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Formatted question description: https://leetcode.ca/all/888.html

# 888. Fair Candy Swap (Easy)

Alice and Bob have candy bars of different sizes: A[i] is the size of the i-th bar of candy that Alice has, and B[j] is the size of the j-th bar of candy that Bob has.

Since they are friends, they would like to exchange one candy bar 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 sizes of candy bars they have.)

Return an integer array ans where ans[0] is the size of the candy bar that Alice must exchange, and ans[1] is the size of the candy bar that Bob must exchange.

If there are multiple answers, you may return any one of them.  It is guaranteed an answer exists.

Example 1:

Input: A = [1,1], B = [2,2]
Output: [1,2]


Example 2:

Input: A = [1,2], B = [2,3]
Output: [1,2]


Example 3:

Input: A = [2], B = [1,3]
Output: [2,3]


Example 4:

Input: A = [1,2,5], B = [2,4]
Output: [5,4]


Note:

• 1 <= A.length <= 10000
• 1 <= B.length <= 10000
• 1 <= A[i] <= 100000
• 1 <= B[i] <= 100000
• It is guaranteed that Alice and Bob have different total amounts of candy.
• It is guaranteed there exists an answer.

Companies:
Fidessa

Related Topics:
Array

## Solution 1.

• class Solution {
public int[] fairCandySwap(int[] A, int[] B) {
int sumA = 0, sumB = 0;
for (int num : A)
sumA += num;
for (int num : B)
sumB += num;
int dif = (sumA - sumB) / 2;
int[] swap = new int[2];
for (int numA : A) {
for (int numB : B) {
if (numA - numB == dif) {
swap[0] = numA;
swap[1] = numB;
return swap;
}
}
}
return 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) {
s2 += b;
}
int diff = (s1 - s2) >> 1;
for (int a : aliceSizes) {
int target = a - diff;
if (s.contains(target)) {
return new int[] {a, target};
}
}
return null;
}
}

• // OJ: https://leetcode.com/problems/fair-candy-swap/
// Time: O(A+B)
// Space: O(B)
class Solution {
public:
vector<int> fairCandySwap(vector<int>& A, vector<int>& B) {
int s1 = accumulate(A.begin(), A.end(), 0);
int s2 = accumulate(B.begin(), B.end(), 0);
int d = (s1 - s2) / 2;
unordered_set<int> s;
for (int b : B) s.insert(b);
for (int a : A) {
if (s.find(a - d) != s.end()) return { a, a - d };
}
}
};

• 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:
target = a - diff
if target in s:
return [a, target]

############

class Solution(object):
def fairCandySwap(self, A, B):
"""
:type A: List[int]
:type B: List[int]
:rtype: List[int]
"""
sum_A, sum_B, set_B = sum(A), sum(B), set(B)
target = (sum_A + sum_B) // 2
for a in A:
expect_b = target - (sum_A - a)
if expect_b in set_B:
return [a, expect_b]
return []

• function fairCandySwap(aliceSizes: number[], bobSizes: number[]): number[] {
let s1 = aliceSizes.reduce((a, c) => a + c, 0);
let s2 = bobSizes.reduce((a, c) => a + c, 0);
let diff = (s1 - s2) >> 1;
for (let num of aliceSizes) {
let target = num - diff;
if (bobSizes.includes(target)) {
return [num, target];
}
}
}