LeetCode 287. Find the Duplicate Number

Given an array nums containing n + 1 integers where each integer is between 1 andn (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.

The most straight forward way is to use hashmap..... However, the note said O(1) space allowed. hmm, after reading the note, there is a clue...

There are a few algorithm complexity less then O(n2). Sort -> nlgn. In this case, we need to sort the inputs first and then use binary search to find the element.

Suppose, we have a sorted array. How to apply binary search is the main trick here.

So, let's make several examples

[1, 2, 2, 2, 3]:

  First Round: left = 0, right = 4, mid = 2, nums[mid] = 2  < mid + 1.... .We thus know that the duplicate number is on the left. Thus, right = mid;

  Second Round: left = 0, right = 2, mid = 1, nums[mid] = 2 == mid + 1, we thus know that the duplicate number is on the right. Thus, left = mid;

  Third Round: left = 1, right 2,  ---> in this case, only two elements left, the duplicate is in them two.We dont need to apply binary search anymore. Since, left(1) have been verified in Secound Round. Thus, Duplicate = nums[right].

[1, 2, 3, 4, 4]

  First Round: left = 0, right = 4, mid = 2, nums[mid]  = 3 == mid + 1, we thus know that the duplicate number is on the right. Thus, left = mid;

  Second Round: left = 2, right = 4, mid = 3, nums[mid] = 4 == mid + 1, we thus know that the duplicate number is on the right. Thus, left = mid;

 Third Round: left = 3, right = 4 --> in this case, there are only two elements left, the duplicate is in them two.Since we know that 3 is not the duplicate from the second round. Then, nums[right] must be the duplicate.

[1, 2, 4, 5, 7, 7, 7, 7, 8]

  First Round: left = 0, right = 8, mid = 4, nums[mid] = 7 > mid + 1, we thus know that the duplicate number must be on the right. thus, left = mid;

  Second Round: left = 4, right = 8, mid = 6, nums[mid] = 7 == mid + 1, we thus know that the duplicate must be on the right. Thus, left = mid;

  Third Round: left = 6, right = 8, mid = 7, nums[mid] = 7 < mid + 1, we thus know that the duplicate must be on the left. Thus, right = mid;

  Fourth Round: left = 6, right = 7. ---> in this case, there are only two elements left.The duplicate must be in them two. Since we know that nums[6] is not the duplicate from the second round,then, nums[right] is the duplicate.

#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

int findDuplicate(vector<int>& nums) {
    if(nums.size() <= 1) return -1;
    sort(nums.begin(), nums.end());   // nlgn time complexity.
 
    // binary search then.
    int left = 0, right = nums.size() - 1;
    while(left < right - 1) {
        int mid = left + (right - left) / 2;
        if(nums[mid] >= mid + 1) left = mid;
        else right = mid;
    }
    return nums[right];
}

int main(void) {
    vector<int> nums{7, 9, 7, 4, 2, 8, 7, 7, 1, 5};
    int target = findDuplicate(nums);
    cout << target << endl;
}