Question

1536. Minimum Swaps to Arrange a Binary Grid

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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

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.

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;
}