Find the Duplicate Number
A medium Arrays interview guide. Task statement, worked examples, intuition, and step-by-step solutions.
- Topic
- Arrays
- Sheets
- 2
- Core for
- 26 roles
The problem
Given an array of n+1 integers where each integer is between 1 and n inclusive, find the duplicate number. There is guaranteed to be exactly one duplicate.
Example 1
- Input
- [2,4,1,3,4]
- Output
- 4
Example 2
- Input
- [3,3,1,4,2]
- Output
- 3
Example 3
- Input
- [1,1]
- Output
- 1
Constraints
- 2 <= n <= 10^5
- 1 <= arr[i] <= n
How to think about it
Updated 2026-09-29Every index pointing to arr[i] forms a directed edge in a functional graph. Because every value is between 1 and n while the array has n + 1 cells, the pigeonhole principle guarantees two different indices point to the same node, creating a cycle whose entry point is precisely the duplicate value.
Approaches, worst first
Visited hash set
time O(n) · space O(n)
Traverse the array and insert each element into a hash set. The first element already present in the set is the duplicate. Correct and quick to verify, but allocates linear extra memory.
In-place sign marking
time O(n) · space O(1)
Treat each value as an index and negate the value at that destination. If a destination is already negative, its index is the duplicate. Runs in linear time and constant space but permanently mutates the underlying input array.
Floyd cycle detectionWrite this one
time O(n) · space O(1)
Treat array values as next pointers: slow advances one step (arr[slow]), fast advances two steps (arr[arr[fast]]). Once they collide, reset slow to index 0 and advance both one step at a time until they meet at the cycle entrance.
Where people lose marks · 3
- Starting Floyd's algorithm at index `slow = 0` and `fast = 0` triggers the termination condition immediately before the pointers even move.
- Attempting in-place sign negation on a read-only or immutable input buffer fails at runtime.
- Binary searching on values instead of indices requires counting elements less than or equal to mid, which takes O(n log n) time and is slower than cycle finding.
Full solution
Floyd cycle detection over the value-as-pointer graph: it is O(n) time and O(1) space without mutating the input, which is what rules out the hash set and the sign-marking trick in an interview.
Python
def find_duplicate(nums: list[int]) -> int:
# values are indices, so nums[i] is a "next" pointer; the duplicate is the cycle entrance
slow = fast = nums[0]
while True:
slow = nums[slow]
fast = nums[nums[fast]]
if slow == fast:
break
# phase 2: the distance from the start to the entrance equals the distance from the meeting point
slow = nums[0]
while slow != fast:
slow = nums[slow]
fast = nums[fast]
return slow
JavaScript
function findDuplicate(nums) {
let slow = nums[0];
let fast = nums[0];
do {
slow = nums[slow];
fast = nums[nums[fast]];
} while (slow !== fast);
slow = nums[0];
while (slow !== fast) {
slow = nums[slow];
fast = nums[fast];
}
return slow;
}
The theory behind it
Arrays — the ground this problem stands on. All Arrays problems
What Arrays is
An array is a row of fixed boxes laid side by side in computer memory, like numbered lockers in a hallway. Because each box occupies identical space and sits directly next to its neighbors, jumping to locker zero or locker ten thousand takes the exact same tiny fraction of time. Every box holds an item of the same type, addressed by an offset number called an index.
When to reach for it
Reach for an array when items arrive in a known sequence and need immediate retrieval by position number. Problems asking for running totals, prefix accumulations, cyclic rotations, or in-place rearrangements signal array mechanics. Whenever constraints require constant-time random lookups or contiguous cache scans across fixed collections, a flat sequence is the default container.
How the pattern works
Visualize a tape with zero-indexed slots stretching from start to end. Keep track of write and read cursors when modifying contents without allocating helper buffers. For running computations, maintain an invariant such as having processed all elements left of the current index while pending elements wait to the right. When modifying entries in place, consider scanning backwards from the end so unread data is not overwritten.
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