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.

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

Binary search.

low = 1, high = n, mid = (low + high) / 2;

count how many elements are smaller or equal to mid

if this count is bigger than mid, the duplicated number must sits between 1 and mid (inclusive)

else the duplicated number must sits between mid + 1 and high(inclusive)

// 1 2 2 n = 2
// low = 1, high = 2, mid = 1
// count = 1
class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        int n = nums.size() - 1;
        int low = 1;
        int high = n;
        while(low != high){
            int mid = (low + high) / 2;
            int count = 0;
            for(int i = 0; i <= n; i++){
                if(nums[i] <= mid) count++;
            }
            if(count > mid){
                //duplicated number is between low and mid
                high = mid;
            }
            else{
                low = mid + 1;
            }
        }
        return low;
    }
};

Another approach is use fast and slow pointers.

start from index 0, since there are n + 1 numbers all ranges in 0 – n. so a circle must exist. We consider nums[i] as the next node of i. It was like node.next in linked list.

Find the entry of ring in linked list.

class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        int fast = 0;
        int slow = 0;
        fast = nums[nums[fast]];
        slow = nums[slow];
        while(fast != slow){
            fast = nums[nums[fast]];
            slow = nums[slow];
        }
        //fast == slow
        fast = 0;
        while(fast != slow){
            fast = nums[fast];
            slow = nums[slow];
        }
        return slow;
    }
};