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

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