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2033. Minimum Operations to Make a Uni-Value Grid

Description

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

Solutions

Solution 1: Greedy

Firstly, to make the grid a single-value grid, the remainder of all elements of the grid with $x$ must be the same.

Therefore, we can first traverse the grid to check whether the remainder of all elements with $x$ is the same. If not, return $-1$. Otherwise, we put all elements into an array, sort the array, take the median, then traverse the array, calculate the difference between each element and the median, divide it by $x$, and add all the differences to get the answer.

The time complexity is $O((m \times n) \times \log (m \times n))$, and the space complexity is $O(m \times n)$. Here, $m$ and $n$ are the number of rows and columns of the grid, respectively.

  • class Solution {
    public int minOperations(int[][] grid, int x) {
        int m = grid.length, n = grid[0].length;
        int[] nums = new int[m * n];
        int mod = grid[0][0] % x;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] % x != mod) {
                    return -1;
                }
                nums[i * n + j] = grid[i][j];
            }
        }
        Arrays.sort(nums);
        int mid = nums[nums.length >> 1];
        int ans = 0;
        for (int v : nums) {
            ans += Math.abs(v - mid) / x;
        }
        return ans;
    }
}

  • class Solution {
public:
    int minOperations(vector<vector<int>>& grid, int x) {
        int m = grid.size(), n = grid[0].size();
        int mod = grid[0][0] % x;
        int nums[m * n];
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] % x != mod) {
                    return -1;
                }
                nums[i * n + j] = grid[i][j];
            }
        }
        sort(nums, nums + m * n);
        int mid = nums[(m * n) >> 1];
        int ans = 0;
        for (int v : nums) {
            ans += abs(v - mid) / x;
        }
        return ans;
    }
};

  • class Solution:
    def minOperations(self, grid: List[List[int]], x: int) -> int:
        nums = []
        mod = grid[0][0] % x
        for row in grid:
            for v in row:
                if v % x != mod:
                    return -1
                nums.append(v)
        nums.sort()
        mid = nums[len(nums) >> 1]
        return sum(abs(v - mid) // x for v in nums)


  • func minOperations(grid [][]int, x int) int {
	mod := grid[0][0] % x
	nums := []int{}
	for _, row := range grid {
		for _, v := range row {
			if v%x != mod {
				return -1
			}
			nums = append(nums, v)
		}
	}
	sort.Ints(nums)
	mid := nums[len(nums)>>1]
	ans := 0
	for _, v := range nums {
		ans += abs(v-mid) / x
	}
	return ans
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

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