diff --git a/Square_SubMat_With_Ones.py b/Square_SubMat_With_Ones.py
new file mode 100644
index 00000000..2e5185de
--- /dev/null
+++ b/Square_SubMat_With_Ones.py
@@ -0,0 +1,47 @@
+from typing import List
+
+class Solution:
+    def countSquares(self, matrix: List[List[int]]) -> int:
+        # Get number of rows and columns
+        r = len(matrix)
+        c = len(matrix[0])
+
+        # dp[i][j] will store the size of the largest square 
+        # whose bottom-right corner is at cell (i, j)
+        dp = [[0] * c for _ in range(r)]
+
+        count = 0  # To store total number of square submatrices
+
+        # Initialize first row
+        # Each 1 in the first row forms a 1x1 square
+        for j in range(c):
+            if matrix[0][j] == 1:
+                dp[0][j] = 1
+                count += 1
+        
+        # Initialize first column
+        # Each 1 in the first column forms a 1x1 square
+        for i in range(1, r):
+            if matrix[i][0] == 1:
+                dp[i][0] = 1
+                count += 1
+
+        # Fill the dp table for the rest of the matrix
+        for i in range(1, r):
+            for j in range(1, c):
+                # If the cell is 0, no square can end here
+                if matrix[i][j] == 0:
+                    continue
+                # If the cell is 1, find the smallest neighboring square
+                # and extend it by 1 to form a bigger square
+                elif dp[i-1][j] == 0 or dp[i-1][j-1] == 0 or dp[i][j-1] == 0:
+                    dp[i][j] = 1  # Only a 1x1 square possible
+                else:
+                    # Minimum of top, left, and top-left + 1
+                    dp[i][j] = min(dp[i-1][j], dp[i-1][j-1], dp[i][j-1]) + 1
+                
+                # Add the number of squares ending at (i, j)
+                count += dp[i][j]
+        
+        # Return total count of all square submatrices with all 1s
+        return count