LeetCode 2033: Minimum Operations to Make a Uni-Value Grid Solution

Master LeetCode problem 2033 (Minimum Operations to Make a Uni-Value Grid), 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.

2033. Minimum Operations to Make a Uni-Value Grid

Medium

Array Math Sorting Matrix

Problem Explanation

Explanation

To solve this problem, we need to find the minimum number of operations required to make all elements in the grid equal. We can achieve this by iterating through all possible target values (which is the value of some element in the grid) and calculating the total number of operations required to make all elements in the grid equal to that target value. The minimum number of operations among all target values will be our answer.

Here is the high-level algorithm:

Iterate through all elements in the grid and calculate the difference between the current element and the target value.
Count the number of operations required to make all elements in the grid equal to the target value.
Update the minimum number of operations seen so far.
Return the minimum number of operations if it is possible to make the grid uni-value, otherwise return -1.

Time Complexity: O(m*n) where m is the number of rows and n is the number of columns in the grid. Space Complexity: O(1) as we are using constant extra space.

Solution Code

class Solution {
    public int minOperations(int[][] grid, int x) {
        int m = grid.length;
        int n = grid[0].length;
        
        List<Integer> list = new ArrayList<>();
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                list.add(grid[i][j]);
            }
        }
        
        Collections.sort(list);
        
        int median = list.get((m * n) / 2);
        int operations = 0;
        for (int num : list) {
            int diff = Math.abs(num - median);
            if (diff % x != 0) {
                return -1;
            }
            operations += diff / x;
        }
        
        return operations;
    }
}

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