#861

Medium

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Score After Flipping Matrix

Maximize the score of a binary matrix by flipping rows and columns using a greedy approach and validation of invariants.

array greedy bit manipulation matrix

Problem Statement

You are given an m x n binary matrix, and your goal is to maximize the score of this matrix by flipping rows or columns. A flip consists of selecting a row or a column and toggling every bit within it (0 to 1 and 1 to 0). The matrix's score is calculated by interpreting each row as a binary number, and the total score is the sum of these binary numbers.

To achieve the maximum score, you must determine which rows or columns to flip based on a greedy approach while validating invariant properties. Given the constraints of the problem, an efficient solution must find the best configuration of flips and compute the highest possible score for the matrix.

Examples

Example 1

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

Output: 39

0b1111 + 0b1001 + 0b1111 = 15 + 9 + 15 = 39

Example 2

Input: grid = [[0]]

Output: 1

Example details omitted.

Constraints

m == grid.length
n == grid[i].length
1 <= m, n <= 20
grid[i][j] is either 0 or 1.

Solution Approach

Greedy Choice for Row Flips

Start by flipping rows so that the leftmost bit in each row is always 1. This ensures that the most significant bits contribute to the total score. After adjusting the rows, the greedy choice is to flip columns to maximize the overall score, considering how the other bits in the rows interact.

Invariant Validation in Columns

After flipping rows, validate the invariant that each column must either remain the same or be flipped entirely to maximize the score. This means if the number of 1's in a column is greater than the number of 0's, the column should be left as is; otherwise, it should be flipped.

Efficient Calculation of the Score

Once the flips have been decided, compute the score by converting the binary values of each row and summing them up. Given the constraints, the solution must calculate the score in O(m * n) time and O(1) space.

Complexity Analysis

Metric	Value
Time	O(m \cdot n)
Space	O(1)

The time complexity is O(m * n) due to iterating through the matrix to flip rows and columns, and the space complexity is O(1) since the solution only requires a constant amount of extra space for calculations.

What Interviewers Usually Probe

Look for an approach that efficiently handles the greedy choice of flipping rows and columns.
Ensure the candidate can correctly apply invariant validation to the matrix after row flips.
The candidate should demonstrate the ability to compute the final score efficiently.

Common Pitfalls or Variants

Common pitfalls

Failing to flip rows initially to ensure the most significant bit is 1.
Not correctly validating column flips based on the number of 1's and 0's.
Overcomplicating the solution by not sticking to the greedy choice plus invariant validation pattern.

Follow-up variants

Consider the scenario where m or n is at the upper limit of 20, testing the scalability of the approach.
Modify the problem to allow flipping only rows or only columns, testing how the solution adapts to different constraints.
Introduce additional constraints on which rows and columns can be flipped based on certain conditions, adding complexity to the problem.

FAQ

How can I maximize the score in the Score After Flipping Matrix problem?

The key is to apply a greedy approach where you flip rows to maximize the leftmost bit, and then flip columns based on the number of 1's in each column.

What is the main algorithmic pattern for solving Score After Flipping Matrix?

The problem follows the greedy choice plus invariant validation pattern, where you make local optimal decisions (flipping rows and columns) and validate the overall configuration.

What is the time complexity of the solution for Score After Flipping Matrix?

The time complexity is O(m * n), where m is the number of rows and n is the number of columns in the matrix.

Can the Score After Flipping Matrix problem be solved with dynamic programming?

No, dynamic programming is not the best approach for this problem; the optimal solution involves greedy choices with row and column flips, followed by invariant validation.

How do I avoid common mistakes in the Score After Flipping Matrix problem?

Focus on flipping rows to ensure the most significant bit is 1, then validate column flips based on the number of 1's and 0's in each column to maximize the score.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def matrixScore(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        for i in range(m):
            if grid[i][0] == 0:
                for j in range(n):
                    grid[i][j] ^= 1
        ans = 0
        for j in range(n):
            cnt = sum(grid[i][j] for i in range(m))
            ans += max(cnt, m - cnt) * (1 << (n - j - 1))
        return ans

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