8.1: triple set ✔️

08. Recursion/8.1. Triple step.md

@@ -0,0 +1,81 @@
+# 8.1. Triple step
+
+> A child is running up a staircase with n steps and can hop either 1 step, 2 steps, or 3 steps at a time. Implement a method to count how many possible ways the child can run up the stairs.
+
+I'm assuming the kid can't go back. Also, the result for n must be the addition/multiplication of the result for n-1.
+
+* Example: 0 1 2
+    - 1x: 1, 012
+    - 2x: 1, 02
+    - 3x: -
+    - total: 2
+* Example: 0 1 2 3
+    - 1x: 1, 0123
+    - 2x: 2, 023, 013
+    - 3x: 1, 03
+    - total: 4
+* Example: 0 1 2 3 4 5
+    - 1x: 1, 01234
+    - 2x: 4, 0245, 0135, 01345, 02345
+    - 3x: 5, 035, 0345, 0145, 0125, 025
+    - total: 10
+* Example: 0 1 2 3 4 5 6
+    - 1x: 1, 0123456
+    - 2x: 6, 0246, 02456, 02356, 01356, 01246, 012346
+    - 3x: 8, 036, 0346, 0356, 03456, 01456, 0146, 01256, 0256
+    - total, 22
+* Example: 0 1 2 3 4 5 6 7
+    - 1x: 1, 0123456
+    - 2x: x, 0134567, 01357, 013457, 013467, 0234567, 02357, 023467, 023457, 024567, 02457, 02467, ?
+
+## Hints
+
+> Approach this from the top down. What is the very last hop the child made?
+
+Either a hop from n-1, n-2, or n-3.
+
+> If we knew the number of paths to each of the steps before step 100, could we compute the number of steps to 100?
+
+That's the point of dynamic programming but I don't see it.
+
+> We can compute the number of steps to 100 by the number of steps to 99, 98, and 97. This corresponds to the child hopping 1, 2, or 3 steps at the end. Do we add those or multiply them?That is: Is it f(100) = f(99) + f(98) + f(97) or f(100) = f(99) * f(98) * f(97)?
+
+Add, either one path or the other is taken. Not all. Multiplying one path with another would signify taking one path and then taking the other.
+
+## Solution
+
+countWays(n) = countWays(n-1) + countWays(n-2) + countWays(n-3). What about base cases? We could say that if n==0, return 1.
+
+```cpp
+int countWays(int n) {
+    if (n < 0) {
+        return 0;
+    } else if (n == 0) {
+        return 1;
+    } else {
+        return countWays(n-1) + countWays(n-2)+ countWays(n-3);
+    }
+}
+```
+
+But this is very inefficient, O(3<sup>n</sup>). Memoization:
+
+```cpp
+int countWays(int n) {
+    int[] memo = new int[n+ 1];
+    Arrays.fill(memo, -1);
+    return countWays(n, memo);
+}
+
+int countWays(int n, int[] memo) {
+
+    if (n < 0) return 0;
+    else if (n == 0) return 1;
+    else if (memo[n] > -1) return memo[n];
+
+    memo[n] = countWays(n-1, memo) + countWays(n-2, memo) + countWays(n-3, memo);
+    return memo[n];
+}
+```
+
+First fill the memo with -1s. When we start with our `n`, n=4 for example, countWays(3), memo is empty so it's created. Memo is created from the bottom up.

08. Recursion/README.md

@@ -1,6 +1,6 @@
 # Chapter 8 Recursion and dynamic programming
 
-Questions p145, solutions p353
+Questions p146, solutions p353
 
 You can know a problem is recursive when it can be built off of subproblems. If a program says: compute the nth, or the first n, or all..., it might be recursive.