What is the best method for finding the minimum common value?

Using a hash set to check existence or a two-pointer scan on sorted arrays ensures the minimum common value is found efficiently.

Can duplicates affect the minimum common value result?

No, duplicates do not change the result; only the smallest integer present in both arrays matters.

How does the two-pointer approach work?

Pointers start at the beginning of each sorted array, advancing the smaller value until a match is found, which is the minimum common value.

Is binary search faster than a hash set in this problem?

Binary search is O(n log m) and can be efficient when one array is much smaller, but a hash set or two-pointer scan is often simpler and faster in practice.

Does GhostInterview guide me through array scanning plus hash lookup patterns?

Yes, it provides detailed step guidance, highlights pitfalls, and ensures your solution handles edge cases effectively.

#2540

Easy

auto_awesome数组·哈希·扫描

LeetCode 题解工作台

最小公共值

给你两个整数数组 nums1 和 nums2 ，它们已经按非降序排序，请你返回两个数组的最小公共整数。如果两个数组 nums1 和 nums2 没有公共整数，请你返回 -1 。如果一个整数在两个数组中都至少出现一次，那么这个整数是数组 nums1 和 nums2 公共的。示例 1：输…

数组哈希表双指针二分查找

题目描述

给你两个整数数组 nums1 和 nums2 ，它们已经按非降序排序，请你返回两个数组的 最小公共整数 。如果两个数组 nums1 和 nums2 没有公共整数，请你返回 -1 。

如果一个整数在两个数组中都 至少出现一次 ，那么这个整数是数组 nums1 和 nums2 公共的。

示例 1：

输入：nums1 = [1,2,3], nums2 = [2,4]
输出：2
解释：两个数组的最小公共元素是 2 ，所以我们返回 2 。

示例 2：

输入：nums1 = [1,2,3,6], nums2 = [2,3,4,5]
输出：2
解释：两个数组中的公共元素是 2 和 3 ，2 是较小值，所以返回 2 。

提示：

1 <= nums1.length, nums2.length <= 10⁵
1 <= nums1[i], nums2[j] <= 10⁹
nums1 和 nums2 都是 非降序 的。

lightbulb

解题思路

方法一：双指针

遍历两个数组，如果两个指针指向的元素相等，则返回该元素；如果两个指针指向的元素不相等，则将指向较小元素的指针右移一位，直到找到相等的元素或者遍历完数组。

时间复杂度 $O(m + n)$ ，其中 $m$ 和 $n$ 分别是两个数组的长度。空间复杂度 $O(1)$ 。

1

2

3

4

5

6

7

8

9

10

11

12

13

class Solution:
    def getCommon(self, nums1: List[int], nums2: List[int]) -> int:
        i = j = 0
        m, n = len(nums1), len(nums2)
        while i < m and j < n:
            if nums1[i] == nums2[j]:
                return nums1[i]
            if nums1[i] < nums2[j]:
                i += 1
            else:
                j += 1
        return -1

speed

复杂度分析

指标	值
时间	complexity is O(n log m) when using binary search or O(n + m) for a hash set or two-pointer method. Space complexity is O(1) for two-pointer scan or O(n) if a set is used.
空间	O(1)

psychology

面试官常问的追问

外企场景

question_mark
Do you recognize the sorted array property that allows two-pointer optimization?
question_mark
Can you identify when a hash set reduces lookup time versus scanning?
question_mark
How do you handle arrays with duplicates without affecting the minimum common value?

warning

常见陷阱

外企场景

error
Returning the first duplicate match instead of the minimum value.
error
Using nested loops leading to O(n*m) time instead of O(n + m) or O(n log m).
error
Neglecting edge cases when one array has no common elements.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Find all common values instead of only the minimum.
arrow_right_alt
Return the maximum common value between two arrays.
arrow_right_alt
Handle arrays that are not sorted and compare trade-offs in approach.

help

常见问题

外企场景

继续练习

#2856 删除数对后的最小数组长度 #3186 施咒的最大总伤害 #1782 统计点对的数目