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.

Example 1:

Input: [1,3,4,2,2]
Output: 2
Example 2:

Input: [3,1,3,4,2]
Output: 3
Note:

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




Solution 1 2 3 
https://leetcode.com/problems/find-the-duplicate-number/solution/


Binary search : 
Nice solution. A quick explanation for these don't understand the idea behind the scenes. The binary search is utilizing the fact that the duplicate number could only occur in the "denser" half of the array (This is only true, we have no missing numbers from 1 to n). So we should set low or high to move towards the denser half. Eventually when low exceeds high we will get the duplicated number.


Input: [1,3,4,2,2]
Output: 2

class Solution {
    public int findDuplicate(int[] nums) {
        int left = 0; // 0, 0, 1， 2 
        int right = nums.length - 1; // 4 , 1 , 1， 1 
        while(left <= right){
            int mid = left + (right - left) / 2; // 2, 0, 1 
            int count = 0; 
            for(int a : nums){
                if(a <= mid) count++; // count = 3, 0  ,1 
            }
            if(count <= mid) left = mid + 1; //  1 , 2 
            else right = mid - 1; // right = mid - 1 = 2 - 1 = 1 
        }
        return left;//  2 
        
    }
}


    Input: [1,3,4,2,2]
Output: 2

 

这个解法用 count 来 数 数组中一共有多少位在mid value 以前， 这样就不用 sort 这个array 了

若是原本有两个在前面， 可是 数出来三个， 就说明 多余的那个 就在前面， otherwise， 在后面

 

This solution is based on binary search.

 

At first the search space is numbers between 1 to n. Each time I select a number mid (which is the one in the middle) and count all the numbers equal to or less than mid. Then if the count is more than mid, the search space will be [1 mid] otherwise [mid+1 n]. I do this until search space is only one number.

 

Let's say n=10 and I select mid=5. Then I count all the numbers in the array which are less than equal mid. If the there are more than 5 numbers that are less than 5, then by Pigeonhole Principle (https://en.wikipedia.org/wiki/Pigeonhole_principle) one of them has occurred more than once. So I shrink the search space from [1 10] to [1 5]. Otherwise the duplicate number is in the second half so for the next step the search space would be [6 10].