5 Best Ways to Find a Duplicate Element in an n+1 Range of 1 to n in Python

Rate this post

πŸ’‘ Problem Formulation: We are tasked to find a duplicate number in a list that contains numbers ranging from 1 to n, with one additional number adding up to n+1 list items. For instance, given a list [1, 3, 4, 2, 2], where the numbers range from 1 to 4 (n), and there are 5 (n+1) elements in total, we want to identify the duplicate number which is ‘2’ in this case.

Method 1: Use a Hash Map (Dictionary)

This method involves using a Python dictionary to count the occurrences of each number. As we iterate through the list, if we encounter a number that has already been added to the dictionary, we know it is the duplicate. This method is fast and efficient, with a linear time complexity of O(n).

Here’s an example:

def find_duplicate(nums):
    seen = {}
    for num in nums:
        if num in seen:
            return num
        seen[num] = 1
print(find_duplicate([1, 2, 3, 4, 4]))

Output: 4

This code creates a dictionary, seen, to track the occurrences of numbers. It then iterates through the list nums, and the first number that appears twice is returned as the duplicate. This is an efficient way to find the duplicate as it only requires a single pass through the list.

Method 2: Sort and Compare

Another approach is to sort the list first, and then look for adjacent elements that are identical. As soon as a duplicate pair is identified, the search can end. Sorting initially introduces an O(n log n) time complexity, making it less efficient than the dictionary method for large datasets.

Here’s an example:

def find_duplicate(nums):
    nums.sort()
    for i in range(1, len(nums)):
        if nums[i] == nums[i-1]:
            return nums[i]
print(find_duplicate([1, 2, 3, 2, 5]))

Output: 2

In this code, we first sort the list and then iterate through the sorted list. If a duplicate element is found (when two consecutive elements are the same), that element is immediately returned. This method is straightforward but less time-efficient due to the sorting step.

Method 3: Summation Formula

The summation formula method leverages the mathematical fact that the sum of the first n numbers is n(n+1)/2. The idea is to subtract this expected sum from the actual sum of the list elements to find the duplicate number. This method is also O(n) but requires care with large numbers due to potential integer overflow.

Here’s an example:

def find_duplicate(nums):
    n = len(nums) - 1
    expected_sum = n * (n + 1) // 2
    actual_sum = sum(nums)
    return actual_sum - expected_sum
print(find_duplicate([1, 3, 4, 2, 2]))

Output: 2

This snippet calculates the expected sum of numbers from 1 to n using the summation formula and then subtracts it from the actual sum of the list elements. The result is the duplicate number. The strength of this approach is its simplicity and fixed O(n) time complexity.

Method 4: Floyd’s Tortoise and Hare (Cycle Detection)

Floyd’s Tortoise and Hare algorithm is an ingenious way to find duplicates based on cycle detection in a linked list. This algorithm uses two pointers moving at different speeds and detects a cycle (which corresponds to the duplicate number) without requiring extra space, yielding O(n) time complexity and O(1) space complexity.

Here’s an example:

def find_duplicate(nums):
    tortoise = nums[0]
    hare = nums[0]
    # Find the intersection point of the two runners.
    while True:
        tortoise = nums[tortoise]
        hare = nums[nums[hare]]
        if tortoise == hare:
            break

    # Find the entrance to the cycle.
    tortoise = nums[0]
    while tortoise != hare:
        tortoise = nums[tortoise]
        hare = nums[hare]

    return hare

print(find_duplicate([1, 3, 4, 2, 2]))

Output: 2

This code snippet uses two pointers, ‘tortoise’ and ‘hare’, that move through the list at different speeds. The point where both pointers meet indicates a cycle and allows us to find the duplicate element by resetting one pointer to the start of the list and then moving both pointers at the same speed until they meet again.

Bonus One-Liner Method 5: Using Set

This bonus one-liner uses set operations to quickly identify the duplicate element. A set in Python cannot have duplicate elements, so converting the list to a set and comparing sizes immediately reveals if there’s a duplicate. However, unlike other methods, it doesn’t identify which number is the duplicate.

Here’s an example:

nums = [1, 2, 3, 4, 4]
find_duplicate = lambda nums: len(nums) != len(set(nums))
print(find_duplicate(nums))

Output: True

The lambda function checks if the length of the original list is different from the length of the set created from the list. A discrepancy in lengths indicates the presence of a duplicate. This method is concise but is only useful for confirming the existence of a duplicate.

Summary/Discussion

  • Method 1: Dictionary. Fast with O(n) complexity. Straightforward, but requires extra memory for the dictionary.
  • Method 2: Sort and Compare. Simple logic. Inefficient for large sets due to O(n log n) complexity.
  • Method 3: Summation Formula. Fast and elegant with O(n) complexity. Must be wary of integer overflow with large n.
  • Method 4: Floyd’s Tortoise and Hare. Efficient with O(n) complexity and O(1) space. Can be tricky to understand.
  • Bonus Method 5: Set Comparison. Extremely simple one-liner. Only indicates the presence of a duplicate; does not find its value.