2160. Minimum Operations To Make A Uni Value Grid¶

LeetCode Problem View on GitHub

2160. Minimum Operations to Make a Uni-Value Grid

Medium

You are given a 2D integer grid of size m x n and an integer x. In one operation, you can add x to or subtract x from any element in the grid.

A uni-value grid is a grid where all the elements of it are equal.

Return the minimum number of operations to make the grid uni-value. If it is not possible, return -1.

Example 1:

Input: grid = [[2,4],[6,8]], x = 2
Output: 4
Explanation: We can make every element equal to 4 by doing the following: 
- Add x to 2 once.
- Subtract x from 6 once.
- Subtract x from 8 twice.
A total of 4 operations were used.

Example 2:

Input: grid = [[1,5],[2,3]], x = 1
Output: 5
Explanation: We can make every element equal to 3.

Example 3:

Input: grid = [[1,2],[3,4]], x = 2
Output: -1
Explanation: It is impossible to make every element equal.

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 105
• 1 <= m * n <= 105
• 1 <= x, grid[i][j] <= 104

Solution¶

class Solution {
    public int minOperations(int[][] grid, int x) {
        int n = grid.length, m = grid[0].length;
        if (n == 1 && m == 1) return 0;
        int mod = grid[0][0] % x;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] % x != mod) return -1;
            }
        }   

        ArrayList<Integer> res = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) res.add(grid[i][j]);
        }
        Collections.sort(res);
        int req = res.get(res.size() / 2);
        int total = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                int current = grid[i][j];
                total += Math.abs(current - req) / x;
            }
        }
        return total;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here