LeetCode 287 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.

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.

Here is a O(nlogn) algorithm. The key idea is to use binary search. Every time we will make a guess, it is the mid value of the current range of candidates. We then compute the following values:

count_lt_mid = #{x: x < guessed value}
count_eq_mid = #{x: x == guessed value}

if count_eq_mid > 1, we need to return guessed value (obvious).

The key observation here is if count_eq_mid >= guessed value,  the duplicated value must be smaller than guessed value because [1 .. guessed value - 1] has only "guessed value - 1" possibilities.

class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        return helper(nums, 1, nums.size()-1);
        
    }
    
    int helper(vector<int>& nums, int start, int end) {
        
        if (start == end) return start;
        
        int guess = (start + end) / 2;
        
        int count_lt_mid = 0;
        int count_eq_mid = 0;
        for (int i = 0; i < nums.size(); ++i) {
            if (nums[i] < guess) ++count_lt_mid;
            if (nums[i] == guess) ++count_eq_mid;
        }
        
        if (count_eq_mid > 1) return guess;
        
        if (count_lt_mid >= guess) return helper(nums, start, guess-1);
        if (count_lt_mid < guess) return helper(nums, guess+1, end);
        
        return -1;

    }
};