Arrays: Contains Duplicate — Kotlin Solution

Problem Info

LeetCode #	217 — Contains Duplicate
Difficulty	Easy
Topic	Arrays, HashSet

Given an integer array nums, return true if any value appears at least twice.

Example:

Input:  nums = [1, 2, 3, 1]
Output: true

Input:  nums = [1, 2, 3, 4]
Output: false

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Approach

Use a HashSet. For each number, check if it’s already in the set. If yes — duplicate found. If no — add it and continue.

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Kotlin Solution

fun containsDuplicate(nums: IntArray): Boolean {
    val seen = HashSet<Int>()

    for (num in nums) {
        if (!seen.add(num)) return true
    }

    return false
}

One-liner version (idiomatic Kotlin)

fun containsDuplicate(nums: IntArray): Boolean {
    return nums.size != nums.toHashSet().size
}

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Why Kotlin Shines Here

seen.add() returns Boolean — HashSet.add() returns false if the element already exists. We use that directly:

if (!seen.add(num)) return true
// No separate contains() check needed

toHashSet() — Kotlin’s collection extension makes the one-liner possible:

nums.toHashSet() // converts IntArray to HashSet<Int> in one call

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Two Approaches Compared

Approach	Time	Space	Notes
HashSet with early exit	O(n)	O(n)	Stops as soon as duplicate found
One-liner toHashSet()	O(n)	O(n)	Clean but no early exit

Use the first approach in interviews — shows you understand early termination.

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Complexity

Time	O(n)
Space	O(n)

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Key Takeaway

HashSet add() returns false on duplicates — use that return value directly instead of a separate contains() check.

🔗 Solve it on LeetCode →

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📚 Kotlin DSA Series

This post is part of the Kotlin DSA series — solving LeetCode problems using idiomatic Kotlin.

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