solve: climbing stairs

Author: sounmind

climbing-stairs/sounmind.js

@@ -0,0 +1,51 @@
+class Solution:
+    def climbStairs(self, n: int) -> int:
+        """
+        Climbing Stairs Problem: You are climbing a staircase with n steps.
+        Each time you can either climb 1 or 2 steps. How many distinct ways can you climb to the top?
+
+        This is essentially a Fibonacci sequence problem:
+        - To reach step n, you can either:
+          1. Take a single step from step (n-1)
+          2. Take a double step from step (n-2)
+        - Therefore, the total ways to reach step n = ways to reach (n-1) + ways to reach (n-2)
+
+        Time Complexity: O(n) - We need to calculate each step once
+        Space Complexity: O(1) - We only store two previous values regardless of input size
+        """
+        # Base cases: There's only 1 way to climb 1 step and 2 ways to climb 2 steps
+        if n == 1:
+            return 1
+        if n == 2:
+            return 2
+
+        # Initialize variables with base cases
+        # For staircase with 1 step, there's only 1 way to climb
+        ways_to_reach_n_minus_2 = 1
+
+        # For staircase with 2 steps, there are 2 ways to climb (1+1 or 2)
+        ways_to_reach_n_minus_1 = 2
+
+        # Variable to store the current calculation
+        ways_to_reach_current_step = 0
+
+        # Start calculating from step 3 up to step n
+        for _ in range(3, n + 1):
+            # To reach current step, we can either:
+            # 1. Take a single step after reaching step (n-1)
+            # 2. Take a double step after reaching step (n-2)
+            # So the total ways = ways to reach (n-1) + ways to reach (n-2)
+            ways_to_reach_current_step = (
+                ways_to_reach_n_minus_1 + ways_to_reach_n_minus_2
+            )
+
+            # Shift our window of calculations forward:
+            # The previous (n-1) step now becomes the (n-2) step for the next iteration
+            ways_to_reach_n_minus_2 = ways_to_reach_n_minus_1
+
+            # The current step calculation becomes the (n-1) step for the next iteration
+            ways_to_reach_n_minus_1 = ways_to_reach_current_step
+
+        # After the final iteration, both ways_to_reach_n_minus_1 and ways_to_reach_current_step
+        # have the same value (the answer for step n)
+        return ways_to_reach_n_minus_1  # Could also return ways_to_reach_current_step