Merge pull request #2071 from hongseoupyun/main

[hongseoupyun] Week 02 solutions

Author: TonyKim9401

climbing-stairs/hongseoupyun.py

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+
+# This problem is a getting the number of cases to climb stairs just taking 1 or 2 steps at a time.
+# The pattern is similar to Fibonacci sequence.
+# As n increases by 1, the number of ways to climb will be 1, 2, 3, 5, 8, 13, 21
+# The Formula is to get number of ways to climb to the nth step is:
+# ways(n) = ways(n-1) + ways(n-2) with base of ways(1) = 1 and ways(2) = 2
+# which means the number of ways to climb is the sum of last two previous ways
+
+
+
+# Time complexity is O(n) - loops n times
+# Space complexity is  O(1) — memorizes only prev1 & prev2
+class Solution:
+    def climbStairs(self, n: int) -> int:
+        # Handling base case
+        if n <= 1:
+            return 1
+        if n == 2:
+            return 2
+        
+        prev1 = 1
+        prev2 = 2
+        for i in range(n,n+1):
+            current = prev2 + prev1
+            prev1 = prev2
+            prev2 = current
+
+        return prev2

climbing-stairs/hongseoupyun.ts

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+// # This problem is a getting the number of cases to climb stairs just taking 1 or 2 steps at a time.
+// # The pattern is similar to Fibonacci sequence.
+// # As n increases by 1, the number of ways to climb will be 1, 2, 3, 5, 8, 13, 21
+// # The Formula is to get number of ways to climb to the nth step is:
+// # ways(n) = ways(n-1) + ways(n-2) with base of ways(1) = 1 and ways(2) = 2
+// # which means the number of ways to climb is the sum of last two previous ways
+
+
+//Time complexity is O(n) - loops n times
+//Space complexity is  O(1) — memorizes only prev1 & prev2
+function climbStairs(n: number): number {
+
+
+    if (n <=1 ) return 1;
+    if (n === 2) return 2;
+
+    // initial settiongs
+    let prev1 = 1
+    let prev2 = 2
+
+    //Loop from when the number of stairs is 3
+    for (let i = 3; i<=n; i++){
+        const current = prev2 + prev1
+        //eg) when n = 3, prev1 = 1, prev2 = 2, current = 3 => prev1 = 2, prev2 = 3, return prev2
+        //eg) when n = 4, prev1 = 2, prev2 = 3, current = 5 => prev1 = 3, preev2 = 5, return prev2
+        prev1 = prev2
+        prev2 = current
+    }
+
+    return prev2
+};