How do you solve Minimum Operations to Make a Uni-Value Grid on LeetCode?

Minimum Operations to Make a Uni-Value Grid (LeetCode #2033) can be solved using Brute Force. The optimal approach is Brute Force with O(n²) time complexity and O(1) space complexity. Key insight: The brute-force approach checks every possible target value in the grid and calculates the number of operations needed to convert all elements to that target. This is straightforward but inefficient for larger grids.

What is the time complexity of Minimum Operations to Make a Uni-Value Grid?

The optimal solution for Minimum Operations to Make a Uni-Value Grid runs in O(n²) time and uses O(1) space. This complexity arises because we check each element in the grid for every unique value, leading to a quadratic time complexity.

What algorithm or data structure is used to solve Minimum Operations to Make a Uni-Value Grid?

Minimum Operations to Make a Uni-Value Grid uses the following concepts: Array, Math, Sorting, Matrix. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Operations to Make a Uni-Value Grid?

Always check for edge cases, such as grids with only one element. Be clear about your assumptions regarding the input, especially regarding the value of x. Explain your thought process clearly while coding to show your understanding.

What are common mistakes when solving Minimum Operations to Make a Uni-Value Grid?

Forgetting to check remainders when determining if conversion is possible. Not considering the median as the target value for minimizing operations.

Minimum Operations to Make a Uni-Value Grid — LeetCode #2033 (Medium)

Tags: Array, Math, Sorting, Matrix

Related patterns: Hash Map, Array

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

The brute-force approach checks every possible target value in the grid and calculates the number of operations needed to convert all elements to that target. This is straightforward but inefficient for larger grids.

This complexity arises because we check each element in the grid for every unique value, leading to a quadratic time complexity.

Step 1: Identify all unique values in the grid.
Step 2: For each unique value, calculate the number of operations needed to convert all elements to this value.
Step 3: Return the minimum operations found or -1 if conversion is not possible.

For grid = [[2,4],[6,8]], x = 2: 1. Unique values: {2, 4, 6, 8} 2. Target = 2: Operations = 3 (2->4, 6->4, 8->4) 3. Target = 4: Operations = 4 (2->4, 6->4, 8->4) 4. Target = 6: Operations = 3 (2->4, 4->6, 8->6) 5. Target = 8: Operations = 4 (2->4, 4->6, 6->8) 6. Min operations = 4 (to make all equal to 4).

Common Mistakes

Forgetting to check remainders when determining if conversion is possible.
Not considering the median as the target value for minimizing operations.

Interview Tips

Always check for edge cases, such as grids with only one element.
Be clear about your assumptions regarding the input, especially regarding the value of x.
Explain your thought process clearly while coding to show your understanding.

#2033

Minimum Operations to Make a Uni-Value Grid

Medium

Array↗Math↗Sorting↗Matrix↗Hash Map↗Array↗

LeetCode ↗

Approaches

💡

Intuition

Time O(n²)Space O(1)

⚙️

Algorithm

3 steps

1Step 1: Identify all unique values in the grid.
2Step 2: For each unique value, calculate the number of operations needed to convert all elements to this value.
3Step 3: Return the minimum operations found or -1 if conversion is not possible.

solution.py14 lines

1# Full working Python code
2
3def minOperations(grid, x):
4    unique_values = set(val for row in grid for val in row)
5    min_operations = float('inf')
6    for target in unique_values:
7        operations = 0
8        for row in grid:
9            for val in row:
10                if (val - target) % x != 0:
11                    return -1
12                operations += abs(val - target) // x
13        min_operations = min(min_operations, operations)
14    return min_operations if min_operations != float('inf') else -1

ℹ

Complexity note: This complexity arises because we check each element in the grid for every unique value, leading to a quadratic time complexity.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.