Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Move from brute-force thinking to an efficient approach using array strategy.
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.lengthn == grid[i].length1 <= m, n <= 1051 <= m * n <= 1051 <= x, grid[i][j] <= 104Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Math
[[2,4],[6,8]] 2
[[1,5],[2,3]] 1
[[1,2],[3,4]] 2
minimum-moves-to-equal-array-elements-ii)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2033: Minimum Operations to Make a Uni-Value Grid
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;
}
}
// Accepted solution for LeetCode #2033: Minimum Operations to Make a Uni-Value Grid
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
}
# Accepted solution for LeetCode #2033: Minimum Operations to Make a Uni-Value Grid
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)
// Accepted solution for LeetCode #2033: Minimum Operations to Make a Uni-Value Grid
/**
* [2033] 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:
* <img alt="" src="https://assets.leetcode.com/uploads/2021/09/21/gridtxt.png" style="width: 164px; height: 165px;" />
* 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:
* <img alt="" src="https://assets.leetcode.com/uploads/2021/09/21/gridtxt-1.png" style="width: 164px; height: 165px;" />
* Input: grid = [[1,5],[2,3]], x = 1
* Output: 5
* Explanation: We can make every element equal to 3.
*
* Example 3:
* <img alt="" src="https://assets.leetcode.com/uploads/2021/09/21/gridtxt-2.png" style="width: 164px; height: 165px;" />
* 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 <= 10^5
* 1 <= m * n <= 10^5
* 1 <= x, grid[i][j] <= 10^4
*
*/
pub struct Solution {}
// problem: https://leetcode.com/problems/minimum-operations-to-make-a-uni-value-grid/
// discuss: https://leetcode.com/problems/minimum-operations-to-make-a-uni-value-grid/discuss/?currentPage=1&orderBy=most_votes&query=
// submission codes start here
impl Solution {
pub fn min_operations(grid: Vec<Vec<i32>>, x: i32) -> i32 {
0
}
}
// submission codes end
#[cfg(test)]
mod tests {
use super::*;
#[test]
#[ignore]
fn test_2033_example_1() {
let grid = vec![vec![2, 4], vec![6, 8]];
let x = 2;
let result = 4;
assert_eq!(Solution::min_operations(grid, x), result);
}
#[test]
#[ignore]
fn test_2033_example_2() {
let grid = vec![vec![1, 5], vec![2, 3]];
let x = 1;
let result = 5;
assert_eq!(Solution::min_operations(grid, x), result);
}
#[test]
#[ignore]
fn test_2033_example_3() {
let grid = vec![vec![1, 2], vec![3, 4]];
let x = 2;
let result = -1;
assert_eq!(Solution::min_operations(grid, x), result);
}
}
// Accepted solution for LeetCode #2033: Minimum Operations to Make a Uni-Value Grid
function minOperations(grid: number[][], x: number): number {
const arr = grid.flat(2);
arr.sort((a, b) => a - b);
const median = arr[Math.floor(arr.length / 2)];
let res = 0;
for (const val of arr) {
const c = Math.abs(val - median) / x;
if (c !== (c | 0)) return -1;
res += c;
}
return res;
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.
Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.