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This approach begins searching from the top-right corner of the matrix. If the current element is equal to the target, return true. If the current element is greater than the target, move left. If the current element is less than the target, move down. This method effectively narrows down the search space, taking advantage of the sorted property of the matrix.
Time Complexity: O(m + n), where m is the number of rows and n is the number of columns.
Space Complexity: O(1), as no extra space is used.
1#include <stdbool.h>
2
3bool searchMatrix(int** matrix, int matrixSize, int* matrixColSize, int target) {
4 int row = 0, col = matrixColSize[0] - 1;
5 while (row < matrixSize && col >= 0) {
6 if (matrix[row][col] == target) {
7 return true;
8 } else if (matrix[row][col] > target) {
9 col--;
10 } else {
11 row++;
12 }
13 }
14 return false;
15}The solution initializes the search at the top-right corner of the matrix. It iteratively checks the current element against the target and adjusts the row and column pointers accordingly. If the current element is larger than the target, it moves left; if smaller, it moves down. It returns true if the target is found and false if the pointers go out of bounds.
This approach uses binary search on each row of the matrix. Since each row is sorted, binary search can efficiently determine if the target exists within the row. If found in any row, the function returns true.
Time Complexity: O(m * log(n)), where m is the number of rows and n is the number of columns.
Space Complexity: O(1).
1def
The Python function defines an internal helper `binary_search` which operates on each row. It iteratively applies binary search to rows, leveraging the sorted property to quickly declutter impossible sections.