LeetCode #1833 — MEDIUM

Maximum Ice Cream Bars

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

It is a sweltering summer day, and a boy wants to buy some ice cream bars.

At the store, there are n ice cream bars. You are given an array costs of length n, where costs[i] is the price of the i^th ice cream bar in coins. The boy initially has coins coins to spend, and he wants to buy as many ice cream bars as possible.

Note: The boy can buy the ice cream bars in any order.

Return the maximum number of ice cream bars the boy can buy with coins coins.

You must solve the problem by counting sort.

Example 1:

Input: costs = [1,3,2,4,1], coins = 7
Output: 4
Explanation: The boy can buy ice cream bars at indices 0,1,2,4 for a total price of 1 + 3 + 2 + 1 = 7.

Example 2:

Input: costs = [10,6,8,7,7,8], coins = 5
Output: 0
Explanation: The boy cannot afford any of the ice cream bars.

Example 3:

Input: costs = [1,6,3,1,2,5], coins = 20
Output: 6
Explanation: The boy can buy all the ice cream bars for a total price of 1 + 6 + 3 + 1 + 2 + 5 = 18.

Constraints:

costs.length == n
1 <= n <= 10⁵
1 <= costs[i] <= 10⁵
1 <= coins <= 10⁸

Step 01

Brute Force Baseline

Problem summary: It is a sweltering summer day, and a boy wants to buy some ice cream bars. At the store, there are n ice cream bars. You are given an array costs of length n, where costs[i] is the price of the ith ice cream bar in coins. The boy initially has coins coins to spend, and he wants to buy as many ice cream bars as possible. Note: The boy can buy the ice cream bars in any order. Return the maximum number of ice cream bars the boy can buy with coins coins. You must solve the problem by counting sort.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Greedy

Example 1

[1,3,2,4,1]
7

Example 2

[10,6,8,7,7,8]
5

Example 3

[1,6,3,1,2,5]
20

Step 02

Core Insight

What unlocks the optimal approach

It is always optimal to buy the least expensive ice cream bar first.
Sort the prices so that the cheapest ice cream bar comes first.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1833: Maximum Ice Cream Bars
class Solution {
    public int maxIceCream(int[] costs, int coins) {
        Arrays.sort(costs);
        int n = costs.length;
        for (int i = 0; i < n; ++i) {
            if (coins < costs[i]) {
                return i;
            }
            coins -= costs[i];
        }
        return n;
    }
}

// Accepted solution for LeetCode #1833: Maximum Ice Cream Bars
func maxIceCream(costs []int, coins int) int {
	sort.Ints(costs)
	for i, c := range costs {
		if coins < c {
			return i
		}
		coins -= c
	}
	return len(costs)
}

# Accepted solution for LeetCode #1833: Maximum Ice Cream Bars
class Solution:
    def maxIceCream(self, costs: List[int], coins: int) -> int:
        costs.sort()
        for i, c in enumerate(costs):
            if coins < c:
                return i
            coins -= c
        return len(costs)

// Accepted solution for LeetCode #1833: Maximum Ice Cream Bars
struct Solution;

impl Solution {
    fn max_ice_cream(mut costs: Vec<i32>, mut coins: i32) -> i32 {
        costs.sort_unstable();
        costs.reverse();
        let mut res = 0;
        while coins > 0 {
            if let Some(cheapest) = costs.pop() {
                if cheapest <= coins {
                    coins -= cheapest;
                    res += 1;
                } else {
                    break;
                }
            } else {
                break;
            }
        }
        res
    }
}

#[test]
fn test() {
    let costs = vec![1, 3, 2, 4, 1];
    let coins = 7;
    let res = 4;
    assert_eq!(Solution::max_ice_cream(costs, coins), res);

    let costs = vec![10, 6, 8, 7, 7, 8];
    let coins = 5;
    let res = 0;
    assert_eq!(Solution::max_ice_cream(costs, coins), res);

    let costs = vec![1, 6, 3, 1, 2, 5];
    let coins = 20;
    let res = 6;
    assert_eq!(Solution::max_ice_cream(costs, coins), res);
}

// Accepted solution for LeetCode #1833: Maximum Ice Cream Bars
function maxIceCream(costs: number[], coins: number): number {
    costs.sort((a, b) => a - b);
    const n = costs.length;
    for (let i = 0; i < n; ++i) {
        if (coins < costs[i]) {
            return i;
        }
        coins -= costs[i];
    }
    return n;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(log n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.