move to 2d directory

Author: akgmage

2D Arrays (Matrix)/searching_in_sorted_array.py

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+'''
+Date:28/6/23
+About:Search in 2D sorted array in Python
+
+Input:
+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.
+
+Time Complexity:
+You must write a solution in O(log(m * n)) time complexity.
+Space Complexity:O(1)
+
+//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
+
+//Explanation
+The method takes two parameters: matrix, which represents the sorted matrix, and target, which is the value we want to find in the matrix.
+
+The code initializes variables nRows and nCols to store the number of rows and columns in the matrix, respectively.
+
+The starting position for the search is set to the bottom-left corner of the matrix (row = nRows - 1, col = 0).
+
+The code enters a while loop that continues as long as the current row index (row) is within the bounds of the matrix (0 to nRows - 1
+and the current column index (col) is within the bounds of the matrix (0 to nCols - 1).
+
+Inside the loop, the code retrieves the value at the current position in the matrix (val = matrix[row][col]).
+
+If the current value (val) == value, the method returns True, indicating that the target is found in the matrix.
+
+If the current value (val) < target value, it means the target can only be found in the rows above the current row. 
+Therefore, the column index (col) is incremented to move to the next column.
+
+If the current value (val) >target value, it means the target can only be found in the columns to the left of the current column. 
+Therefore, the row index (row) is decremented to move to the previous row.
+
+If the loop completes without finding the target value, the method returns False.
+
+'''
+class Solution(object):
+    def searchMatrix(self, matrix, target):
+        nRows = len(matrix)
+        nCols = len(matrix[0])
+        row = nRows - 1
+        col = 0
+        while 0 <= row < nRows and 0 <= col < nCols:
+            val = matrix[row][col]
+            if val == target:
+                return True
+            elif val < target:
+                col += 1
+            else:
+                row -= 1
+        return False
+
+
+# Create an instance of the Solution class
+solution = Solution()
+
+# Define the matrix and target value
+matrix = [[1, 3, 5, 7], [10, 11, 16, 20], [23, 30, 34, 60]]
+target = 3
+
+# Call the searchMatrix method and print the result
+result = solution.searchMatrix(matrix, target)
+print(result)
+
+
+
+
+ 
+