SetEasy

Contains Duplicate

SetEasy

Step 1 of 8 · Understand

1. Understand the problem

Given a list of integers, return true if any value appears more than once in the list, or false if every value is unique.

Input

An array of integers `nums`.

Output

A boolean — `true` if any number appears at least twice, `false` if all numbers are distinct.

Examples

input → nums = [1, 2, 3, 1]

output → true

1 appears at index 0 and index 3.

input → nums = [1, 2, 3, 4]

output → false

All values are different.

input → nums = [1, 1, 1, 3, 3, 4, 3, 2, 4, 2]

output → true

Multiple values repeat.

1. Understand the problem

Given a list of integers, return true if any value appears more than once in the list, or false if every value is unique.

Input

An array of integers `nums`.

Output

A boolean — `true` if any number appears at least twice, `false` if all numbers are distinct.

Examples

input → nums = [1, 2, 3, 1]

output → true

1 appears at index 0 and index 3.

input → nums = [1, 2, 3, 4]

output → false

All values are different.

input → nums = [1, 1, 1, 3, 3, 4, 3, 2, 4, 2]

output → true

Multiple values repeat.

2. Brute force idea

For each number, scan the rest of the array to see if that same value appears again. The moment you find a match, return true.

Brute force — the slow one

O(n²)

for i in 0 .. len(nums) - 1:
    for j in i + 1 .. len(nums) - 1:    # ← the nested loop is the trap
        if nums[i] == nums[j]:
            return true
return false

Why this is too slow

Every element is compared against every element after it. For n numbers that's roughly n × n / 2 comparisons — O(n²). A list of 100,000 numbers means up to 5 billion checks.

Why this is not ideal for interviews

Nested loops are the sign that you haven't spotted the better structure yet. Interviewers want to see you jump to the set-based O(n) approach — it's the canonical answer for 'have I seen this before?' problems.

Time: O(n²)

Space: O(1)

5. The optimized approach

Walk through the array once. For each number, check if it's already in a set. If it is, you have a duplicate — return true. If not, add it to the set and keep going. If you finish the loop, return false.

Pseudocode

seen = empty set

for n in nums:
    if n in seen:
        return true
    seen.add(n)

return false

6. Dry run, row by row

Trace it with the first example — nums = [1, 2, 3, 1].

step	i	nums[i]	in seen?	seen (after)
1	0	1	no	{1}
2	1	2	no	{1, 2}
3	2	3	no	{1, 2, 3}
4	3	1	yes ✓	—

step 1: 1 is new — add it to seen.

step 2: 2 is new — add it.

step 3: 3 is new — add it.

step 4: 1 is already in seen — duplicate found, return true.

8. Review & common mistakes

Common mistakes

Sorting first then comparing neighbors
Sorting works (O(n log n)) and is fine to mention, but the set approach is faster and uses the same 'have I seen this' intuition that carries to harder problems. Know both, prefer the set.
Adding to the set before checking
If you add n before checking, you'd need to verify the count is > 1 instead of simply doing 'is it already there?' — it's more error-prone. Check first, add second.
Comparing set size to array length
`new Set(nums).size < nums.length` is a cute one-liner but it builds the entire set before telling you anything. The early-return loop stops as soon as it finds a duplicate, which is faster on average.

Try next (when these problems ship)

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Contains Duplicate

1. Understand the problem

1. Understand the problem

2. Brute force idea

3. Spot the pattern

4. Climb the hint ladder

5. The optimized approach

6. Dry run, row by row

7. The code

8. Review & common mistakes