LeetCode #888 — EASY

Fair Candy Swap

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

Alice and Bob have a different total number of candies. You are given two integer arrays aliceSizes and bobSizes where aliceSizes[i] is the number of candies of the i^th box of candy that Alice has and bobSizes[j] is the number of candies of the j^th box of candy that Bob has.

Since they are friends, they would like to exchange one candy box each so that after the exchange, they both have the same total amount of candy. The total amount of candy a person has is the sum of the number of candies in each box they have.

Return an integer array answer where answer[0] is the number of candies in the box that Alice must exchange, and answer[1] is the number of candies in the box that Bob must exchange. If there are multiple answers, you may return any one of them. It is guaranteed that at least one answer exists.

Example 1:

Input: aliceSizes = [1,1], bobSizes = [2,2]
Output: [1,2]

Example 2:

Input: aliceSizes = [1,2], bobSizes = [2,3]
Output: [1,2]

Example 3:

Input: aliceSizes = [2], bobSizes = [1,3]
Output: [2,3]

Constraints:

1 <= aliceSizes.length, bobSizes.length <= 10⁴
1 <= aliceSizes[i], bobSizes[j] <= 10⁵
Alice and Bob have a different total number of candies.
There will be at least one valid answer for the given input.

Step 01

Brute Force Baseline

Problem summary: Alice and Bob have a different total number of candies. You are given two integer arrays aliceSizes and bobSizes where aliceSizes[i] is the number of candies of the ith box of candy that Alice has and bobSizes[j] is the number of candies of the jth box of candy that Bob has. Since they are friends, they would like to exchange one candy box each so that after the exchange, they both have the same total amount of candy. The total amount of candy a person has is the sum of the number of candies in each box they have. Return an integer array answer where answer[0] is the number of candies in the box that Alice must exchange, and answer[1] is the number of candies in the box that Bob must exchange. If there are multiple answers, you may return any one of them. It is guaranteed that at least one answer exists.

Baseline thinking

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

Pattern signal: Array · Hash Map · Binary Search

Example 1

[1,1]
[2,2]

Example 2

[1,2]
[2,3]

Example 3

[2]
[1,3]

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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 #888: Fair Candy Swap
class Solution {
    public int[] fairCandySwap(int[] aliceSizes, int[] bobSizes) {
        int s1 = 0, s2 = 0;
        Set<Integer> s = new HashSet<>();
        for (int a : aliceSizes) {
            s1 += a;
        }
        for (int b : bobSizes) {
            s.add(b);
            s2 += b;
        }
        int diff = (s1 - s2) >> 1;
        for (int a : aliceSizes) {
            int b = a - diff;
            if (s.contains(b)) {
                return new int[] {a, b};
            }
        }
        return null;
    }
}

// Accepted solution for LeetCode #888: Fair Candy Swap
func fairCandySwap(aliceSizes []int, bobSizes []int) []int {
	s1, s2 := 0, 0
	s := map[int]bool{}
	for _, a := range aliceSizes {
		s1 += a
	}
	for _, b := range bobSizes {
		s2 += b
		s[b] = true
	}
	diff := (s1 - s2) / 2
	for _, a := range aliceSizes {
		if b := a - diff; s[b] {
			return []int{a, b}
		}
	}
	return nil
}

# Accepted solution for LeetCode #888: Fair Candy Swap
class Solution:
    def fairCandySwap(self, aliceSizes: List[int], bobSizes: List[int]) -> List[int]:
        diff = (sum(aliceSizes) - sum(bobSizes)) >> 1
        s = set(bobSizes)
        for a in aliceSizes:
            if (b := (a - diff)) in s:
                return [a, b]

// Accepted solution for LeetCode #888: Fair Candy Swap
struct Solution;

use std::collections::HashSet;

impl Solution {
    fn fair_candy_swap(a: Vec<i32>, b: Vec<i32>) -> Vec<i32> {
        let sum_a: i32 = a.iter().sum();
        let sum_b: i32 = b.iter().sum();
        let hs: HashSet<i32> = b.into_iter().collect();
        for x in a {
            let y = x + (sum_b - sum_a) / 2;
            if hs.contains(&y) {
                return vec![x, y];
            }
        }
        unreachable!();
    }
}

#[test]
fn test() {
    let a = vec![1, 2];
    let b = vec![2, 3];
    let res = vec![1, 2];
    assert_eq!(Solution::fair_candy_swap(a, b), res);
    let a = vec![2];
    let b = vec![1, 3];
    let res = vec![2, 3];
    assert_eq!(Solution::fair_candy_swap(a, b), res);
    let a = vec![1, 2, 5];
    let b = vec![2, 4];
    let res = vec![5, 4];
    assert_eq!(Solution::fair_candy_swap(a, b), res);
}

// Accepted solution for LeetCode #888: Fair Candy Swap
function fairCandySwap(aliceSizes: number[], bobSizes: number[]): number[] {
    const s1 = aliceSizes.reduce((acc, cur) => acc + cur, 0);
    const s2 = bobSizes.reduce((acc, cur) => acc + cur, 0);
    const diff = (s1 - s2) >> 1;
    const s = new Set(bobSizes);
    for (const a of aliceSizes) {
        const b = a - diff;
        if (s.has(b)) {
            return [a, b];
        }
    }
    return [];
}

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(m + n)

Space

O(n)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.