2033. Minimum Operations to Make a Uni-Value Grid LeetCode Solution
In this guide, you will get 2033. Minimum Operations to Make a Uni-Value Grid LeetCode Solution with the best time and space complexity. The solution to Minimum Operations to Make a Uni-Value Grid problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Problem Statement of Minimum Operations to Make a Uni-Value Grid
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
Time Complexity: O(mn) (C++), O(mn\log mn) (Java/Python)
Space Complexity: O(mn)
2033. Minimum Operations to Make a Uni-Value Grid LeetCode Solution in C++
class Solution {
public:
int minOperations(vector<vector<int>>& grid, int x) {
vector<int> arr;
for (const vector<int>& row : grid)
arr.insert(arr.end(), row.begin(), row.end());
if (ranges::any_of(arr, [&](int a) { return (a - arr[0]) % x; }))
return -1;
int ans = 0;
nth_element(arr.begin(), arr.begin() + arr.size() / 2, arr.end());
for (const int a : arr)
ans += abs(a - arr[arr.size() / 2]) / x;
return ans;
}
}; /* code provided by PROGIEZ */
2033. Minimum Operations to Make a Uni-Value Grid LeetCode Solution in Java
class Solution {
public int minOperations(int[][] grid, int x) {
final int m = grid.length;
final int n = grid[0].length;
int[] arr = new int[m * n];
for (int i = 0; i < m; ++i)
for (int j = 0; j < n; ++j)
arr[i * n + j] = grid[i][j];
if (Arrays.stream(arr).anyMatch(a -> (a - arr[0]) % x != 0))
return -1;
int ans = 0;
Arrays.sort(arr);
for (final int a : arr)
ans += Math.abs(a - arr[arr.length / 2]) / x;
return ans;
}
} // code provided by PROGIEZ
2033. Minimum Operations to Make a Uni-Value Grid LeetCode Solution in Python
class Solution:
def minOperations(self, grid: list[list[int]], x: int) -> int:
arr = sorted([a for row in grid for a in row])
if any((a - arr[0]) % x for a in arr):
return -1
ans = 0
for a in arr:
ans += abs(a - arr[len(arr) // 2]) // x
return ans # code by PROGIEZ
