LeetCode 1330: Reverse Subarray To Maximize Array Value Solution

Master LeetCode problem 1330 (Reverse Subarray To Maximize Array Value), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1330. Reverse Subarray To Maximize Array Value

Hard

Array Math Greedy

Problem Explanation

Explanation:

To solve this problem, we can iterate through the array and calculate the contribution of each pair of neighboring elements to the array value. Then, we can find the subarray with the maximum contribution that can be reversed to maximize the array value.

We can calculate the contribution of each pair of neighboring elements by considering four cases:

If both elements are in their original order (nums[i] <= nums[i+1]), the contribution is nums[i+1] - nums[i].
If both elements are in reverse order (nums[i] > nums[i+1]), the contribution is nums[i] - nums[i+1].
If we reverse the subarray between these two elements, the contribution is 2 * (nums[i] - nums[i+1]).
If we reverse the subarray before nums[i] or after nums[i+1], the contribution is 0.

We can iterate through the array and calculate the total array value and the maximum contribution that can be obtained by reversing a subarray. The final answer will be the total array value plus the maximum contribution.

Time Complexity: O(N) Space Complexity: O(1)

Solution Code

class Solution {
    public int maxValueAfterReverse(int[] nums) {
        int n = nums.length;
        
        int totalValue = 0;
        int maxDiff = 0;
        
        for (int i = 0; i < n - 1; i++) {
            totalValue += Math.abs(nums[i] - nums[i + 1]);
            
            if (i > 0) {
                maxDiff = Math.max(maxDiff, Math.abs(nums[i] - nums[0]) - Math.abs(nums[i] - nums[i + 1]));
                maxDiff = Math.max(maxDiff, Math.abs(nums[i - 1] - nums[n - 1]) - Math.abs(nums[i] - nums[i + 1]));
            }
        }
        
        int res = Math.max(maxDiff, 0);
        
        for (int i = 1; i < n - 1; i++) {
            res = Math.max(res, Math.max(
                Math.abs(nums[0] - nums[i + 1]) - Math.abs(nums[i] - nums[i + 1]),
                Math.abs(nums[i - 1] - nums[n - 1]) - Math.abs(nums[i] - nums[i + 1])
            ));
        }
        
        return totalValue + res;
    }
}

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