Added climbing-stairs solution, with description

dynamic_programming/climbing-stairs.rb

@@ -0,0 +1,58 @@
+#You are climbing a staircase. It takes n steps to reach the top.
+#Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?
+
+#Example 1:
+#Input: n = 2
+#Output: 2
+#Explanation: There are two ways to climb to the top.
+#1. 1 step + 1 step
+#2. 2 steps
+
+#Example 2:
+#Input: n = 3
+#Output: 3
+#Explanation: There are three ways to climb to the top.
+#1. 1 step + 1 step + 1 step
+#2. 1 step + 2 steps
+#3. 2 steps + 1 step
+
+#Constraints:
+#1 <= n <= 45
+
+
+
+
+#Dynamic Programming, Recursive Bottom Up Approach - O(n) Time / O(n) Space
+#Init memoization hash (only 1 parameter)
+#Set base cases which are memo[0] = 1 and memo[1] = 1, since there are only 1 way to get to each stair
+#Iterate from 2..n and call recurse(n, memo) for each value n. 
+#Return memo[n].
+
+#recurse(n, memo) - Recurrence Relation is n = (n - 1) + (n - 2)
+#return memo[n] if memo[n] exists. 
+#otherwise, memo[n] = recurse(n - 1, memo) + recurse(n - 2, memo)               
+
+
+# @param {Integer} n
+# @return {Integer}
+def climb_stairs(n)
+    memo = Hash.new
+    
+    memo[0] = 1
+    memo[1] = 1
+    
+    return memo[n] if n <= 1 && n >= 0
+    
+    (2..n).each do |n|
+      recurse(n, memo)
+    end
+    
+    memo[n]
+end
+
+
+def recurse(n, memo)
+    return memo[n] if memo[n]
+    
+    memo[n] = recurse(n - 1, memo) + recurse(n - 2, memo)
+end