README_EN.md

16.21. Sum Swap

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Description

Given two arrays of integers, find a pair of values (one value from each array) that you can swap to give the two arrays the same sum.

Return an array, where the first element is the element in the first array that will be swapped, and the second element is another one in the second array. If there are more than one answers, return any one of them. If there is no answer, return an empty array.

Example:

Input: array1 = [4, 1, 2, 1, 1, 2], array2 = [3, 6, 3, 3]

Output: [1, 3]

Example:

Input: array1 = [1, 2, 3], array2 = [4, 5, 6]

Output: []

Note:

• 1 <= array1.length, array2.length <= 100000

Solutions

Solution 1: Hash Table

We first calculate the sum of the two arrays, and then calculate the difference $diff$ between the sums. If $diff$ is odd, it means that the sums of the two arrays cannot be equal, so we directly return an empty array.

If $diff$ is even, we can traverse one of the arrays. Suppose the current element being traversed is $a$, then we need to find an element $b$ in the other array such that $a - b = diff / 2$, i.e., $b = a - diff / 2$. We can use a hash table to quickly check whether $b$ exists. If it exists, it means that we have found a pair of elements that meet the conditions, and we can return them directly.

The time complexity is $O(m + n)$, and the space complexity is $O(n)$. Here, $m$ and $n$ are the lengths of the two arrays.

class Solution:
    def findSwapValues(self, array1: List[int], array2: List[int]) -> List[int]:
        diff = sum(array1) - sum(array2)
        if diff & 1:
            return []
        diff >>= 1
        s = set(array2)
        for a in array1:
            if (b := (a - diff)) in s:
                return [a, b]
        return []

class Solution {
    public int[] findSwapValues(int[] array1, int[] array2) {
        long s1 = 0, s2 = 0;
        Set<Integer> s = new HashSet<>();
        for (int x : array1) {
            s1 += x;
        }
        for (int x : array2) {
            s2 += x;
            s.add(x);
        }
        long diff = s1 - s2;
        if (diff % 2 != 0) {
            return new int[0];
        }
        diff /= 2;
        for (int a : array1) {
            int b = (int) (a - diff);
            if (s.contains(b)) {
                return new int[] {a, b};
            }
        }
        return new int[0];
    }
}

class Solution {
public:
    vector<int> findSwapValues(vector<int>& array1, vector<int>& array2) {
        long long s1 = accumulate(array1.begin(), array1.end(), 0LL);
        long long s2 = accumulate(array2.begin(), array2.end(), 0LL);
        long long diff = s1 - s2;
        if (diff & 1) {
            return {};
        }
        diff >>= 1;
        unordered_set<int> s(array2.begin(), array2.end());
        for (int x : array1) {
            int y = x - diff;
            if (s.count(y)) {
                return {x, y};
            }
        }
        return {};
    }
};

func findSwapValues(array1 []int, array2 []int) []int {
	s1, s2 := 0, 0
	s := map[int]bool{}
	for _, a := range array1 {
		s1 += a
	}
	for _, b := range array2 {
		s2 += b
		s[b] = true
	}
	diff := s1 - s2
	if (diff & 1) == 1 {
		return []int{}
	}
	diff >>= 1
	for _, a := range array1 {
		if b := a - diff; s[b] {
			return []int{a, b}
		}
	}
	return []int{}
}

function findSwapValues(array1: number[], array2: number[]): number[] {
    const s1 = array1.reduce((a, b) => a + b, 0);
    const s2 = array2.reduce((a, b) => a + b, 0);
    let diff = s1 - s2;
    if (diff & 1) {
        return [];
    }
    diff >>= 1;
    const s: Set<number> = new Set(array2);
    for (const x of array1) {
        const y = x - diff;
        if (s.has(y)) {
            return [x, y];
        }
    }
    return [];
}