algorithms/leetcode/3239-MinimumNumberOfFlipsToMakeBinaryGridPalindromicI.go at master · freewu/algorithms · GitHub

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package main

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

// Example 1:
// Input: grid = [[1,0,0],[0,0,0],[0,0,1]]
// Output: 2
// Explanation:
// <img src="https://assets.leetcode.com/uploads/2024/07/07/screenshot-from-2024-07-08-00-20-10.png" />
// Flipping the highlighted cells makes all the rows palindromic.

// Example 2:
// Input: grid = [[0,1],[0,1],[0,0]]
// Output: 1
// Explanation:
// <img src="https://assets.leetcode.com/uploads/2024/07/07/screenshot-from-2024-07-08-00-31-23.png" />
// Flipping the highlighted cell makes all the columns palindromic.

// Example 3:
// Input: grid = [[1],[0]]
// Output: 0
// Explanation:
// All rows are already palindromic.

// Constraints:
//     m == grid.length
//     n == grid[i].length
//     1 <= m * n <= 2 * 10^5
//     0 <= grid[i][j] <= 1

import "fmt"

func minFlips(grid [][]int) int {
    m, l, r, c := 0, 0, len(grid), len(grid[0])
    for i := 0; i < r; i++ {
        for j := 0; j < c / 2; j++ {
            if grid[i][j] != grid[i][c-j-1] {
                m++
            }
        }
    }
    for i := 0; i < c; i++ {
        for j := 0; j < r / 2; j++ {
            if grid[j][i] != grid[r-1-j][i] {
                l++
            }
        }
    }
    if m < l {
        return m
    }
    return l
}

func minFlips1(grid [][]int) int {
    m, n, x, y := len(grid), len(grid[0]), 0, 0
    for i := 0; i < m; i++ {
        j, k := 0, n - 1
        for j < k {
            if grid[i][j] != grid[i][k] { x++ }
            j++
            k--
        }
    }
    for i := 0; i < n; i++ {
        j, k := 0, m - 1
        for j < k {
            if grid[j][i] != grid[k][i] { y++ }
            j++
            k--
        }
    }
    min := func (x, y int) int { if x < y { return x; }; return y; }
    return min(x, y)
}

func minFlips2(grid [][]int) int {
    rn, cn := 0, 0
    for y, row := range grid {
        for x, num := range row {
            if x < len(row)/2 {
                if num != row[len(row) - x - 1] {
                    rn++
                }
            }
            if y < len(grid)/2 {
                if num != grid[len(grid)-y-1][x] {
                    cn++
                }
            }
        }
    }
    min := func (x, y int) int { if x < y { return x; }; return y; }
    return min(rn, cn)
}

func main() {
    // Example 1:
    // Input: grid = [[1,0,0],[0,0,0],[0,0,1]]
    // Output: 2
    // Explanation:
    // <img src="https://assets.leetcode.com/uploads/2024/07/07/screenshot-from-2024-07-08-00-20-10.png" />
    // Flipping the highlighted cells makes all the rows palindromic.
    fmt.Println(minFlips([][]int{{1,0,0},{0,0,0},{0,0,1}})) // 2
    // Example 2:
    // Input: grid = [[0,1],[0,1],[0,0]]
    // Output: 1
    // Explanation:
    // <img src="https://assets.leetcode.com/uploads/2024/07/07/screenshot-from-2024-07-08-00-31-23.png" />
    // Flipping the highlighted cell makes all the columns palindromic.
    fmt.Println(minFlips([][]int{{0,1},{0,1},{0,0}})) // 1
    // Example 3:
    // Input: grid = [[1],[0]]
    // Output: 0
    // Explanation:
    // All rows are already palindromic.
    fmt.Println(minFlips([][]int{{1},{0}})) // 0

    fmt.Println(minFlips1([][]int{{1,0,0},{0,0,0},{0,0,1}})) // 2
    fmt.Println(minFlips1([][]int{{0,1},{0,1},{0,0}})) // 1
    fmt.Println(minFlips1([][]int{{1},{0}})) // 0

    fmt.Println(minFlips2([][]int{{1,0,0},{0,0,0},{0,0,1}})) // 2
    fmt.Println(minFlips2([][]int{{0,1},{0,1},{0,0}})) // 1
    fmt.Println(minFlips2([][]int{{1},{0}})) // 0
}