TheAlgorithms · cclauss · Oct 26, 2022 · Oct 26, 2022 · Oct 26, 2022 · Oct 26, 2022
diff --git a/matrix/largest_square_area_in_matrix.py b/matrix/largest_square_area_in_matrix.py
@@ -0,0 +1,191 @@
+"""
+Question:
+Given a binary matrix mat of size n * m, find out the maximum size square
+sub-matrix with all 1s.
+
+---
+Example 1:
+
+Input:
+n = 2, m = 2
+mat = [[1, 1],
+       [1, 1]]
+
+Output:
+2
+
+Explanation: The maximum size of the square
+sub-matrix is 2. The matrix itself is the
+maximum sized sub-matrix in this case.
+---
+Example 2
+
+Input:
+n = 2, m = 2
+mat = [[0, 0],
+       [0, 0]]
+Output: 0
+
+Explanation: There is no 1 in the matrix.
+
+
+Approach:
+We initialize another matrix (dp) with the same dimensions
+as the original one initialized with all 0’s.
+
+dp_array(i,j) represents the side length of the maximum square whose
+bottom right corner is the cell with index (i,j) in the original matrix.
+
+Starting from index (0,0), for every 1 found in the original matrix,
+we update the value of the current element as
+
+dp_array(i,j)=dp_array(dp(i−1,j),dp_array(i−1,j−1),dp_array(i,j−1)) + 1.
+"""
+
+
+def largest_square_area_in_matrix_top_down_approch(
+    rows: int, cols: int, mat: list[list[int]]
+) -> int:
+    """
+    Function updates the largest_square_area[0], if recursive call found
+    square with maximum area.
+
+    We aren't using dp_array here, so the time complexity would be exponential.
+
+    >>> largest_square_area_in_matrix_top_down_approch(2, 2, [[1,1], [1,1]])
+    2
+    >>> largest_square_area_in_matrix_top_down_approch(2, 2, [[0,0], [0,0]])
+    0
+    """
+
+    def update_area_of_max_square(row: int, col: int) -> int:
+
+        # BASE CASE
+        if row >= rows or col >= cols:
+            return 0
+
+        right = update_area_of_max_square(row, col + 1)
+        diagonal = update_area_of_max_square(row + 1, col + 1)
+        down = update_area_of_max_square(row + 1, col)
+
+        if mat[row][col]:
+            sub_problem_sol = 1 + min([right, diagonal, down])
+            largest_square_area[0] = max(largest_square_area[0], sub_problem_sol)
+            return sub_problem_sol
+        else:
+            return 0
+
+    largest_square_area = [0]
+    update_area_of_max_square(0, 0)
+    return largest_square_area[0]
+
+
+def largest_square_area_in_matrix_top_down_approch_with_dp(
+    rows: int, cols: int, mat: list[list[int]]
+) -> int:
+    """
+    Function updates the largest_square_area[0], if recursive call found
+    square with maximum area.
+
+    We are using dp_array here, so the time complexity would be O(N^2).
+
+    >>> largest_square_area_in_matrix_top_down_approch_with_dp(2, 2, [[1,1], [1,1]])
+    2
+    >>> largest_square_area_in_matrix_top_down_approch_with_dp(2, 2, [[0,0], [0,0]])
+    0
+    """
+
+    def update_area_of_max_square_using_dp_array(
+        row: int, col: int, dp_array: list[list[int]]
+    ) -> int:
+        if row >= rows or col >= cols:
+            return 0
+        if dp_array[row][col] != -1:
+            return dp_array[row][col]
+
+        right = update_area_of_max_square_using_dp_array(row, col + 1, dp_array)
+        diagonal = update_area_of_max_square_using_dp_array(row + 1, col + 1, dp_array)
+        down = update_area_of_max_square_using_dp_array(row + 1, col, dp_array)
+
+        if mat[row][col]:
+            sub_problem_sol = 1 + min([right, diagonal, down])
+            largest_square_area[0] = max(largest_square_area[0], sub_problem_sol)
+            dp_array[row][col] = sub_problem_sol
+            return sub_problem_sol
+        else:
+            return 0
+
+    largest_square_area = [0]
+    dp_array = [[-1] * cols for _ in range(rows)]
+    update_area_of_max_square_using_dp_array(0, 0, dp_array)
+
+    return largest_square_area[0]
+
+
+def largest_square_area_in_matrix_bottom_up(
+    rows: int, cols: int, mat: list[list[int]]
+) -> int:
+    """
+    Function updates the largest_square_area, using bottom up approach.
+
+    >>> largest_square_area_in_matrix_bottom_up(2, 2, [[1,1], [1,1]])
+    2
+    >>> largest_square_area_in_matrix_bottom_up(2, 2, [[0,0], [0,0]])
+    0
+
+    """
+    dp_array = [[0] * (cols + 1) for _ in range(rows + 1)]
+    largest_square_area = 0
+    for row in range(rows - 1, -1, -1):
+        for col in range(cols - 1, -1, -1):
+
+            right = dp_array[row][col + 1]
+            diagonal = dp_array[row + 1][col + 1]
+            bottom = dp_array[row + 1][col]
+
+            if mat[row][col] == 1:
+                dp_array[row][col] = 1 + min(right, diagonal, bottom)
+                largest_square_area = max(dp_array[row][col], largest_square_area)
+            else:
+                dp_array[row][col] = 0
+
+    return largest_square_area
+
+
+def largest_square_area_in_matrix_bottom_up_space_optimization(
+    rows: int, cols: int, mat: list[list[int]]
+) -> int:
+    """
+    Function updates the largest_square_area, using bottom up
+    approach. with space optimization.
+
+    >>> largest_square_area_in_matrix_bottom_up_space_optimization(2, 2, [[1,1], [1,1]])
+    2
+    >>> largest_square_area_in_matrix_bottom_up_space_optimization(2, 2, [[0,0], [0,0]])
+    0
+    """
+    current_row = [0] * (cols + 1)
+    next_row = [0] * (cols + 1)
+    largest_square_area = 0
+    for row in range(rows - 1, -1, -1):
+        for col in range(cols - 1, -1, -1):
+
+            right = current_row[col + 1]
+            diagonal = next_row[col + 1]
+            bottom = next_row[col]
+
+            if mat[row][col] == 1:
+                current_row[col] = 1 + min(right, diagonal, bottom)
+                largest_square_area = max(current_row[col], largest_square_area)
+            else:
+                current_row[col] = 0
+        next_row = current_row
+
+    return largest_square_area
+
+
+if __name__ == "__main__":
+    import doctest
+
+    doctest.testmod()
+    print(largest_square_area_in_matrix_bottom_up(2, 2, [[1, 1], [1, 1]]))