Find The Duplicate Number
Detailed guide and Python implementation for the 'Find The Duplicate Number' problem.
1. Concept Overview
The 'Find The Duplicate Number' problem is a key challenge in the Linked List section.
This implementation focuses on easy-level logic in Python.
We prioritize technical accuracy and code readability in our provided solutions.
2. Real-World Applications
3. Visual Intuition
Visualizing the logic flow for Find The Duplicate Number.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Find The Duplicate Number carefully.
2. Formulate brute force
Draft a simple iterative solution.
3. Identify inefficiency
Look for redundant calculations.
4. Optimize search path
Use hashing or sorting to speed up the process.
5. Final Implementation
Clean up the code for production standards.
Problem Statement
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 using only constant extra space.
Implement a function findDuplicate(nums: list) -> int that returns the duplicate number.
- •1 <= n <= 100000
- •nums.length == n + 1
- •1 <= nums[i] <= n
- •There is only one repeated number in nums, but it could be repeated more than once
Examples
[1,3,4,2,2]
2
The duplicate number is 2. It appears twice in the array.
[3,1,3,4,2]
3
The duplicate number is 3. It appears twice in the array.
[3,3,3,3,3]
3
The duplicate number is 3. It appears five times.
Need a Hint?
Edge Cases to Watch
- Empty input structures
- Single element inputs
- Large numerical bounds
Ready to Solve?
Open the problem in PyRun's browser-based Python editor. Your code runs fully offline — no server required.
Interview Insights & Variations
Complexity Analysis Breakdown
Why Time: Directly evaluates all possibilities.
Why Space: Uses standard local memory.
Why Time: Optimized paths reduce total operations.
Why Space: May trade memory for speed.
Optimized Solution Python Code
Optimized Solution Python Code
def find_duplicate_opt(nums: list[int]) -> int:
slow, fast = 0, 0
while True:
slow = nums[slow]
fast = nums[nums[fast]]
if slow == fast:
break
slow2 = 0
while True:
slow = nums[slow]
slow2 = nums[slow2]
if slow == slow2:
return slowBrute Force Code (Spoiler Guarded)
Brute Force Code (Spoiler Guarded)
def find_duplicate_brute(nums: list[int]) -> int:
nums.sort()
for i in range(len(nums) - 1):
if nums[i] == nums[i + 1]:
return nums[i]
return -1Algorithm Pattern Checklist
When dealing with Linked List data patterns.
Core Prerequisites
Revision Key Notes
Common Mistakes & Pitfalls
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