Merge pull request #115 from BrianLusina/feat/arrays-lucky-number

feat(arrays, puzzles, matrix): lucky number in a matrix

Author: BrianLusina

DIRECTORY.md

@@ -484,6 +484,8 @@
       * [Test Longest Increasing Subsequence](https://github.com/BrianLusina/PythonSnips/blob/master/puzzles/arrays/longest_increasing_subsequence/test_longest_increasing_subsequence.py)
     * Longest Subarray Of Ones
       * [Test Longest Subarray](https://github.com/BrianLusina/PythonSnips/blob/master/puzzles/arrays/longest_subarray_of_ones/test_longest_subarray.py)
+    * Lucky Numbers In A Matrix
+      * [Test Lucky Numbers In A Matrix](https://github.com/BrianLusina/PythonSnips/blob/master/puzzles/arrays/lucky_numbers_in_a_matrix/test_lucky_numbers_in_a_matrix.py)
     * Max Consecutive Ones
       * [Test Max Consecutive Ones](https://github.com/BrianLusina/PythonSnips/blob/master/puzzles/arrays/max_consecutive_ones/test_max_consecutive_ones.py)
     * Max Number Of Ksum Pairs

puzzles/arrays/lucky_numbers_in_a_matrix/README.md

@@ -0,0 +1,146 @@
+# Lucky Numbers in a Matrix
+
+Given an m × n matrix of distinct numbers, return the lucky number in the matrix.
+
+> A lucky number is an element of the matrix such that it is the smallest element in its row and largest in its column.
+
+Constraints
+
+- m = `matrix.length`
+- n = `matrix[i].length`
+- 1 <= m, n <= 50
+- 1 <= `matrix[i][j]` <= 10^5
+- All elements in the matrix are distinct.
+
+## Examples
+
+![Example 1](./images/examples/lucky_numbers_in_a_matrix_example_1.png)
+![Example 2](./images/examples/lucky_numbers_in_a_matrix_example_2.png)
+![Example 3](./images/examples/lucky_numbers_in_a_matrix_example_3.png)
+
+## Solution: Greedy
+
+The core idea behind the solution is to recognize that there can be, at most, one lucky number in the matrix. This is 
+proven by contradiction, as having two such numbers would violate the unique conditions for being a lucky number.
+
+> **Proof by contradiction:**
+> 
+> Suppose we have an integer x located at row r1 and column c1 in a matrix. The integer x is the smallest value in its 
+> row and the largest value in its column, making it a lucky number. 
+> Now, assume another integer y exists in row r2 and column c2. For the sake of argument, let’s assume y is also a 
+> lucky number, meaning it is the smallest value in its row and the largest in its column. 
+> We assess these assumptions using the following steps:
+> 
+> 1. As `y` is a lucky number, the smallest value is in row r2 and the largest value is in column c2. Let’s denote the 
+> integer at position (r2,c2) as `a`
+>    - Then, `y` < `a` because `y` is the minimum in its row
+>    - Then, `x` > `a` because `x` is the maximum in it column.
+> 
+>   Therefore `y` < `x`
+> 
+> 2. Next, let's consider the integer at position (r1, c2), which we'll call `b`
+>    - Then, `y` > `b` because `y` is the maximum in its column
+>    - Then, `x` < `b` because `x` is the minimum in its row
+> 
+>   Therefore `y` > `x`
+> 
+> This leads to a contradiction, as we deduced `y<x` and `y>x`. This inconsistency implies that our initial assumption—
+> that y is a lucky number—is incorrect. Therefore, only x can be the lucky number in this configuration.
+> 
+> Visually, this looks like this:
+> 
+> ![Proof by contradiction](./images/solutions/lucky_numbers_proof_of_contradiction.png)
+
+This problem can be solved using a greedy algorithm that analyzes the matrix row by row and column by column.
+
+We start by iterating over the rows to find the minimum values. Out of those minimum values, we choose the largest 
+minimum value and store it in r_largest_min. Similarly, we calculate the maximum values in columns and after finding 
+all the largest values, we choose the smallest of them and store them in c_smallest_max. Once we have found the values 
+in both rows and columns, we match them to see if they are the same. If they are, we return either of the values; 
+r_largest_min or c_smallest_max. Otherwise if now matching value is found, we return an empty matrix.
+
+Following are the detailed steps of the algorithm that we have just discussed:
+
+1. We define two variables, r_largest_min and c_smallest_max:
+
+   - r_largest_min is set to negative infinity (float('-inf')) to ensure any row’s minimum value can be updated.
+   - c_smallest_max is set to positive infinity (float('inf')) to ensure any column’s maximum value can be updated.
+
+2. For each row in the matrix:
+
+   - We calculate the minimum value of the row (r_min).
+   - Then, we update r_largest_min to the maximum of r_largest_min and r_min.
+   - The steps above ensure we consider only minimum values in their rows, narrowing the candidate set for a lucky number.
+
+3. For each column in the matrix:
+
+   - We calculate the maximum value of the column (c_max) by iterating over all rows.
+   - Next, we update c_max_min to the minimum of c_smallest_max and c_max.
+   - The above steps ensure we consider only maximum values in their columns, further narrowing the candidate set for a 
+     lucky number.
+
+4. Finally, we compare whether r_largest_min equals c_smallest_max. If TRUE, we return the value stored in [r_largest_min]. 
+   Otherwise, we return an empty array []. The comparison ensures that the identified value satisfies both conditions of
+   being the minimum in its row and the maximum in its column, making it a valid lucky number.
+
+![Solution 1](./images/solutions/lucky_numbers_in_a_matrix_solution_1.png)
+![Solution 2](./images/solutions/lucky_numbers_in_a_matrix_solution_2.png)
+![Solution 3](./images/solutions/lucky_numbers_in_a_matrix_solution_3.png)
+![Solution 4](./images/solutions/lucky_numbers_in_a_matrix_solution_4.png)
+![Solution 5](./images/solutions/lucky_numbers_in_a_matrix_solution_5.png)
+![Solution 6](./images/solutions/lucky_numbers_in_a_matrix_solution_6.png)
+![Solution 7](./images/solutions/lucky_numbers_in_a_matrix_solution_7.png)
+![Solution 8](./images/solutions/lucky_numbers_in_a_matrix_solution_8.png)
+![Solution 9](./images/solutions/lucky_numbers_in_a_matrix_solution_9.png)
+![Solution 10](./images/solutions/lucky_numbers_in_a_matrix_solution_10.png)
+![Solution 11](./images/solutions/lucky_numbers_in_a_matrix_solution_11.png)
+![Solution 12](./images/solutions/lucky_numbers_in_a_matrix_solution_12.png)
+![Solution 13](./images/solutions/lucky_numbers_in_a_matrix_solution_13.png)
+![Solution 14](./images/solutions/lucky_numbers_in_a_matrix_solution_14.png)
+![Solution 15](./images/solutions/lucky_numbers_in_a_matrix_solution_15.png)
+![Solution 16](./images/solutions/lucky_numbers_in_a_matrix_solution_16.png)
+
+### Time Complexity
+
+The time complexity of the solution is O(m×n), where m is the number of columns in the matrix and n is the number of 
+rows in the matrix.
+
+### Space Complexity
+
+The solution’s space complexity is O(1) as no extra space is required apart from the few variables.
+
+## Solution: Simulation
+
+We are given a matrix of size MXN with distinct integers. We need to return the list of lucky numbers in the matrix. 
+An integer in the matrix is lucky if it is the maximum integer in its column and it is the minimum value in its row.
+
+In this approach, we will simulate the process by iterating over each integer in the matrix, checking if it is the 
+maximum in its row and the minimum in its column. If it meets both criteria, we will add it to the list of lucky numbers, 
+luckyNumbers.
+
+The naive approach to check the criteria for each integer involves iterating over each integer in the current row and 
+column to verify the minimum and maximum criteria, requiring M+N operations per integer. A more efficient method is to 
+precompute the minimum of each row and the maximum of each column before processing the matrix. This allows us to check 
+the criteria for each integer in constant time. We iterate over each row to store the minimum in rowMin and each column 
+to store the maximum in colMax.
+
+### Algorithm
+
+1. Iterate over each row and store the minimum of the ith row at the ith position in the list rowMin.
+2. Iterate over each column and store the maximum of the ith column at the ith position in the list colMax.
+3. Iterate over each integer in the matrix and for each integer at (i, j), check if the integer is equal to rowMin[i] 
+   and colMax[j]. If yes, add it to the list luckyNumbers.
+4. Return luckyNumbers.
+
+### Complexity Analysis
+
+Here, N is the number of rows in the matrix and M is the number of columns in the matrix.
+
+#### Time complexity: O(N*M).
+
+To store the maximum of each row, we require N*M operations and the same for strong the maximum of each column. In the 
+end, to find the lucky numbers we again iterate over each integer. Hence, the total time complexity is equal to O(N*M).
+
+#### Space complexity: O(N+M).
+
+We require two lists, rowMin and colMax of size N and M respectively. Hence the total space complexity is equal to O(N+M).

puzzles/arrays/lucky_numbers_in_a_matrix/__init__.py

@@ -0,0 +1,99 @@
+from typing import List
+
+
+def lucky_numbers(matrix: List[List[int]]) -> List[int]:
+    """
+    This function takes a matrix as input and returns a list containing the lucky number(s) if they exist.
+
+    A lucky number is a number that is the maximum of the minimum values from each row and the minimum of the maximum
+    values from each column.
+
+    If a lucky number exists, the function returns a list containing that number. Otherwise, it returns an empty list.
+
+    Time Complexity: O(m * n) where m is the number of columns and n is the number of rows in the matrix.
+    Space Complexity: O(1) as we are using a constant amount of extra space.
+
+    Args:
+        matrix (List[List[int]]): The input matrix.
+
+    Returns:
+        List[int]: A list containing the lucky number(s) if they exist, otherwise an empty list.
+    """
+    row_length = len(matrix)
+    col_length = len(matrix[0])
+
+    # initialize a variable to keep track of the maximum of the minimum values from each row
+    r_largest_min = float("-inf")
+
+    # We start by iterating over the rows to find the minimum values. Out of those minimum values, we choose the
+    # largest minimum value and store it in r_largest_min
+    for i in range(row_length):
+        # find the minimum value in current row
+        row_min = min(matrix[i])
+        # update r_largest_min to be the maximum of current r_largest_min and the current row minimum
+        r_largest_min = max(r_largest_min, row_min)
+
+    # initialize a variable to keep track of the minimum of the maximum values from each column
+    c_smallest_max = float("inf")
+    # Similarly, we calculate the maximum values in columns and after finding all the largest values, we choose the
+    # smallest of them and store them in c_smallest_max
+    for c in range(col_length):
+        # find the maximum value in the current row
+        col_max = max(matrix[r][c] for r in range(row_length))
+        # update c_smallest_max to be the minium of the current c_smallest_max and the current column maximum
+        c_smallest_max = min(c_smallest_max, col_max)
+
+    # If they are, we return either of the values; r_largest_min or c_smallest_max. Otherwise if now matching value is
+    # found, we return an empty matrix.
+
+    # check if the maximum of row minima is equal to the minimum of column maxima
+    if r_largest_min == c_smallest_max:
+        # if they are equal, return a list containing the luky number
+        return [r_largest_min]
+    # Otherwise, return an empty list indicating no luky number exists
+    return []
+
+
+def lucky_numbers_simulation(matrix: List[List[int]]) -> List[int]:
+    """
+    This function takes a matrix as input and returns a list containing all the lucky number(s) if they exist.
+
+    A lucky number is a number that is the maximum of the minimum values from each row and the minimum of the maximum
+    values from each column.
+
+    The function first calculates the minimum values from each row and the maximum values from each column, then compares
+    these two lists to find the lucky number(s).
+
+    Time Complexity: O(m * n) where m is the number of columns and n is the number of rows in the matrix.
+    Space Complexity: O(m + n) as we are using two lists of size m and n respectively.
+
+    Args:
+        matrix (List[List[int]]): The input matrix.
+
+    Returns:
+        List[int]: A list containing all the lucky number(s) if they exist, otherwise an empty list.
+    """
+    row_length = len(matrix)
+    col_length = len(matrix[0])
+
+    row_min = []
+    for i in range(row_length):
+        r_min = float('inf')
+        for j in range(col_length):
+            r_min = min(r_min, matrix[i][j])
+        row_min.append(r_min)
+
+    col_max = []
+    for i in range(col_length):
+        c_max = float('-inf')
+        for j in range(row_length):
+            c_max = max(c_max, matrix[j][i])
+        col_max.append(c_max)
+
+    result = []
+    for i in range(row_length):
+        for j in range(col_length):
+            if matrix[i][j] == row_min[i] and matrix[i][j] == col_max[j]:
+                result.append(matrix[i][j])
+
+    return result