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.

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(n2).
4. There is only one duplicate number in the array, but it could be repeated more than once.

Solution:

Method 1: Floyd Loop Detection

This is a classic interview question. Donald Knuth took 24 hours to come up with an O(n) time and O(1) space solution.

If the input is [4, 2, 3, 5, 1, 3], and the duplicate number is 3.

We iterate the array by using the value as index. So the output of the above array is:

nums[0] = 4
nums[4] = 1
nums[1] = 2
nums[2] = 3
nums[3] = 5
nums[5] = 3
nums[3] = 5
nums[5] = 3
...

We write these numbers down.

4 --> 1 --> 2 --> 3 --> 5
                           ^       |
                           |       v
                            ------

We found a cycle!

Now the problem becomes the Floyd loop detection problem and we use two pointers to find the entry of the loop.



Code:

public class Solution {
    public int findDuplicate(int[] nums) {
        int fast = 0;
        int slow = 0;
        do {
            fast = nums[nums[fast]];
            slow = nums[slow];
        } while(slow != fast);
        
        int find = 0;
        while (find != slow) {
            slow = nums[slow];
            find = nums[find];
        }
        return find;
    }
}

Method 2: Brute Force, O(n²) time and O(1) space

Method 3: HashSet, O(n) time and O(n) space

Method 4: Binary Search O(nlogn) time and O(1) space