LeetCode 3122 - Minimum Number of Operations to Satisfy Conditions

The problem presents a two-dimensional matrix grid of size m x n with integer values in the range 0 to 9. The task is to transform this matrix using the minimum number of operations, where each operation allows you to change the value of any cell to any non-negative integer.

leetcodemediumarraydynamic-programmingmatrix

LeetCode Problem 3122

Difficulty: 🟡 Medium
Topics: Array, Dynamic Programming, Matrix

Solution

Problem Understanding

The problem presents a two-dimensional matrix grid of size m x n with integer values in the range 0 to 9. The task is to transform this matrix using the minimum number of operations, where each operation allows you to change the value of any cell to any non-negative integer. After all operations, the matrix must satisfy two conditions: vertically, each cell must equal the cell directly below it (if it exists), and horizontally, each cell must differ from the cell directly to its right (if it exists).

The input is a standard 2D array, and the output is a single integer representing the minimum number of operations required. The constraints indicate that m and n can be as large as 1000, so a naive brute-force approach that examines all possible value combinations for the matrix is computationally infeasible. Edge cases include matrices with only one row or one column, or matrices where all values are already uniform.

The key insight is that the problem has a structured dependency: the vertical condition requires uniformity in each column, while the horizontal condition requires adjacent cells in each row to differ. This naturally suggests a dynamic programming approach to efficiently compute the minimum operations while maintaining these constraints.

Approaches

The brute-force approach would attempt to assign every possible value to each cell while checking both vertical and horizontal constraints. For a matrix of size m x n and a value range of 0-9, this results in 10^(m*n) combinations. This guarantees correctness but is computationally intractable for the input constraints.

The optimal approach leverages dynamic programming. Since vertical equality is required, we can treat each column independently by deciding a "column value" for the entire column. For horizontal inequality, the column values must alternate row by row. Using DP, we compute the minimum operations required for each column value while maintaining the horizontal constraints from the previous column. This approach drastically reduces the number of computations by collapsing the matrix into manageable states per column.

Approach Time Complexity Space Complexity Notes
Brute Force O(10^(m*n)) O(1) Try every combination of numbers for all cells.
Optimal O(m * n * 10 * 10) O(10) DP per column using all 0-9 values and previous column values to maintain horizontal constraints.

Algorithm Walkthrough

  1. Define a DP array where dp[val] stores the minimum operations to transform the previous column so that all its cells are equal to val. Initialize all values to infinity except for the first column.
  2. For each column from left to right, consider each candidate value (0-9) that can be assigned to all cells in that column.
  3. Compute the operations required for the current column to have all cells equal to the candidate value. This is done by comparing the candidate value with each cell in the column.
  4. For each candidate value in the current column, check all previous column values and add the DP cost from the previous column if the horizontal constraint (current != previous) is satisfied.
  5. Update the DP array with the minimum operations for each candidate value.
  6. After processing all columns, the minimum value in the DP array represents the minimum number of operations required to satisfy all constraints.

Why it works: This approach guarantees that all vertical and horizontal constraints are respected because each DP state explicitly considers vertical uniformity and horizontal adjacency constraints. By iterating column by column and maintaining minimal operations, we ensure a globally optimal solution.

Python Solution

from typing import List

class Solution:
    def minimumOperations(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        INF = float('inf')
        prev_dp = [0] * 10  # Cost for previous column
        for j in range(n):
            curr_dp = [INF] * 10
            for val in range(10):
                # Compute operations to set column j to val
                ops = sum(1 for i in range(m) if grid[i][j] != val)
                for prev_val in range(10):
                    if j == 0 or prev_val != val:  # horizontal constraint
                        curr_dp[val] = min(curr_dp[val], prev_dp[prev_val] + ops)
            prev_dp = curr_dp
        return min(prev_dp)

The solution first initializes a DP array to track costs for each column. For each column, it iterates over all candidate values (0-9) and computes the cost to make the entire column equal to that value. It then updates the DP using the previous column’s values, enforcing the horizontal constraint. After processing all columns, the minimum value in the DP array is the result.

Go Solution

func minimumOperations(grid [][]int) int {
    m, n := len(grid), len(grid[0])
    INF := int(1e9)
    prevDP := make([]int, 10)
    for j := 0; j < n; j++ {
        currDP := make([]int, 10)
        for i := 0; i < 10; i++ {
            currDP[i] = INF
        }
        for val := 0; val < 10; val++ {
            ops := 0
            for row := 0; row < m; row++ {
                if grid[row][j] != val {
                    ops++
                }
            }
            for prevVal := 0; prevVal < 10; prevVal++ {
                if j == 0 || prevVal != val {
                    if prevDP[prevVal]+ops < currDP[val] {
                        currDP[val] = prevDP[prevVal] + ops
                    }
                }
            }
        }
        prevDP = currDP
    }
    minOps := INF
    for _, v := range prevDP {
        if v < minOps {
            minOps = v
        }
    }
    return minOps
}

In Go, the implementation mirrors the Python version. Arrays are initialized with a high INF value to simulate infinity. The main differences involve explicit array initialization and integer handling. The logic for computing column operations and updating DP remains identical.

Worked Examples

Example 1:

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

Column Candidate Operations DP Update
0 1 0 prevDP[1] = 0
1 0 0 prevDP[0] = 0
2 2 0 prevDP[2] = 0

Output: 0

Example 2:

Input: [[1,1,1],[0,0,0]]

Column Candidate Operations DP Update
0 1 1 DP[1] = 1
0 0 1 DP[0] = 1
1 0 1 DP[0] = 2
1 1 1 DP[1] = 2
2 1 1 DP[1] = 3
2 0 1 DP[0] = 3

Output: 3

Example 3:

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

Single column, minimum operations to make all rows identical is 2. Output: 2.

Complexity Analysis

Measure Complexity Explanation
Time O(m * n * 100) = O(m * n) For each column, we consider 10 candidate values and 10 previous column values, checking all m rows.
Space O(10) DP array of size 10 is reused for each column.

The time complexity is feasible for the given constraints (m, n <= 1000) because 1000 * 1000 * 100 is acceptable for modern computation. Space usage is minimal as only DP arrays are maintained.

Test Cases

# Provided examples
assert Solution().minimumOperations([[1,0,2],[1,0,2]]) == 0
assert Solution().minimumOperations([[1,1,1],[0,0,0]]) == 3
assert Solution().minimumOperations([[1],[2],[3]]) == 2

# Edge cases
assert Solution().minimumOperations([[5]]) == 0  # Single cell, no operation needed
assert Solution().minimumOperations([[1,1]]) == 1  # Two cells in a row need to differ
assert Solution().minimumOperations([[0],[0]]) == 0  # Single column, same value, no operation needed
assert Solution().minimumOperations([[1,2,3],[4,5,6],[7,8,9]]) == 6  # Multiple changes needed
Test Why
[[1,0,2],[1,0,2]] Already satisfies both conditions
[[1,1,1],[0,0,0]] Requires multiple changes to satisfy horizontal and vertical constraints
[[1],[2],[3]] Single column, vertical equality only
[[5]] Minimal matrix, single element
[[1,1]] Horizontal difference required
[[0],[0]] Single column, already uniform
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