refactor(work): restructure solution directories with bucket level and add .gitkeep placeholders

work/1/Medium/Array/74.search-a-2d-matrix.py

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+"""
+74. Search a 2D Matrix
+Difficulty: Medium
+https://leetcode.com/problems/search-a-2d-matrix/
+
+──────────────────────────────────────────────────
+
+You are given an m x n integer matrix matrix with the following two
+properties:
+
+        • Each row is sorted in non-decreasing order.
+
+• The first integer of each row is greater than the last integer of
+the previous row.
+
+Given an integer target, return true if target is in matrix or false
+otherwise.
+
+You must write a solution in O(log(m * n)) time complexity.
+
+
+
+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
+"""
+
+
+class Solution:
+    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
+        ROWS, COLS = len(matrix), len(matrix[0])
+
+        top, bot = 0, ROWS - 1
+        while top <= bot:
+            row = (top + bot) // 2
+            if target > matrix[row][-1]:
+                top = row + 1
+            elif target < matrix[row][0]:
+                bot = row - 1
+            else:
+                break
+
+        if not (top <= bot):
+            return False
+
+        row = (top + bot) // 2
+        l, r = 0, COLS - 1
+        while l <= r:
+            m = (l + r) // 2
+            if target > matrix[row][m]:
+                l = m + 1
+            elif target < matrix[row][m]:
+                r = m - 1
+            else:
+                return True
+        return False