README_EN.md

287. Find the Duplicate Number

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Description

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?

Solutions

Python3

class Solution:
    def findDuplicate(self, nums: List[int]) -> int:
        left, right = 1, len(nums) - 1
        while left < right:
            mid = (left + right) >> 1
            cnt = sum(v <= mid for v in nums)
            if cnt > mid:
                right = mid
            else:
                left = mid + 1
        return left

Java

class Solution {
    public int findDuplicate(int[] nums) {
        int left = 1, right = nums.length - 1;
        while (left < right) {
            int mid = (left + right) >> 1;
            int cnt = 0;
            for (int v : nums) {
                if (v <= mid) {
                    ++cnt;
                }
            }
            if (cnt > mid) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

C++

class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        int left = 1, right = nums.size() - 1;
        while (left < right)
        {
            int mid = (left + right) >> 1;
            int cnt = 0;
            for (int& v : nums)
                if (v <= mid)
                    ++cnt;
            if (cnt > mid) right = mid;
            else left = mid + 1;
        }
        return left;
    }
};

Go

func findDuplicate(nums []int) int {
	left, right := 1, len(nums)-1
	for left < right {
		mid := (left + right) >> 1
		cnt := 0
		for _, v := range nums {
			if v <= mid {
				cnt++
			}
		}
		if cnt > mid {
			right = mid
		} else {
			left = mid + 1
		}
	}
	return left
}

JavaScript

/**
 * @param {number[]} nums
 * @return {number}
 */
var findDuplicate = function (nums) {
    let left = 1,
        right = nums.length - 1;
    while (left < right) {
        const mid = (left + right) >> 1;
        let cnt = 0;
        for (let v of nums) {
            if (v <= mid) {
                ++cnt;
            }
        }
        if (cnt > mid) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return left;
};

TypeScript

function findDuplicate(nums: number[]): number {
    let left = 1,
        right = nums.length - 1;
    while (left < right) {
        const mid = (left + right) >> 1;
        let cnt = 0;
        for (let v of nums) {
            if (v <= mid) {
                ++cnt;
            }
        }
        if (cnt > mid) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return left;
}

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