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LeetCode 2612: Minimum Reverse Operations Solution

Master LeetCode problem 2612 (Minimum Reverse Operations), 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.

2612. Minimum Reverse Operations

Hard
ArrayBreadth-First SearchOrdered Set

Problem Explanation

Explanation:

To solve this problem, we can iterate through each position in the array and calculate the minimum number of operations needed to move the single 1 to that position. We can do this by considering the possible reversals of subarrays that do not contain any banned positions.

Here is the high-level algorithm:

  1. Create a helper function that checks if a given position is banned.
  2. Iterate through each position i in the array:
    • If position i is banned, set the answer for position i to -1.
    • Otherwise, calculate the minimum number of operations needed to move the single 1 to position i by considering all possible subarrays that can be reversed.
  3. Return the array of minimum operations.

Time Complexity: O(n^2) where n is the length of the array Space Complexity: O(n) for the result array

:

Solution Code

class Solution {
    public int[] minOperations(int n, int p, int[] banned, int k) {
        int[] result = new int[n];
        Set<Integer> bannedSet = new HashSet<>();
        for (int b : banned) {
            bannedSet.add(b);
        }
        
        for (int i = 0; i < n; i++) {
            if (bannedSet.contains(i)) {
                result[i] = -1;
            } else {
                int minOps = Math.abs(p - i);
                int reverseOps = Math.abs(n - k);
                int ops = (minOps + reverseOps - 1) / k;
                result[i] = ops;
            }
        }
        
        return result;
    }
}

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