Merge pull request #17 from xchux/feature/74-search-a-2d-matrix

✨ (Solution): Problem 74. Search a 2D Matrix

Author: xchux

solutions/74. Search a 2D Matrix/README.md

@@ -0,0 +1,61 @@
+---
+comments: true
+difficulty: medium
+# Follow `Topics` tags
+tags:
+    - Array
+    - Binary Search
+    - Matrix
+---
+
+# [74. Search a 2D Matrix](https://leetcode.com/problems/search-a-2d-matrix/description/)
+
+## Description
+
+You are given a 2D integer matrix `matrix` of size `m x n` that satisfies the following conditions:
+
+Each row is sorted in non-decreasing order.
+
+The first number of each row is greater than the last number of the previous row.
+
+Given an integer `target`, determine whether it exists in the matrix. Return `true` if it is found, otherwise return `false`.
+
+Your solution must run in **O(log(m * n))** time.
+
+
+**Example 1:**
+```
+Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
+Output: true
+```
+
+**Example 2:**
+```
+Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
+Output: false
+```
+
+**Constraints:**
+
+* `m == matrix.length`
+* `n == matrix[i].length`
+* `1 <= m, n <= 100`
+* `-10^4 <= matrix[i][j], target <= 10^4`
+
+## Solution
+
+
+```java
+```
+
+```python
+```
+
+## Complexity
+
+- Time complexity: $$O(log(m * n))$$
+<!-- Add time complexity here, e.g. $$O(n)$$ -->
+
+- Space complexity: $$O(1)$$
+<!-- Add space complexity here, e.g. $$O(n)$$ -->
+

solutions/74. Search a 2D Matrix/Solution.java

@@ -0,0 +1,21 @@
+class Solution {
+    public boolean searchMatrix(int[][] matrix, int target) {
+        int rows = matrix.length;
+        int cols = matrix[0].length;
+        int start = 0;
+        int end = rows * cols - 1;
+        while (start <= end) {
+            int mid = (start + end) / 2;
+            int row = mid / cols;
+            int col = mid % cols;
+            if (matrix[row][col] == target) {
+                return true;
+            } else if (matrix[row][col] < target) {
+                start = mid + 1;
+            } else {
+                end = mid - 1;
+            }
+        }
+        return false;
+    }
+}

solutions/74. Search a 2D Matrix/Solution.py

@@ -0,0 +1,14 @@
+class Solution:
+    def searchMatrix(self, matrix: list[list[int]], target: int) -> bool:
+        rows, cols = len(matrix), len(matrix[0])
+        start, end = 0, rows * cols - 1
+        while start <= end:
+            mid = (start + end) // 2
+            row, col = divmod(mid, cols)
+            if matrix[row][col] == target:
+                return True
+            elif matrix[row][col] < target:
+                start = mid + 1
+            else:
+                end = mid - 1
+        return False