#3122

Medium

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Minimum Number of Operations to Satisfy Conditions

This problem asks to find the minimum number of operations on a 2D grid using state transition dynamic programming efficiently.

array dynamic programming matrix

Problem Statement

You are given an m x n matrix grid with non-negative integers. In one operation, you can change any cell to any non-negative number. The goal is to perform operations so that each cell satisfies a specific condition based on problem rules.

Return the minimum number of operations required to make all cells satisfy the conditions. Constraints are 1 <= m, n <= 1000 and 0 <= grid[i][j] <= 9. Example: grid = [[1,0,2],[1,0,2]] outputs 0 because all cells already meet the requirements.

Examples

Example 1

Input: grid = [[1,0,2],[1,0,2]]

Output: 0

All the cells in the matrix already satisfy the properties.

Example 2

Input: grid = [[1,1,1],[0,0,0]]

Output: 3

The matrix becomes [[1,0,1],[1,0,1]] which satisfies the properties, by doing these 3 operations:

Example 3

Input: grid = [[1],[2],[3]]

Output: 2

There is a single column. We can change the value to 1 in each cell using 2 operations.

Constraints

1 <= n, m <= 1000
0 <= grid[i][j] <= 9

Solution Approach

Define DP State

Create a DP table where dp[i][v] represents the minimum operations to make row i valid if the last cell has value v. This state captures both the row progression and possible cell values efficiently.

Row-wise Transition

Iterate through each row and update dp[i][v] based on dp[i-1][prev] and the operations needed to change the current cell to value v. Ensure transitions follow the required conditions to avoid invalid states.

Compute Minimum Operations

After processing all rows, the answer is the minimum value in the last row of dp. This captures the total minimum operations needed while considering all possible value transitions across rows.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity depends on m, n, and the number of possible values per cell, typically O(m n k^2) for k value options. Space complexity is O(n*k) if optimized row by row.

What Interviewers Usually Probe

Expect candidates to define DP states clearly for each row and value.
Look for correct handling of cell transitions while minimizing operations.
Watch for approaches that avoid redundant computation across large grids.

Common Pitfalls or Variants

Common pitfalls

Ignoring state dependencies between rows, which leads to overcounted operations.
Failing to optimize space usage, storing full DP for all rows unnecessarily.
Misapplying operations count when a cell already satisfies the conditions.

Follow-up variants

Change grid conditions, e.g., increase-decrease patterns or parity requirements.
Restrict the allowed cell values to a smaller subset, affecting DP transitions.
Allow diagonal or adjacent cells to affect validity, increasing state complexity.

FAQ

What is the best approach for Minimum Number of Operations to Satisfy Conditions?

Use state transition dynamic programming row by row, tracking the minimal operations for each possible cell value.

Can this problem be solved without dynamic programming?

Brute-force approaches are possible but inefficient; DP ensures minimal operations without redundant computation.

How do I handle large grids efficiently?

Optimize space by storing only the previous row's DP values and prune impossible transitions.

Why is state definition critical in this DP problem?

Incorrect state leads to double-counting operations or missing valid transitions, yielding wrong minimum counts.

Does changing a single cell affect the whole DP solution?

Yes, each cell influences possible next states, so proper transition accounting is essential for correctness.

terminal

Solution

Solution 1: Dynamic Programming

We notice that the values in the cells of the matrix only have 10 possibilities. The problem requires us to find the minimum number of operations for each column to have the same number, and the numbers in adjacent columns are different. Therefore, we only need to consider the case of modifying the number to 0 to 9.

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class Solution:
    def minimumOperations(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        f = [[inf] * 10 for _ in range(n)]
        for i in range(n):
            cnt = [0] * 10
            for j in range(m):
                cnt[grid[j][i]] += 1
            if i == 0:
                for j in range(10):
                    f[i][j] = m - cnt[j]
            else:
                for j in range(10):
                    for k in range(10):
                        if k != j:
                            f[i][j] = min(f[i][j], f[i - 1][k] + m - cnt[j])
        return min(f[-1])

Continue Practicing

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