added Find the Duplicate Number

Author: minhyeong-joe

Medium/FindDuplicateNumber/Driver.java

@@ -0,0 +1,19 @@
+package Medium.FindDuplicateNumber;
+
+import java.util.Arrays;
+
+public class Driver {
+ public static void main(String[] args) {
+ Solution sol = new Solution();
+ int[] arr = {1,3,4,2,2};
+ System.out.println("Input: " + Arrays.toString(arr));
+ int dup = sol.findDuplicate(arr);
+ System.out.println("Output: " + dup);
+
+ System.out.println();
+ int[] arr2 = {3,1,3,4,2};
+ System.out.println("Input: " + Arrays.toString(arr2));
+ int dup2 = sol.findDuplicate(arr2);
+ System.out.println("Output: " + dup2);
+ }
+}

Medium/FindDuplicateNumber/README.md

@@ -0,0 +1,25 @@
+
+# Find the Duplicate Number
+[Leetcode Link](https://leetcode.com/problems/find-the-duplicate-number/)
+
+## Problem:
+
+Given an array nums containing `n + 1` integers where each integer is between `1` and `n` (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.
+
+## Example:
+
+```
+Input: [1,3,4,2,2]
+Output: 2
+```
+```
+Input: [3,1,3,4,2]
+Output: 3
+```
+
+## Note:
+
+1. You **must not** modify the array (assume the array is read only).
+2. You must use only constant, O(1) extra space.
+3. Your runtime complexity should be less than O(n^2).
+4. There is only one duplicate number in the array, but it could be repeated more than once.

Medium/FindDuplicateNumber/Solution.java

@@ -0,0 +1,27 @@
+package Medium.FindDuplicateNumber;
+
+// Use Floyd's Tortoise and Hare Algorithm to detect cycle's entry point
+// https://leetcode.com/problems/find-the-duplicate-number/solution/
+// See approach 3 for detailed explanation.
+// Key ideas
+// Two phases:
+// Phase 1: Hare runs double speed as tortoise, and where they meet is intersection in the cycle.
+// Phase 2: Tortoise start from beginning, hare runs at tortoise's pace, where they meet again is the cycle's entry point.
+
+class Solution {
+ public int findDuplicate(int[] nums) {
+ int slow = nums[nums[0]],
+ fast = nums[slow];
+ while (slow != fast) {
+ slow = nums[slow];
+ fast = nums[nums[fast]];
+ }
+ // move slower traversal back to starting
+ slow = nums[0];
+ while (slow != fast) {
+ slow = nums[slow];
+ fast = nums[fast];
+ }
+ return slow;
+ }
+}

README.md

@@ -15,8 +15,9 @@ Languages used: Java and Python
  - [Intersection of Two Arrays](Easy/IntersectionOfTwoArrays)
 - Medium
  - [Minimum Add to Make Parentheses Valid](Medium/MinimumAddtoMakeParenthesesValid)
- - [Distribute Coins in Binary Tree](#Medium/DistributionCoinsInBinaryTree)
- - [Find Minimum Number of Fibonacci Numbers Whose Sum is K](#Medium/FindMinNumFibNumSumK)
+ - [Distribute Coins in Binary Tree](Medium/DistributionCoinsInBinaryTree)
+ - [Find Minimum Number of Fibonacci Numbers Whose Sum is K](Medium/FindMinNumFibNumSumK)
+ - [Find the Duplicate Number](Medium/FindDuplicateNumber)
 - Hard
 
 ---