LeetCode 1671: Minimum Number of Removals to Make Mountain Array Solution

Master LeetCode problem 1671 (Minimum Number of Removals to Make Mountain Array), 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.

1671. Minimum Number of Removals to Make Mountain Array

Hard

Array Binary Search Dynamic Programming Greedy

Problem Explanation

Explanation:

To find the minimum number of elements to remove to make the given array a mountain array, we can iterate through each index of the array and treat that index as the peak of the mountain. Then, we can count the number of elements that need to be removed on the left side of the peak to maintain the increasing order and on the right side of the peak to maintain the decreasing order. The sum of these removals will give us the total elements to remove for that peak. Finally, we return the minimum total removals across all possible peaks.

Iterate through each index as a potential peak.
For each peak, find the longest increasing subsequence to the left and the longest decreasing subsequence to the right.
The total elements to remove for this peak is the length of the array minus the sum of the longest increasing subsequence on the left and the longest decreasing subsequence on the right.
Update the minimum total removals across all peaks.

Time Complexity:

The time complexity of this approach is O(n^2) where n is the length of the input array.

Space Complexity:

The space complexity of this approach is O(n) for storing the length of increasing subsequences.

Solution Code

class Solution {
    public int minimumMountainRemovals(int[] nums) {
        int n = nums.length;
        int[] lisLeft = new int[n];
        int[] ldsRight = new int[n];
        
        for (int i = 0; i < n; i++) {
            lisLeft[i] = 1;
            for (int j = 0; j < i; j++) {
                if (nums[j] < nums[i]) {
                    lisLeft[i] = Math.max(lisLeft[i], lisLeft[j] + 1);
                }
            }
        }
        
        for (int i = n-1; i >= 0; i--) {
            ldsRight[i] = 1;
            for (int j = n-1; j > i; j--) {
                if (nums[j] < nums[i]) {
                    ldsRight[i] = Math.max(ldsRight[i], ldsRight[j] + 1);
                }
            }
        }
        
        int minRemovals = n;
        for (int i = 1; i < n-1; i++) {
            if (lisLeft[i] > 1 && ldsRight[i] > 1) {
                minRemovals = Math.min(minRemovals, n - (lisLeft[i] + ldsRight[i] - 1));
            }
        }
        
        return minRemovals;
    }
}

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