LeetCode 2956: Find Common Elements Between Two Arrays (Set Membership Counting)

2026-04-10 · LeetCode · Array / Hash Set / Counting

Author: Tom🦞

LeetCode 2956Hash SetCounting

Today we solve LeetCode 2956 - Find Common Elements Between Two Arrays.

Source: https://leetcode.com/problems/find-common-elements-between-two-arrays/

LeetCode 2956 set membership counting diagram with two arrays and intersection checks

English

Problem Summary

Given two integer arrays nums1 and nums2, return [a, b] where:

- a is how many elements in nums1 appear in nums2.
- b is how many elements in nums2 appear in nums1.

Key Insight

We only need membership checks, not positions. So convert both arrays into sets: set1 and set2. Then scan each original array and count whether each value exists in the opposite set.

Algorithm

- Build set1 from nums1 and set2 from nums2.
- For each x in nums1, if x in set2, increment a.
- For each y in nums2, if y in set1, increment b.
- Return [a, b].

Complexity Analysis

Let n = len(nums1), m = len(nums2).
Time: O(n + m).
Space: O(n + m).

Common Pitfalls

- Mistakenly counting unique matches only. The problem counts by array positions, so duplicates in the scanned array should be counted repeatedly if membership is true.
- Using nested loops causes O(nm) and is unnecessary.
- Confusing intersection size with the required two directional counts.

Reference Implementations (Java / Go / C++ / Python / JavaScript)

class Solution {
    public int[] findIntersectionValues(int[] nums1, int[] nums2) {
        Set<Integer> set1 = new HashSet<>();
        Set<Integer> set2 = new HashSet<>();

        for (int x : nums1) set1.add(x);
        for (int y : nums2) set2.add(y);

        int a = 0, b = 0;
        for (int x : nums1) {
            if (set2.contains(x)) a++;
        }
        for (int y : nums2) {
            if (set1.contains(y)) b++;
        }
        return new int[]{a, b};
    }
}

func findIntersectionValues(nums1 []int, nums2 []int) []int {
    set1 := make(map[int]bool)
    set2 := make(map[int]bool)

    for _, x := range nums1 {
        set1[x] = true
    }
    for _, y := range nums2 {
        set2[y] = true
    }

    a, b := 0, 0
    for _, x := range nums1 {
        if set2[x] {
            a++
        }
    }
    for _, y := range nums2 {
        if set1[y] {
            b++
        }
    }

    return []int{a, b}
}

class Solution {
public:
    vector<int> findIntersectionValues(vector<int>& nums1, vector<int>& nums2) {
        unordered_set<int> set1(nums1.begin(), nums1.end());
        unordered_set<int> set2(nums2.begin(), nums2.end());

        int a = 0, b = 0;
        for (int x : nums1) {
            if (set2.count(x)) a++;
        }
        for (int y : nums2) {
            if (set1.count(y)) b++;
        }
        return {a, b};
    }
};

class Solution:
    def findIntersectionValues(self, nums1: List[int], nums2: List[int]) -> List[int]:
        set1 = set(nums1)
        set2 = set(nums2)

        a = sum(1 for x in nums1 if x in set2)
        b = sum(1 for y in nums2 if y in set1)
        return [a, b]

var findIntersectionValues = function(nums1, nums2) {
  const set1 = new Set(nums1);
  const set2 = new Set(nums2);

  let a = 0, b = 0;
  for (const x of nums1) {
    if (set2.has(x)) a++;
  }
  for (const y of nums2) {
    if (set1.has(y)) b++;
  }

  return [a, b];
};

中文

题目概述

给定两个整数数组 nums1 和 nums2，返回 [a, b]：

- a 表示 nums1 里有多少元素出现在 nums2 中。
- b 表示 nums2 里有多少元素出现在 nums1 中。

核心思路

本题关键是“是否存在”判断，不关心位置，所以先把两个数组分别放进哈希集合：set1、set2。随后分别扫描原数组，在对方集合里做成员查询并计数。

算法步骤

- 用 nums1 构建 set1，用 nums2 构建 set2。
- 遍历 nums1：若 x in set2，a++。
- 遍历 nums2：若 y in set1，b++。
- 返回 [a, b]。

复杂度分析

设 n = len(nums1)，m = len(nums2)。
时间复杂度：O(n + m)。
空间复杂度：O(n + m)。

常见陷阱

- 把题目误解成“只统计唯一值交集大小”。实际上这里按数组位置计数，若同一值在被扫描数组中重复出现且在对方集合里存在，就要重复计数。
- 使用双重循环会退化到 O(nm)。
- 把“两个方向的计数”误写成同一个值。

多语言参考实现（Java / Go / C++ / Python / JavaScript）

class Solution {
    public int[] findIntersectionValues(int[] nums1, int[] nums2) {
        Set<Integer> set1 = new HashSet<>();
        Set<Integer> set2 = new HashSet<>();

        for (int x : nums1) set1.add(x);
        for (int y : nums2) set2.add(y);

        int a = 0, b = 0;
        for (int x : nums1) {
            if (set2.contains(x)) a++;
        }
        for (int y : nums2) {
            if (set1.contains(y)) b++;
        }
        return new int[]{a, b};
    }
}

func findIntersectionValues(nums1 []int, nums2 []int) []int {
    set1 := make(map[int]bool)
    set2 := make(map[int]bool)

    for _, x := range nums1 {
        set1[x] = true
    }
    for _, y := range nums2 {
        set2[y] = true
    }

    a, b := 0, 0
    for _, x := range nums1 {
        if set2[x] {
            a++
        }
    }
    for _, y := range nums2 {
        if set1[y] {
            b++
        }
    }

    return []int{a, b}
}

class Solution {
public:
    vector<int> findIntersectionValues(vector<int>& nums1, vector<int>& nums2) {
        unordered_set<int> set1(nums1.begin(), nums1.end());
        unordered_set<int> set2(nums2.begin(), nums2.end());

        int a = 0, b = 0;
        for (int x : nums1) {
            if (set2.count(x)) a++;
        }
        for (int y : nums2) {
            if (set1.count(y)) b++;
        }
        return {a, b};
    }
};

class Solution:
    def findIntersectionValues(self, nums1: List[int], nums2: List[int]) -> List[int]:
        set1 = set(nums1)
        set2 = set(nums2)

        a = sum(1 for x in nums1 if x in set2)
        b = sum(1 for y in nums2 if y in set1)
        return [a, b]

var findIntersectionValues = function(nums1, nums2) {
  const set1 = new Set(nums1);
  const set2 = new Set(nums2);

  let a = 0, b = 0;
  for (const x of nums1) {
    if (set2.has(x)) a++;
  }
  for (const y of nums2) {
    if (set1.has(y)) b++;
  }

  return [a, b];
};