1536. Minimum Swaps to Arrange a Binary Grid

SourceWeekly Contest 200 Q3DifficultyMediumRating1880

Description

Given an n x n binary grid, in one step you can choose two adjacent rows of the grid and swap them.

A grid is said to be valid if all the cells above the main diagonal are zeros.

Return the minimum number of steps needed to make the grid valid, or -1 if the grid cannot be valid.

The main diagonal of a grid is the diagonal that starts at cell (1, 1) and ends at cell (n, n).

Example 1:

Input: grid = [[0,0,1],[1,1,0],[1,0,0]]
Output: 3

Example 2:

Input: grid = [[0,1,1,0],[0,1,1,0],[0,1,1,0],[0,1,1,0]]
Output: -1
Explanation: All rows are similar, swaps have no effect on the grid.

Example 3:

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

Constraints:

n == grid.length == grid[i].length
1 <= n <= 200
grid[i][j] is either 0 or 1

Solutions

Solution 1: Greedy

Thinking

Adjacent row swaps must make every row \(i\) all zeros to the right of the diagonal. Only the rightmost \(1\) in a row matters: it is legal for row \(i\) iff that index is at most \(i\). \(n\le 200\) allows a per-row greedy.

For each \(i\), among unused rows take the first whose rightmost \(1\) is at most \(i\), then bubble it to position \(i\) at cost \(k-i\). If none exists the instance is impossible. Satisfying upper rows first does not block later ones, which require fewer trailing zeros.

We process row by row. For the \(i\)-th row, the position of the last '1' must be less than or equal to \(i\). We find the first row that meets the condition in \([i, n)\), denoted as \(k\). Then, starting from the \(k\)-th row, we swap the adjacent two rows upwards until the \(i\)-th row.

The time complexity is \(O(n^2)\), and the space complexity is \(O(n)\). Here, \(n\) is the side length of the grid.

Python3JavaC++GoTypeScript

class Solution:
    def minSwaps(self, grid: List[List[int]]) -> int:
        n = len(grid)
        pos = [-1] * n
        for i in range(n):
            for j in range(n - 1, -1, -1):
                if grid[i][j] == 1:
                    pos[i] = j
                    break
        ans = 0
        for i in range(n):
            k = -1
            for j in range(i, n):
                if pos[j] <= i:
                    ans += j - i
                    k = j
                    break
            if k == -1:
                return -1
            while k > i:
                pos[k], pos[k - 1] = pos[k - 1], pos[k]
                k -= 1
        return ans

class Solution {
    public int minSwaps(int[][] grid) {
        int n = grid.length;
        int[] pos = new int[n];
        Arrays.fill(pos, -1);
        for (int i = 0; i < n; ++i) {
            for (int j = n - 1; j >= 0; --j) {
                if (grid[i][j] == 1) {
                    pos[i] = j;
                    break;
                }
            }
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            int k = -1;
            for (int j = i; j < n; ++j) {
                if (pos[j] <= i) {
                    ans += j - i;
                    k = j;
                    break;
                }
            }
            if (k == -1) {
                return -1;
            }
            for (; k > i; --k) {
                int t = pos[k];
                pos[k] = pos[k - 1];
                pos[k - 1] = t;
            }
        }
        return ans;
    }
}

class Solution {
public:
    int minSwaps(vector<vector<int>>& grid) {
        int n = grid.size();
        vector<int> pos(n, -1);
        for (int i = 0; i < n; ++i) {
            for (int j = n - 1; j >= 0; --j) {
                if (grid[i][j] == 1) {
                    pos[i] = j;
                    break;
                }
            }
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            int k = -1;
            for (int j = i; j < n; ++j) {
                if (pos[j] <= i) {
                    ans += j - i;
                    k = j;
                    break;
                }
            }
            if (k == -1) {
                return -1;
            }
            for (; k > i; --k) {
                swap(pos[k], pos[k - 1]);
            }
        }
        return ans;
    }
};

func minSwaps(grid [][]int) (ans int) {
    n := len(grid)
    pos := make([]int, n)
    for i := range pos {
        pos[i] = -1
    }
    for i := 0; i < n; i++ {
        for j := n - 1; j >= 0; j-- {
            if grid[i][j] == 1 {
                pos[i] = j
                break
            }
        }
    }
    for i := 0; i < n; i++ {
        k := -1
        for j := i; j < n; j++ {
            if pos[j] <= i {
                ans += j - i
                k = j
                break
            }
        }
        if k == -1 {
            return -1
        }
        for ; k > i; k-- {
            pos[k], pos[k-1] = pos[k-1], pos[k]
        }
    }
    return
}

function minSwaps(grid: number[][]): number {
    const n = grid.length;
    const pos: number[] = Array(n).fill(-1);
    for (let i = 0; i < n; ++i) {
        for (let j = n - 1; ~j; --j) {
            if (grid[i][j] === 1) {
                pos[i] = j;
                break;
            }
        }
    }
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        let k = -1;
        for (let j = i; j < n; ++j) {
            if (pos[j] <= i) {
                ans += j - i;
                k = j;
                break;
            }
        }
        if (k === -1) {
            return -1;
        }
        for (; k > i; --k) {
            [pos[k], pos[k - 1]] = [pos[k - 1], pos[k]];
        }
    }
    return ans;
}

1536. Minimum Swaps to Arrange a Binary Grid

Description

Solutions

Solution 1: Greedy

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