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[LeetCode 287] Find the Duplicate Number

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.

Notice

You must not modify the array (assume the array is read only).
You must use only constant, O(1) extra space.
Your runtime complexity should be less than O(n^2).
There is only one duplicate number in the array, but it could be repeated more than once.

Example

Given nums = [5,5,4,3,2,1] return 5
Given nums = [5,4,4,3,2,1] return 4

Solution 1. O(n * log n) runtime, O(1) space, Binary Search on the max possible value n

 1 public class Solution {
 2     public int findDuplicate(int[] nums) {
 3         if(nums == null || nums.length < 2){
 4             return -1;
 5         }
 6         int start = 1, end = nums.length - 1;
 7         while(start + 1 < end){
 8             int mid = start + (end - start) / 2;
 9             if(countSmallerAndEqual(mid, nums) <= mid){
10                 start = mid;
11             }
12             else{
13                 end = mid;
14             }
15         }
16         if(countSmallerAndEqual(start, nums) <= start){
17             return end;
18         }
19         return start;
20     }
21     private int countSmallerAndEqual(int mid, int[] nums){
22         int cnt = 0;
23         for(int i = 0; i < nums.length; i++){
24             if(nums[i] <= mid){
25                 cnt++;
26             }
27         }
28         return cnt;
29     }
30 }

Solution 2. O(n) runtime, O(1) extra space, Linked List cycle detection, two pointers, slow and fast pointers

 1 public class Solution {
 2     public int findDuplicate(int[] nums) {
 3         if(nums == null || nums.length < 2){
 4             return -1;
 5         }
 6         
 7         int slow = nums[0];
 8         int fast = nums[nums[0]];
 9         while(slow != fast){
10             slow = nums[slow];
11             fast = nums[nums[fast]];
12         }
13         fast = 0;
14         while(slow != fast){
15             slow = nums[slow];
16             fast = nums[fast];
17         }
18         return slow;
19     }
20 }