Arrays: Two Sum — Kotlin Solution

Problem Info

LeetCode #	1 — Two Sum
Difficulty	Easy
Topic	Arrays, HashMap

Given an array of integers nums and an integer target, return the indices of the two numbers that add up to target.

Example:

Input:  nums = [2, 7, 11, 15], target = 9
Output: [0, 1]  // nums[0] + nums[1] = 2 + 7 = 9

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Approach

The brute force way is O(n²) — check every pair. We can do better.

Key insight: For each number x, we need target - x. Instead of searching for it, store every number we’ve seen in a HashMap. Then for each new number, just check if its complement already exists.

One pass. O(n) time.

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Kotlin Solution

fun twoSum(nums: IntArray, target: Int): IntArray {
    val seen = HashMap<Int, Int>() // value -> index

    for ((index, num) in nums.withIndex()) {
        val complement = target - num

        if (complement in seen) {
            return intArrayOf(seen[complement]!!, index)
        }

        seen[num] = index
    }

    return intArrayOf() // no solution found
}

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Why Kotlin Shines Here

withIndex() — clean destructuring of index and value, no manual i counter:

for ((index, num) in nums.withIndex())
// vs Java: for (int i = 0; i < nums.length; i++)

in operator — readable HashMap lookup:

if (complement in seen)
// vs Java: if (seen.containsKey(complement))

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Complexity

Time	O(n) — single pass
Space	O(n) — HashMap stores up to n elements

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Key Takeaway

Store what you’ve seen. Check if the complement exists. One pass beats two loops every time.

🔗 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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