, sum - minPreSum);
            curMin = Math.min(curMin, sum - maxPreSum);
            
            rightMax[i] = curMax;
            rightMin[i] = curMin;
            
            minPreSum = Math.min(minPreSum, sum);
            maxPreSum = Math.max(maxPreSum, sum);
        }
        
        int max = Integer.MIN_VALUE;
        for (int i = 0; i < n - 1; i++) {
            max = Math.max(max, Math.max(Math.abs(leftMax[i] - rightMin[i + 1]), Math.abs(rightMax[i + 1] - leftMin[i])));
        }
        
        return max;
        
    }
}