#287

Medium

auto_awesomeBinary search over the valid answer space

LeetCode Problem Workspace

Find the Duplicate Number

The problem involves finding the duplicate number in an array of integers, using a binary search approach over the valid answer space.

array two pointers binary search bit manipulation

Problem Statement

You are given an array of integers, nums, of size n + 1, where each integer is in the range [1, n] inclusive. The array contains exactly one duplicate number. Your task is to identify the duplicate number in the array. The solution must meet the constraints of not modifying the array and using only constant extra space.

The challenge involves finding an efficient way to detect the duplicate without using extra space or altering the array itself. A binary search approach can be applied to narrow down the valid answer space, and this method uses the two pointers technique to optimize performance.

Examples

Example 1

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

Output: 2

Example details omitted.

Example 2

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

Output: 3

Example details omitted.

Example 3

Input: nums = [3,3,3,3,3]

Output: 3

Example details omitted.

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.

Solution Approach

Binary Search Over Valid Answer Space

The main idea is to perform a binary search over the range of possible values (from 1 to n). By comparing the count of numbers less than or equal to the middle point, we can adjust our search range to focus on the half that likely contains the duplicate number.

Two Pointers for Detecting Cycles

Using the two pointers method (also known as Floyd's Tortoise and Hare), we detect a cycle in the numbers. The cycle indicates that a duplicate exists and can help pinpoint the exact duplicate number. The two pointers move at different speeds to find the meeting point where the cycle starts.

Constant Space Solution

The solution adheres to the constraint of constant extra space by avoiding the use of additional data structures such as hashmaps or sets. The binary search combined with two pointers ensures that we can solve the problem in O(n) time with O(1) space.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of the solution is O(n log n), where n is the length of the array. The binary search narrows down the potential range of duplicates in logarithmic time, while the two pointers technique efficiently detects the duplicate. The space complexity is O(1), as the solution does not require additional storage beyond a few pointers.

What Interviewers Usually Probe

Look for the candidate's ability to recognize binary search as the key approach to narrowing down the duplicate.
Test the candidate's understanding of constant space solutions and their ability to adapt algorithms to fit stringent space constraints.
Assess the candidate's knowledge of the two pointers technique and how they apply it to detect cycles in arrays.

Common Pitfalls or Variants

Common pitfalls

Not adhering to the constant space requirement by using extra data structures like hashmaps or sets.
Failing to recognize the binary search approach as the optimal solution, opting for brute force methods instead.
Misunderstanding the two pointers technique, leading to incorrect cycle detection or failure to identify the duplicate number.

Follow-up variants

Apply the same technique to find the duplicate in a linked list instead of an array.
Modify the problem to allow more than one duplicate number and adapt the algorithm.
Change the problem to include multiple cycles and test the candidate's ability to handle more complex scenarios.

FAQ

How do I approach the 'Find the Duplicate Number' problem using binary search?

Use binary search on the range from 1 to n. Count the numbers less than or equal to the middle value to adjust your search space towards the half that contains the duplicate.

What are the main constraints in the 'Find the Duplicate Number' problem?

The solution must be found using constant extra space and without modifying the array. The time complexity should be efficient enough to handle large arrays.

How can I avoid using extra space in this problem?

The key to avoiding extra space is using a binary search combined with two pointers to find the duplicate number in-place, without additional data structures.

What is the time complexity of the solution for the 'Find the Duplicate Number' problem?

The time complexity is O(n log n), where n is the length of the array, due to the binary search process combined with the cycle detection technique.

How does the two pointers technique work in this problem?

The two pointers technique detects a cycle in the array, which leads to identifying the duplicate number. One pointer moves slowly, while the other moves quickly to find the meeting point.

terminal

Solution

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 $[x+1,..n]$.

1

2

3

4

5

6

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)

Continue Practicing

#268 missing number #349 intersection of two arrays #350 intersection of two arrays ii