3239. Minimum Number of Flips to Make Binary Grid Palindromic I

Problem Description

You are given an m x n binary matrix grid.

A row or column is considered palindromic if its values read the same forward and backward.

You can flip any number of cells in grid from 0 to 1, or from 1 to 0.

Return the minimum number of cells that need to be flipped to make either all rows palindromic or all columns palindromic.

Intuition

To solve this problem, the goal is to minimize the number of changes required to make all rows or all columns palindromic.

Understanding a Palindrome: A palindrome reads the same forward and backward. For each row or column, this means the first element must match the last, the second must match the second-last, and so on.
Row and Column Checking: We can break down the problem into two separate tasks: making all rows palindromic and making all columns palindromic.
- For each row, count the number of mismatches between pairs: row[j] and row[n-j-1]. If they are not equal, one flip is necessary to make them equal.
- Similarly, for each column, count the number of mismatches between pairs: grid[i][j] and grid[m-i-1][j].
Optimize Flips: After counting the flips required to make rows palindromic (cnt1) and columns palindromic (cnt2), the solution takes the minimum between these two values. This ensures the fewest total flips are needed to satisfy one of the two conditions.

This counting approach ensures that we efficiently calculate the minimum number of cell flips required to achieve the desired format in either dimension.

Learn more about Two Pointers patterns.

Solution Approach

The solution employs a straightforward counting technique to determine the minimum number of flips required to make either all rows or all columns palindromic. Here's a detailed walkthrough of the implementation:

Initialize Variables:
- We define m and n as the number of rows and columns in the matrix grid, respectively.
- Two counters, cnt1 and cnt2, are initialized to zero. They will hold the number of flips needed for rows and columns, respectively.
Count Row Flips (cnt1):
- For each row in grid, iterate through the first half of the row's elements.
- Compare each element row[j] with its corresponding element from the end row[n-j-1].
- If these elements differ, increment cnt1 by 1, indicating a required flip to make them equal.
Count Column Flips (cnt2):
- For each column index j, iterate through the first half of the column's elements.
- Compare each element grid[i][j] with its corresponding element from the bottom grid[m-i-1][j].
- If they differ, increment cnt2 by 1, requiring a flip to make the column element a palindrome.
Determine Minimum Flips:
- The solution returns the minimum of cnt1 or cnt2, using the expression return min(cnt1, cnt2). This ensures we choose the dimension (rows or columns) that requires the fewest total flips to make all entries palindromic.

This counting approach efficiently computes the necessary flips, leveraging simple iteration and comparison while keeping operations within linear complexity relative to the number of elements in the grid.

Ready to land your dream job?

Unlock your dream job with a 2-minute evaluator for a personalized learning plan!

Start Evaluator

Example Walkthrough

Let's consider a simple 3x3 binary matrix grid:

1 0 1
0 1 0
1 0 0

We want to make either all rows or all columns palindromic with the minimum number of flips. We'll walk through the solution approach:

Count Row Flips (cnt1):
- For the first row [1 0 1], compare:
  - 1 and 1 (first and last) – no flip needed.
  - The row is already palindromic, no flips required.
- Second row [0 1 0], compare:
  - 0 and 0 (first and last) – no flip needed.
  - This row is palindromic as well.
- Third row [1 0 0], compare:
  - 1 and 0 – flip is needed to make them equal.
  - One flip required for this row.
Total row flips (cnt1) = 1 flip (only the third row).
Count Column Flips (cnt2):
- For the first column [1 0 1], compare:
  - 1 and 1 (top and bottom) – no flip needed.
- Second column [0 1 0], compare:
  - 0 and 0 – no flip needed.
- Third column [1 0 0], compare:
  - 1 and 0 – flip is needed to make them equal.
  - One flip required for this column.
Total column flips (cnt2) = 1 flip (only the third column).
Determine Minimum Flips:
- We take the minimum of cnt1 and cnt2, which is min(1, 1) = 1.

Thus, the minimum number of flips needed to make either all rows or all columns palindromic in the matrix is 1. This simple example demonstrates how we assess flips at both row and column levels, and select the option with fewer flips.

Solution Implementation

1from typing import List
2
3class Solution:
4    def minFlips(self, grid: List[List[int]]) -> int:
5        m, n = len(grid), len(grid[0])  # m is the number of rows, n is the number of columns
6      
7        # Initialize counters for flip operations
8        count_horizontal_flips = 0
9        count_vertical_flips = 0
10      
11        # Count the number of flips needed to make each row a palindrome
12        for row in grid:
13            for j in range(n // 2):
14                # Check if the element is not the same as its symmetric counterpart
15                if row[j] != row[n - j - 1]:
16                    count_horizontal_flips += 1
17      
18        # Count the number of flips needed to make each column a palindrome
19        for j in range(n):
20            for i in range(m // 2):
21                # Check if the element is not the same as its symmetric counterpart
22                if grid[i][j] != grid[m - i - 1][j]:
23                    count_vertical_flips += 1
24      
25        # Return the minimum number of flips needed by comparing both scenarios
26        return min(count_horizontal_flips, count_vertical_flips)
27

1class Solution {
2    public int minFlips(int[][] grid) {
3        int rows = grid.length;   // Number of rows in the grid
4        int cols = grid[0].length; // Number of columns in the grid
5      
6        int cnt1 = 0; // Count for horizontal flips
7        int cnt2 = 0; // Count for vertical flips
8      
9        // Calculate flips needed to make rows palindromic
10        for (int[] row : grid) {
11            for (int j = 0; j < cols / 2; ++j) {
12                // If the mirror image elements are not identical, increment the flip count
13                if (row[j] != row[cols - j - 1]) {
14                    ++cnt1;
15                }
16            }
17        }
18      
19        // Calculate flips needed to make columns palindromic
20        for (int j = 0; j < cols; ++j) {
21            for (int i = 0; i < rows / 2; ++i) {
22                // If the mirror image elements are not identical, increment the flip count
23                if (grid[i][j] != grid[rows - i - 1][j]) {
24                    ++cnt2;
25                }
26            }
27        }
28      
29        // Return the minimum number of flips needed between the two methods
30        return Math.min(cnt1, cnt2);
31    }
32}
33

1#include <vector>
2#include <algorithm> // for std::min
3using namespace std;
4
5class Solution {
6public:
7    int minFlips(vector<vector<int>>& grid) {
8        int numRows = grid.size(); // Number of rows in the grid
9        int numCols = grid[0].size(); // Number of columns in the grid
10        int flipCount1 = 0; // Count for flips needed by symmetry in rows
11        int flipCount2 = 0; // Count for flips needed by symmetry in columns
12
13        // Check for horizontal symmetry (row-wise)
14        for (const auto& row : grid) {
15            for (int j = 0; j < numCols / 2; ++j) {
16                if (row[j] != row[numCols - j - 1]) {
17                    ++flipCount1; // Increment count if symmetric elements differ
18                }
19            }
20        }
21
22        // Check for vertical symmetry (column-wise)
23        for (int j = 0; j < numCols; ++j) {
24            for (int i = 0; i < numRows / 2; ++i) {
25                if (grid[i][j] != grid[numRows - i - 1][j]) {
26                    ++flipCount2; // Increment count if symmetric elements differ
27                }
28            }
29        }
30
31        // Return the minimum number of flips needed to make the grid symmetric
32        return min(flipCount1, flipCount2);
33    }
34};
35

1function minFlips(grid: number[][]): number {
2    // Dimensions of the grid
3    const [m, n] = [grid.length, grid[0].length];
4  
5    // Counters for mismatches in horizontal and vertical directions
6    let [horizontalMismatches, verticalMismatches] = [0, 0];
7
8    // Traverse each row to count horizontal mismatches 
9    for (const row of grid) {
10        for (let j = 0; j < Math.floor(n / 2); ++j) {
11            // Compare element with its horizontal symmetric counterpart
12            if (row[j] !== row[n - 1 - j]) {
13                ++horizontalMismatches;
14            }
15        }
16    }
17
18    // Traverse each column to count vertical mismatches
19    for (let j = 0; j < n; ++j) {
20        for (let i = 0; i < Math.floor(m / 2); ++i) {
21            // Compare element with its vertical symmetric counterpart
22            if (grid[i][j] !== grid[m - 1 - i][j]) {
23                ++verticalMismatches;
24            }
25        }
26    }
27
28    // Return the minimum of horizontal and vertical mismatches
29    return Math.min(horizontalMismatches, verticalMismatches);
30}
31

Time and Space Complexity

The time complexity of the code is O(m * n). This is because the algorithm processes each element of the grid twice: once for horizontal symmetry checking and once for vertical symmetry checking, resulting in a full traversal of the grid.

The space complexity of the code is O(1), as only a fixed number of integer variables (cnt1 and cnt2) are used for counting, and no additional data structures are employed that scale with the input size.

Learn more about how to find time and space complexity quickly.

Discover Your Strengths and Weaknesses: Take Our 2-Minute Quiz to Tailor Your Study Plan:

Question 1 out of 10

You are given an array of intervals where intervals[i] = [start_i, end_i] represent the start and end of the ith interval. You need to merge all overlapping intervals and return an array of the non-overlapping intervals that cover all the intervals in the input.

Sort the intervals by their start time, then iterate through the sorted intervals, merging any overlapping intervals.

Brute force through all interval combinations and find the answer with the least total intervals.

Iterate through the intervals in their original order, merging any overlapping intervals.

Use a priority queue to keep track of the intervals, merging any overlapping intervals.

I don’t know