LeetCode 3362: Zero Array Transformation III Solution

Master LeetCode problem 3362 (Zero Array Transformation III), a medium 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.

3362. Zero Array Transformation III

Medium

Array Greedy Sorting Heap (Priority Queue)Prefix Sum

Problem Explanation

Explanation:

To solve this problem, we can iterate through the queries and update the count of operations required to convert the array to all zeros. We can track the maximum count of operations that can be removed. If at any point, the remaining operations are not sufficient to convert the array to all zeros, we return -1.

Initialize a variable opsCount to keep track of the total operations needed to convert the array to all zeros.
Iterate through the queries and update the count of operations required for each query.
If at any point, the remaining operations are not enough to convert the array to all zeros, return -1.
Calculate the maximum count of operations that can be removed by subtracting the minimum of the current count of operations and the number of operations needed for the current query.
Return the maximum count of operations that can be removed.

Time Complexity: O(n) where n is the length of the array nums. Space Complexity: O(1)

Solution Code

class Solution {
    public int minOperations(int[] nums, int[][] queries) {
        int opsCount = 0;
        
        for (int[] query : queries) {
            int minVal = Integer.MAX_VALUE;
            for (int i = query[0]; i <= query[1]; i++) {
                minVal = Math.min(minVal, nums[i]);
            }
            opsCount += minVal;
        }
        
        int maxOpsRemoved = 0;
        for (int[] query : queries) {
            int opsRemoved = Math.min(opsCount, query[1] - query[0] + 1);
            maxOpsRemoved = Math.max(maxOpsRemoved, opsRemoved);
        }
        
        return opsCount - maxOpsRemoved;
    }
}

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