Arrays: Best Time to Buy and Sell Stock — Kotlin Solution

Problem Info

LeetCode #	121 — Best Time to Buy and Sell Stock
Difficulty	Easy
Topic	Arrays, Greedy, Sliding Window

Given an array prices where prices[i] is the price on day i, return the maximum profit from a single buy-sell transaction.

Example:

Input:  prices = [7, 1, 5, 3, 6, 4]
Output: 5  // Buy at 1, sell at 6

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Approach

We can’t sell before we buy — so we track the minimum price seen so far and at each step compute potential profit.

No nested loops needed. Single pass.

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Kotlin Solution

fun maxProfit(prices: IntArray): Int {
    var minPrice = Int.MAX_VALUE
    var maxProfit = 0

    for (price in prices) {
        if (price < minPrice) {
            minPrice = price
        } else {
            maxProfit = maxOf(maxProfit, price - minPrice)
        }
    }

    return maxProfit
}

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Why Kotlin Shines Here

maxOf() — cleaner than Math.max():

maxProfit = maxOf(maxProfit, price - minPrice)
// vs Java: maxProfit = Math.max(maxProfit, price - minPrice);

for (price in prices) — no index needed when you don’t need it:

for (price in prices)
// vs Java: for (int price : prices)

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Complexity

Time	O(n) — single pass through prices
Space	O(1) — only two variables

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Key Takeaway

Track the minimum price seen so far. At each step, profit = current price − minimum so far. Keep the max of all profits.

🔗 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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