Leetcode 1582: Special Positions in a Binary Matrix

Input: You are given a matrix 'mat' of size m x n, where each element is either 0 or 1.

Example: Input: mat = [[1, 0, 0], [0, 0, 1], [1, 0, 0]]

Constraints:

• 1 <= m, n <= 100

• mat[i][j] is either 0 or 1

Output: Return an integer representing the number of special positions in the matrix.

Example: Output: 1

Constraints:

Goal: To find special positions, we need to check each 1 in the matrix and verify that all other elements in its row and column are 0.

Steps:

• 1. Traverse the entire matrix to calculate how many 1's are present in each row and column.

• 2. For each element in the matrix, check if it's 1, and if the row and column of that element contain no other 1's.

• 3. Count the number of such positions that meet the condition.

Goal: The matrix can have up to 100 rows and columns, so the solution must be efficient.

Steps:

• Matrix dimensions are between 1 and 100 (inclusive).

• Each element in the matrix is either 0 or 1.

Assumptions:

• The matrix will not be empty, and the elements are constrained to 0 or 1.

• Input: Example 1: mat = [[1, 0, 0], [0, 0, 1], [1, 0, 0]]

• Explanation: In this case, only the position (1, 2) is special because mat[1][2] == 1 and all other elements in row 1 and column 2 are 0. Therefore, the output is 1.

• Input: Example 2: mat = [[1, 0, 0], [0, 1, 0], [0, 0, 1]]

• Explanation: Here, the positions (0, 0), (1, 1), and (2, 2) are all special. Therefore, the output is 3.

Approach: To solve this problem, we need to determine the special positions by first counting how many 1's are present in each row and column, then checking each 1 in the matrix to ensure there are no other 1's in its row and column.

Observations:

• A brute force approach would involve checking each element in the matrix for the special condition, which could lead to a time complexity of O(m * n^2).

• A more efficient approach would involve counting the number of 1's in each row and column first, then performing a single check for each element, which would improve the solution's efficiency.

Steps:

• 1. Create arrays to store the number of 1's in each row and column.

• 2. Traverse the matrix to populate these arrays.

• 3. For each element mat[i][j], if mat[i][j] == 1, check if row[i] and col[j] both contain only 1's, if so increment the result count.

• 4. Return the final result count.

Empty Inputs:

• Matrix dimensions are guaranteed to be at least 1x1.

Large Inputs:

• The solution should handle large matrices efficiently, with dimensions up to 100x100.

Special Values:

• When the matrix consists entirely of 0's, the output will be 0.

Constraints:

• Ensure the solution is efficient enough to handle the maximum input size.

int numSpecial(vector<vector<int>>& mat) {

    int m = mat.size(), n = mat[0].size();
    vector<int> row(m, 0), col(n, 0);
    int res = 0;
    for(int i = 0; i < m; i++)
    for(int j = 0; j < n; j++) {
        if(mat[i][j] == 1) row[i]++, col[j]++;
    }

    for(int i = 0; i < m; i++)
    for(int j = 0; j < n; j++) {
        if(row[i] == 1 && col[j] == 1 && mat[i][j] == 1) {
            res++;
        }
    }
    
    return res;
}

1 : Function Definition

int numSpecial(vector<vector<int>>& mat) {

Define the function `numSpecial` that takes a matrix `mat` as input, where each cell can either be 0 or 1.

2 : Variable Initialization

    int m = mat.size(), n = mat[0].size();

Initialize variables `m` and `n` to represent the number of rows and columns in the matrix, respectively.

3 : Variable Initialization

    vector<int> row(m, 0), col(n, 0);

Create two vectors `row` and `col`, initialized to zero. These vectors will track the count of 1s in each row and each column.

4 : Variable Initialization

    int res = 0;

Initialize `res` to 0, which will store the result (the number of special positions in the matrix).

5 : Outer Loop - Row Iteration

    for(int i = 0; i < m; i++)

Start the first loop to iterate over each row in the matrix.

6 : Inner Loop - Column Iteration

    for(int j = 0; j < n; j++) {

Start the second loop to iterate over each column in the current row.

7 : Update Row and Column Counts

        if(mat[i][j] == 1) row[i]++, col[j]++;

If the current cell is 1, increment the count for the corresponding row (`row[i]`) and column (`col[j]`), indicating that a 1 has been found.

8 : Second Loop - Checking Special Positions

    for(int i = 0; i < m; i++)

Start the third loop to iterate over each row again, this time to check for special positions.

9 : Second Inner Loop - Checking Columns

    for(int j = 0; j < n; j++) {

Start the fourth loop to iterate over each column in the current row and check if the position is special.

10 : Condition Check - Special Position

        if(row[i] == 1 && col[j] == 1 && mat[i][j] == 1) {

Check if the current cell is 1, and if both its row and column contain exactly one 1.

11 : Increment Result

            res++;

If the condition is true, increment the result `res` by 1, as this cell is a special position.

12 : Return Result

    return res;

Return the final result `res`, which represents the number of special positions in the matrix.

Best Case: O(m * n)

Average Case: O(m * n)

Worst Case: O(m * n)

Description: The time complexity is O(m * n) because we process each element in the matrix once and use auxiliary arrays to store the row and column counts.

Best Case: O(m + n)

Worst Case: O(m + n)

Description: The space complexity is O(m + n) because we use two arrays to store the row and column counts.

Leetcode 1582: Special Positions in a Binary Matrix

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