1582. Special Positions in a Binary Matrix LeetCode Solution

In this guide, you will get 1582. Special Positions in a Binary Matrix LeetCode Solution with the best time and space complexity. The solution to Special Positions in a Binary Matrix problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

1. Problem Statement
2. Complexity Analysis
3. Special Positions in a Binary Matrix solution in C++
4. Special Positions in a Binary Matrix solution in Java
5. Special Positions in a Binary Matrix solution in Python
6. Additional Resources

Problem Statement of Special Positions in a Binary Matrix

Given an m x n binary matrix mat, return the number of special positions in mat.
A position (i, j) is called special if mat[i][j] == 1 and all other elements in row i and column j are 0 (rows and columns are 0-indexed).

Example 1:

Input: mat = [[1,0,0],[0,0,1],[1,0,0]]
Output: 1
Explanation: (1, 2) is a special position because mat[1][2] == 1 and all other elements in row 1 and column 2 are 0.

Example 2:

Input: mat = [[1,0,0],[0,1,0],[0,0,1]]
Output: 3
Explanation: (0, 0), (1, 1) and (2, 2) are special positions.

Constraints:

m == mat.length
n == mat[i].length
1 <= m, n <= 100
mat[i][j] is either 0 or 1.

Complexity Analysis

• Time Complexity:
• Space Complexity:

1582. Special Positions in a Binary Matrix LeetCode Solution in C++

class Solution {
 public:
  int numSpecial(vector<vector<int>>& mat) {
    const int m = mat.size();
    const int n = mat[0].size();
    int ans = 0;
    vector<int> rowOnes(m);
    vector<int> colOnes(n);

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        if (mat[i][j] == 1) {
          ++rowOnes[i];
          ++colOnes[j];
        }

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        if (mat[i][j] == 1 && rowOnes[i] == 1 && colOnes[j] == 1)
          ++ans;

    return ans;
  }
};
/* code provided by PROGIEZ */

1582. Special Positions in a Binary Matrix LeetCode Solution in Java

class Solution {
  public int numSpecial(int[][] mat) {
    final int m = mat.length;
    final int n = mat[0].length;
    int ans = 0;
    int[] rowOnes = new int[m];
    int[] colOnes = new int[n];

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        if (mat[i][j] == 1) {
          ++rowOnes[i];
          ++colOnes[j];
        }

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        if (mat[i][j] == 1 && rowOnes[i] == 1 && colOnes[j] == 1)
          ++ans;

    return ans;
  }
}
// code provided by PROGIEZ

1582. Special Positions in a Binary Matrix LeetCode Solution in Python

class Solution:
  def numSpecial(self, mat: list[list[int]]) -> int:
    m = len(mat)
    n = len(mat[0])
    ans = 0
    rowOnes = [0] * m
    colOnes = [0] * n

    for i in range(m):
      for j in range(n):
        if mat[i][j] == 1:
          rowOnes[i] += 1
          colOnes[j] += 1

    for i in range(m):
      for j in range(n):
        if mat[i][j] == 1 and rowOnes[i] == 1 and colOnes[j] == 1:
          ans += 1

    return ans
 # code by PROGIEZ

Additional Resources

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