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287. Find the Duplicate Number

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

Solution 1: Binary Search

We can observe that if the number of elements in $[1,..x]$ is greater than $x$, then the duplicate number must be in $[1,..x]$, otherwise the duplicate number must be in $[x+1,..n]$.

Therefore, we can use binary search to find $x$, and check whether the number of elements in $[1,..x]$ is greater than $x$ at each iteration. This way, we can determine which interval the duplicate number is in, and narrow down the search range until we find the duplicate number.

The time complexity is $O(n \times \log n)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$.

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

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

  • class Solution:
    def findDuplicate(self, nums: List[int]) -> int:
        def f(x: int) -> bool:
            return sum(v <= x for v in nums) > x

        return bisect_left(range(len(nums)), True, key=f)


  • func findDuplicate(nums []int) int {
	return sort.Search(len(nums), func(x int) bool {
		cnt := 0
		for _, v := range nums {
			if v <= x {
				cnt++
			}
		}
		return cnt > x
	})
}

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


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


  • impl Solution {
    #[allow(dead_code)]
    pub fn find_duplicate(nums: Vec<i32>) -> i32 {
        let mut left = 0;
        let mut right = nums.len() - 1;

        while left < right {
            let mid = (left + right) >> 1;
            let cnt = nums
                .iter()
                .filter(|x| **x <= (mid as i32))
                .count();
            if cnt > mid {
                right = mid;
            } else {
                left = mid + 1;
            }
        }

        left as i32
    }
}

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