Solution Explanation: Find the Duplicate Number

This problem asks to find the duplicate number in an array of integers where each integer is within the range [1, n] (inclusive), and the array has n+1 elements. The constraint is to solve this without modifying the input array and using only constant extra space.

The most efficient solution uses Binary Search. Here's a breakdown:

Core Idea:

The solution leverages the fact that if a number x is the duplicate, there must be more numbers less than or equal to x than x itself. This allows us to use binary search to efficiently find the duplicate.

Algorithm:

1. Initialization: Set left to 1 and right to n (the upper bound of the range).

2. Binary Search Iteration: While left is less than right:

   • Calculate the middle point mid as (left + right) / 2.
   • Count the number of elements in the array that are less than or equal to mid. Let's call this count count.
   • Decision:
     • If count > mid, it means there are more numbers less than or equal to mid than the possible unique values (1 to mid). This implies the duplicate must be within the range [1, mid], so we update right to mid.
     • Otherwise, count <= mid, meaning the duplicate is in the range [mid + 1, n], so we update left to mid + 1.

3. Return Result: Once the loop terminates (left == right), left (or right) will hold the duplicate number.

Time Complexity: O(n log n)

The binary search performs log n iterations. In each iteration, we need to scan the entire array (size n) to count the elements. Therefore, the overall time complexity is O(n log n).

Space Complexity: O(1)

The algorithm uses only a few variables to store left, right, mid, and count. It does not use any additional data structures that scale with the input size; therefore, the space complexity is constant.

Code Examples (Python):

import bisect
 
class Solution:
    def findDuplicate(self, nums: List[int]) -> int:
        def count_less_equal(x):
            return sum(1 for num in nums if num <= x)
 
        left, right = 1, len(nums) - 1
        while left < right:
            mid = (left + right) // 2
            if count_less_equal(mid) > mid:
                right = mid
            else:
                left = mid + 1
        return left
 
#Alternative using bisect_left for slightly improved performance in python
 
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)
 

The code in other languages (Java, C++, Go, TypeScript, Rust, Javascript) provided earlier follow the same logic with minor syntactic differences for each language. They all maintain the same O(n log n) time and O(1) space complexities.