287. Find the Duplicate Number

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287. Find the Duplicate Number

Description

Given an array of integers nums containingn + 1 integers where each integer is in the range [1, n] inclusive.

There is only one repeated number in nums, return this repeated number.

You must solve the problem without modifying the array numsand using only constant extra space.

Example 1:

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

Example 2:

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

Example 3:

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Input: nums = [3,3,3,3,3]
Output: 3

Constraints:

  • 1 <= n <= 10^5
  • nums.length == n + 1
  • 1 <= nums[i] <= n
  • All the integers in nums appear only once except for precisely one integer which appears two or more times.

Follow up:

  • How can we prove that at least one duplicate number must exist in nums?
  • Can you solve the problem in linear runtime complexity?

Hints/Notes

  • 2025/01/05
  • binary search and count
  • fast and slow pointers
  • Leetcode solution(checked)

Solution

Language: C++

binary search:

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class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        int left = 0, right = nums.size();
        // the value is within (left, right)
        // check(right) is always true
        // check(left) is always false
        while (left + 1 < right) {
            int mid = (left + right) / 2;
            if (check(mid, nums)) {
                right = mid;
            } else {
                left = mid;
            }

        }
        return right;
    }

    bool check(int mid, vector<int>& nums) {
        int count = 0;
        for (int num : nums) {
            if (num <= mid) {
                count++;
            }
        }
        return count > mid;
    }
};

two pointers

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