Find the Duplicate Number LeetCode Solution

Problem – 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.

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?

Find the Duplicate Number LeetCode Solution in Java

    public static int findDuplicate_2loops(int[] nums) {
        int len = nums.length;
        for (int i = 0; i < len; i++) {
            for (int j = i + 1; j < len; j++) {
                if (nums[i] == nums[j]) {
                    return nums[i];
                }
            }
        }

        return len;
    }

Find the Duplicate Number LeetCode Solution in C++

class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        int low = 1, high = nums.size() - 1, cnt;
        
        while(low <=  high)
        {
            int mid = low + (high - low) / 2;
            cnt = 0;
            // cnt number less than equal to mid
            for(int n : nums)
            {
                if(n <= mid)
                    ++cnt;
            }
            // binary search on left
            if(cnt <= mid)
                low = mid + 1;
            else
            // binary search on right
                high = mid - 1;
            
        }
        return low;
    }
};

Find the Duplicate Number LeetCode Solution in Python

class Solution:
    def findDuplicate(self, nums):
        slow, fast = nums[0], nums[0]
        while True:
            slow, fast = nums[slow], nums[nums[fast]]
            if slow == fast: break
           
        slow = nums[0];
        while slow != fast:
            slow, fast = nums[slow], nums[fast]
        return slow
Find the Duplicate Number LeetCode Solution Review:

In our experience, we suggest you solve this Find the Duplicate Number LeetCode Solution and gain some new skills from Professionals completely free and we assure you will be worth it.

If you are stuck anywhere between any coding problem, just visit Queslers to get the Find the Duplicate Number LeetCode Solution

Find on LeetCode

Conclusion:

I hope this Find the Duplicate Number LeetCode Solution would be useful for you to learn something new from this problem. If it helped you then don’t forget to bookmark our site for more Coding Solutions.

This Problem is intended for audiences of all experiences who are interested in learning about Data Science in a business context; there are no prerequisites.

Keep Learning!

More Coding Solutions >>

LeetCode Solutions

Hacker Rank Solutions

CodeChef Solutions

Leave a Reply

Your email address will not be published.