What is the main approach to solving 'Find the Integer Added to Array I'?

The main approach is sorting both arrays and comparing their corresponding elements to determine the integer added to nums1.

How do I handle edge cases where nums1 and nums2 are identical?

If nums1 and nums2 are identical, the integer added is 0, and no transformation is needed.

Can the solution be optimized for larger arrays?

Yes, by utilizing in-place sorting, you can reduce the space complexity. However, sorting will still have a time complexity of O(n log n).

What should I focus on to avoid common pitfalls in this problem?

Ensure you handle edge cases where the arrays are already equal and focus on sorting to simplify the solution.

How does GhostInterview assist with solving this problem?

GhostInterview helps you understand the array-driven solution strategy and guides you through potential pitfalls, ensuring a clear and efficient solution.

#3131

Easy

auto_awesome数组·driven

LeetCode 题解工作台

找出与数组相加的整数 I

给你两个长度相等的数组 nums1 和 nums2 。数组 nums1 中的每个元素都与变量 x 所表示的整数相加。如果 x 为负数，则表现为元素值的减少。在与 x 相加后， nums1 和 nums2 相等。当两个数组中包含相同的整数，并且这些整数出现的频次相同时，两个数组相等。返回整…

数组

题目描述

给你两个长度相等的数组 nums1 和 nums2。

数组 nums1 中的每个元素都与变量 x 所表示的整数相加。如果 x 为负数，则表现为元素值的减少。

在与 x 相加后，nums1 和 nums2 相等。当两个数组中包含相同的整数，并且这些整数出现的频次相同时，两个数组相等。

返回整数 x 。

示例 1:

输入：nums1 = [2,6,4], nums2 = [9,7,5]

输出：3

解释：

与 3 相加后，nums1 和 nums2 相等。

示例 2:

输入：nums1 = [10], nums2 = [5]

输出：-5

解释：

与 -5 相加后，nums1 和 nums2 相等。

示例 3:

输入：nums1 = [1,1,1,1], nums2 = [1,1,1,1]

输出：0

解释：

与 0 相加后，nums1 和 nums2 相等。

提示：

1 <= nums1.length == nums2.length <= 100
0 <= nums1[i], nums2[i] <= 1000
测试用例以这样的方式生成：存在一个整数 x，使得 nums1 中的每个元素都与 x 相加后，nums1 与 nums2 相等。

lightbulb

解题思路

方法一：求最小值差

我们可以分别求出两个数组的最小值，然后返回两个最小值的差值即可。

时间复杂度 $O(n)$ ，其中 $n$ 为数组的长度。空间复杂度 $O(1)$ 。

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class Solution:
    def addedInteger(self, nums1: List[int], nums2: List[int]) -> int:
        return min(nums2) - min(nums1)

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
The candidate demonstrates proficiency with sorting techniques in array-based problems.
question_mark
The candidate is able to efficiently identify and leverage the one-to-one correspondence between sorted arrays.
question_mark
The candidate handles edge cases, such as arrays being identical, with clarity and correctness.

warning

常见陷阱

外企场景

error
Failing to account for edge cases, such as when nums1 and nums2 are already identical.
error
Overcomplicating the solution with unnecessary steps instead of using sorting for direct comparison.
error
Incorrectly assuming a transformation is not possible when nums1 is already equal to nums2.

swap_horiz

进阶变体

外企场景

arrow_right_alt
What happens if the arrays are very large and sorting takes longer than expected?
arrow_right_alt
How would this solution change if the transformation was a more complex operation, not just addition?
arrow_right_alt
Can this solution be adapted for arrays with non-integer elements?

help

常见问题

外企场景

继续练习

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