solve: uniquePaths

Author: GaggleHelena

unique-paths/yolophg.py

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+# Time Complexity: O(N * M) : iterate through a grid of size m x n once.
+# Space Complexity: O(N * M) : use a DP table of size (m+1) x (n+1) to store the number of paths.
+
+class Solution:
+    def uniquePaths(self, m: int, n: int) -> int:
+        # create a dp table and add extra rows and columns for easier indexing
+        table = [[0 for x in range(n+1)] for y in range(m+1)]
+        
+        # set the starting point, one way to be at the start
+        table[1][1] = 1
+
+        # iterate the grid to calculate paths
+        for i in range(1, m+1):
+            for j in range(1, n+1):
+                # add paths going down
+                if i+1 <= m:
+                    table[i+1][j] += table[i][j] 
+                # add paths going right
+                if j+1 <= n:
+                    table[i][j+1] += table[i][j]
+        
+        return table[m][n]