Question

1252. Cells with Odd Values in a Matrix

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Description

There is an m x n matrix that is initialized to all 0's. There is also a 2D array indices where each indices[i] = [ri, ci] represents a 0-indexed location to perform some increment operations on the matrix.

For each location indices[i], do both of the following:

1. Increment all the cells on row ri.
2. Increment all the cells on column ci.

Given m, n, and indices, return _the number of odd-valued cells in the matrix after applying the increment to all locations in _indices.

 

Example 1:

Input: m = 2, n = 3, indices = [[0,1],[1,1]]
Output: 6
Explanation: Initial matrix = [[0,0,0],[0,0,0]].
After applying first increment it becomes [[1,2,1],[0,1,0]].
The final matrix is [[1,3,1],[1,3,1]], which contains 6 odd numbers.

Example 2:

Input: m = 2, n = 2, indices = [[1,1],[0,0]]
Output: 0
Explanation: Final matrix = [[2,2],[2,2]]. There are no odd numbers in the final matrix.

 

Constraints:

• 1 <= m, n <= 50
• 1 <= indices.length <= 100
• 0 <= ri < m
• 0 <= ci < n

 

Follow up: Could you solve this in O(n + m + indices.length) time with only O(n + m) extra space?

Solutions

Solution 1: Simulation

We create a matrix $g$ to store the results of the operations. For each pair $(r_i, c_i)$ in $indices$, we add $1$ to all elements in the $r_i$th row and the $c_i$th column of the matrix.

After the simulation, we traverse the matrix and count the number of odd numbers.

The time complexity is $O(\text{indices.length} \times (m+n) + mn)$, and the space complexity is $O(mn)$.

class Solution:
  def oddCells(self, m: int, n: int, indices: List[List[int]]) -> int:
    g = [[0] * n for _ in range(m)]
    for r, c in indices:
      for i in range(m):
        g[i][c] += 1
      for j in range(n):
        g[r][j] += 1
    return sum(v % 2 for row in g for v in row)

class Solution {
  public int oddCells(int m, int n, int[][] indices) {
    int[][] g = new int[m][n];
    for (int[] e : indices) {
      int r = e[0], c = e[1];
      for (int i = 0; i < m; ++i) {
        g[i][c]++;
      }
      for (int j = 0; j < n; ++j) {
        g[r][j]++;
      }
    }
    int ans = 0;
    for (int[] row : g) {
      for (int v : row) {
        ans += v % 2;
      }
    }
    return ans;
  }
}

class Solution {
public:
  int oddCells(int m, int n, vector<vector<int>>& indices) {
    vector<vector<int>> g(m, vector<int>(n));
    for (auto& e : indices) {
      int r = e[0], c = e[1];
      for (int i = 0; i < m; ++i) ++g[i][c];
      for (int j = 0; j < n; ++j) ++g[r][j];
    }
    int ans = 0;
    for (auto& row : g)
      for (int v : row) ans += v % 2;
    return ans;
  }
};

func oddCells(m int, n int, indices [][]int) int {
  g := make([][]int, m)
  for i := range g {
    g[i] = make([]int, n)
  }
  for _, e := range indices {
    r, c := e[0], e[1]
    for i := 0; i < m; i++ {
      g[i][c]++
    }
    for j := 0; j < n; j++ {
      g[r][j]++
    }
  }
  ans := 0
  for _, row := range g {
    for _, v := range row {
      ans += v % 2
    }
  }
  return ans
}

Solution 2: Space Optimization

We use row array $row$ and column array $col$ to record the number of times each row and column are increased. For each pair $(r_i, c_i)$ in $indices$, we add $1$ to $row[r_i]$ and $col[c_i]$ respectively.

After the operation, we can calculate that the count at position $(i, j)$ is $row[i] + col[j]$. We traverse the matrix and count the number of odd numbers.

The time complexity is $O(\text{indices.length} + mn)$, and the space complexity is $O(m+n)$.

class Solution:
  def oddCells(self, m: int, n: int, indices: List[List[int]]) -> int:
    row = [0] * m
    col = [0] * n
    for r, c in indices:
      row[r] += 1
      col[c] += 1
    return sum((i + j) % 2 for i in row for j in col)

class Solution {
  public int oddCells(int m, int n, int[][] indices) {
    int[] row = new int[m];
    int[] col = new int[n];
    for (int[] e : indices) {
      int r = e[0], c = e[1];
      row[r]++;
      col[c]++;
    }
    int ans = 0;
    for (int i : row) {
      for (int j : col) {
        ans += (i + j) % 2;
      }
    }
    return ans;
  }
}

class Solution {
public:
  int oddCells(int m, int n, vector<vector<int>>& indices) {
    vector<int> row(m);
    vector<int> col(n);
    for (auto& e : indices) {
      int r = e[0], c = e[1];
      row[r]++;
      col[c]++;
    }
    int ans = 0;
    for (int i : row)
      for (int j : col) ans += (i + j) % 2;
    return ans;
  }
};

func oddCells(m int, n int, indices [][]int) int {
  row := make([]int, m)
  col := make([]int, n)
  for _, e := range indices {
    r, c := e[0], e[1]
    row[r]++
    col[c]++
  }
  ans := 0
  for _, i := range row {
    for _, j := range col {
      ans += (i + j) % 2
    }
  }
  return ans
}

Solution 3: Mathematical Optimization

We notice that only when exactly one of $row[i]$ and $col[j]$ is odd, the number at position $(i, j)$ in the matrix will be odd.

We count the number of odd numbers in $row$, denoted as $cnt1$; the number of odd numbers in $col$, denoted as $cnt2$. Then the final number of odd numbers is $cnt1 \times (n-cnt2) + cnt2 \times (m-cnt1)$.

The time complexity is $O(\text{indices.length} + m + n)$, and the space complexity is $O(m+n)$.

class Solution:
  def oddCells(self, m: int, n: int, indices: List[List[int]]) -> int:
    row = [0] * m
    col = [0] * n
    for r, c in indices:
      row[r] += 1
      col[c] += 1
    cnt1 = sum(v % 2 for v in row)
    cnt2 = sum(v % 2 for v in col)
    return cnt1 * (n - cnt2) + cnt2 * (m - cnt1)

class Solution {
  public int oddCells(int m, int n, int[][] indices) {
    int[] row = new int[m];
    int[] col = new int[n];
    for (int[] e : indices) {
      int r = e[0], c = e[1];
      row[r]++;
      col[c]++;
    }
    int cnt1 = 0, cnt2 = 0;
    for (int v : row) {
      cnt1 += v % 2;
    }
    for (int v : col) {
      cnt2 += v % 2;
    }
    return cnt1 * (n - cnt2) + cnt2 * (m - cnt1);
  }
}

class Solution {
public:
  int oddCells(int m, int n, vector<vector<int>>& indices) {
    vector<int> row(m);
    vector<int> col(n);
    for (auto& e : indices) {
      int r = e[0], c = e[1];
      row[r]++;
      col[c]++;
    }
    int cnt1 = 0, cnt2 = 0;
    for (int v : row) cnt1 += v % 2;
    for (int v : col) cnt2 += v % 2;
    return cnt1 * (n - cnt2) + cnt2 * (m - cnt1);
  }
};

func oddCells(m int, n int, indices [][]int) int {
  row := make([]int, m)
  col := make([]int, n)
  for _, e := range indices {
    r, c := e[0], e[1]
    row[r]++
    col[c]++
  }
  cnt1, cnt2 := 0, 0
  for _, v := range row {
    cnt1 += v % 2
  }
  for _, v := range col {
    cnt2 += v % 2
  }
  return cnt1*(n-cnt2) + cnt2*(m-cnt1)
}