LeetCode 1582: Special Positions in a Binary Matrix Solution
Master LeetCode problem 1582 (Special Positions in a Binary Matrix), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
Problem Explanation
Explanation:
To solve this problem, we need to iterate through each cell of the matrix and check if it satisfies the conditions to be a special position. For a position (i, j) to be special, it should contain 1 and all other elements in its row i and column j should be 0.
We can achieve this by iterating through each cell and keeping track of the counts of ones in each row and column. Then, we check if the current cell contains 1 and if the counts of ones in its row and column are both equal to 1. If all conditions are met, we increment a counter for special positions.
- Start with initializing two arrays to keep track of the counts of ones in each row and column.
- Iterate through each cell of the matrix and update the counts of ones in rows and columns accordingly.
- For each cell, check if it is a special position by verifying the conditions.
- Increment the counter for special positions if the conditions are met.
Time Complexity: O(m * n) where m is the number of rows and n is the number of columns in the matrix. Space Complexity: O(m + n) for the arrays to store counts of ones in each row and column.
:
Solution Code
class Solution {
public int numSpecial(int[][] mat) {
int m = mat.length;
int n = mat[0].length;
int[] rowOnes = new int[m];
int[] colOnes = new int[n];
int specialCount = 0;
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) {
specialCount++;
}
}
}
return specialCount;
}
}Try It Yourself
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