Recursion and Dynamic Programming

Author: turingfly

src/chapter08RecursionAndDynamicProgramming/RobotInAGrid.java

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+package chapter08RecursionAndDynamicProgramming;
+
+/**
+ * 
+ * Problem:
+ *
+ */
+public class RobotInAGrid {
+
+}

src/chapter08RecursionAndDynamicProgramming/TripleSteps.java

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+package chapter08RecursionAndDynamicProgramming;
+
+import java.util.Arrays;
+
+/**
+ * 
+ * Problem: 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.
+ *
+ */
+public class TripleSteps {
+	/**
+	 * Method 1: Brute Force Recursion
+	 * 
+	 * Time Complexity: O(3N)
+	 */
+	public int countWays1(int n) {
+		if (n < 0) {
+			return 0;
+		} else if (n == 0) {
+			return 1;
+		} else {
+			return countWays1(n - 1) + countWays1(n - 2) + countWays1(n - 3);
+		}
+	}
+
+	/**
+	 * Method 1: Brute Force Recursion. Number of ways will quickly overflow the
+	 * bounds of an integer. n = 37; Time Complexity: O(N)
+	 */
+	public int countWays2(int n) {
+		int[] memo = new int[n + 1];
+		Arrays.fill(memo, -1);
+		return countWays2(n, memo);
+	}
+
+	private int countWays2(int n, int[] memo) {
+		if (n < 0) {
+			return 0;
+		} else if (n == 0) {
+			return 1;
+		} else if (memo[n] > -1) {
+			return memo[n];
+		} else {
+			memo[n] = countWays2(n - 1, memo) + countWays2(n - 2, memo) + countWays2(n - 3, memo);
+			return memo[n];
+		}
+	}
+}