Find the Duplicate Number LeetCode Solution | Easy Approach

Find the Duplicate Number Given an array of integers nums containing n + 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 nums and uses only constant extra space.

Example 1:

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

Example 2:

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

Constraints:

• 1 <= n <= 105
• 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.

Find the Duplicate Number Solutions

✅Time: O(n)
✅Space: O(1)

C++

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

    while (slow != fast) {
      slow = nums[slow];
      fast = nums[nums[fast]];
    }

    slow = nums[0];

    while (slow != fast) {
      slow = nums[slow];
      fast = nums[fast];
    }

    return slow;
  }
};

Java

 
class Solution {
  public int findDuplicate(int[] nums) {
    int slow = nums[nums[0]];
    int fast = nums[nums[nums[0]]];

    while (slow != fast) {
      slow = nums[slow];
      fast = nums[nums[fast]];
    }

    slow = nums[0];

    while (slow != fast) {
      slow = nums[slow];
      fast = nums[fast];
    }

    return slow;
  }
}

Python

 class Solution:
  def findDuplicate(self, nums: List[int]) -> int:
    slow = nums[nums[0]]
    fast = nums[nums[nums[0]]]

    while slow != fast:
      slow = nums[slow]
      fast = nums[nums[fast]]

    slow = nums[0]

    while slow != fast:
      slow = nums[slow]
      fast = nums[fast]

    return slow

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