#3131

Easy

auto_awesomeArray-driven solution strategy

LeetCode Problem Workspace

Find the Integer Added to Array I

Find the integer added to nums1 to make it equal to nums2 using an array-driven strategy.

Problem Statement

You are given two arrays, nums1 and nums2, both of equal length. The elements of nums1 have been altered by adding (or subtracting) the same integer, x, to each element. As a result, nums1 becomes identical to nums2.

Your task is to determine the integer, x, that was added to every element of nums1 to transform it into nums2. Both arrays will be of the same length, and the transformation always results in an integer x that makes nums1 and nums2 equal.

Examples

Example 1

Input: nums1 = [2,6,4], nums2 = [9,7,5]

Output: 3

The integer added to each element of nums1 is 3.

Example 2

Input: nums1 = [10], nums2 = [5]

Output: -5

The integer added to each element of nums1 is -5.

Example 3

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

Output: 0

The integer added to each element of nums1 is 0.

Constraints

1 <= nums1.length == nums2.length <= 100
0 <= nums1[i], nums2[i] <= 1000
The test cases are generated in a way that there is an integer x such that nums1 can become equal to nums2 by adding x to each element of nums1.

Solution Approach

Sort and Compare

By sorting both nums1 and nums2, you can easily find the integer added by comparing corresponding elements in the sorted arrays. The difference between any pair of elements from nums1 and nums2 will give you the integer x.

Efficient Element Matching

After sorting, simply subtract each element in nums1 from the corresponding element in nums2. The result will be the same for every index, revealing the value of x that was added to all elements of nums1.

Edge Case Handling

Ensure that you handle cases where nums1 and nums2 are identical without requiring any transformation, in which case the integer x will be 0.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is O(n log n) due to the sorting of both arrays. Space complexity is O(1) if in-place sorting is used, or O(n) if additional space is required for sorting.

What Interviewers Usually Probe

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

Common Pitfalls or Variants

Common pitfalls

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

Follow-up variants

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

FAQ

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.

terminal

Solution

Solution 1: Calculate Minimum Difference

We can find the minimum value of each array, then return the difference between the two minimum values.

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

Continue Practicing

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