Description of the topic: Given a two-dimensional matrix consisting of 0s and 1s, the goal is to find the largest square region that contains only 1s.
For example:
1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0
The correct answer in this case is 4, which corresponds to a square of size 2x2 (area 4).
Problem-solving approach: Dynamic programming is an efficient method for solving this problem. The idea is to define a DP table where dp[i][j] represents the side length of the largest square whose bottom-right corner is at position (i, j).
We iterate through the matrix from top-left to bottom-right. For each cell (i, j), if the value is '1', we check the three neighboring cells: (i-1, j), (i, j-1), and (i-1, j-1). The minimum value among these three determines how large a square can be formed ending at (i, j). This is because the smallest square among those three limits the possible size of the new square.
If all three neighboring cells are part of a square, then dp[i][j] = min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]) + 1. Otherwise, if any of the three is zero, then dp[i][j] = 0 since no square can be formed.
During the iteration, we keep track of the maximum value found in the DP table, which gives us the side length of the largest square. Finally, we return the square of this value as the area of the largest square found.
Here is the implementation in C++:
class Solution {
public:
int maximalSquare(vector
if (matrix.empty()) return 0;
int rows = matrix.size(), cols = matrix[0].size();
int maxSide = 0;
vector
// Initialize the first row
for (int j = 0; j < cols; ++j)
dp[j] = matrix[0][j] - '0';
maxSide = *max_element(dp.begin(), dp.end());
// Process the rest of the rows
for (int i = 1; i < rows; ++i) {
dp[0] = matrix[i][0] - '0';
for (int j = 1; j < cols; ++j) {
if (matrix[i][j] == '1') {
int minVal = min(dp[j-1], dp[j]);
dp[j] = minVal + 1;
} else {
dp[j] = 0;
}
}
maxSide = max(maxSide, *max_element(dp.begin(), dp.end()));
}
return maxSide * maxSide;
}
};
This solution efficiently computes the largest square of 1s using dynamic programming with a time complexity of O(rows × cols) and a space complexity of O(cols), making it suitable for large matrices.
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