#3132

Medium

auto_awesomeTwo-pointer scanning with invariant tracking

LeetCode Problem Workspace

Find the Integer Added to Array II

Given two arrays nums1 and nums2, determine the integer added to nums1 to make it equal to nums2 after removing two elements.

array two pointers sorting enumeration

Problem Statement

You are given two integer arrays, nums1 and nums2. In nums1, two elements have been removed, and the remaining elements have been increased (or decreased for negative values) by a single integer, x. This transformation results in nums1 becoming identical to nums2.

Your task is to find the value of x, the integer added to nums1. You are guaranteed that nums2 is two elements shorter than nums1, and that there is an integer x such that nums1 can be transformed into nums2 by removing two elements and adding x to each of the remaining elements.

Examples

Example 1

Input: nums1 = [4,20,16,12,8], nums2 = [14,18,10]

Output: -2

After removing elements at indices [0,4] and adding -2, nums1 becomes [18,14,10] .

Example 2

Input: nums1 = [3,5,5,3], nums2 = [7,7]

Output: 2

After removing elements at indices [0,3] and adding 2, nums1 becomes [7,7] .

Constraints

3 <= nums1.length <= 200
nums2.length == nums1.length - 2
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 removing two elements and adding x to each element of nums1.

Solution Approach

Two-Pointer Scanning

Use a two-pointer technique to scan both arrays. The first pointer will iterate through nums1 while comparing it to nums2, while the second pointer will track possible positions for the removed elements. Keep track of the invariant of matching elements in nums2 as you adjust for potential removals.

Invariant Tracking

Ensure that the invariant holds throughout the process: after removal and adjustment by x, nums1 must match nums2. By tracking the elements that must remain and adjusting the difference between nums1 and nums2, you can identify the correct value for x.

Enumeration of Possibilities

Try all possible combinations of removing two elements from nums1, adjusting the remaining elements, and comparing with nums2. This exhaustive approach guarantees that the correct transformation is found.

Complexity Analysis

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

The time complexity depends on the method of element removal and the scanning strategy, but typically, a two-pointer technique or enumeration of possibilities can achieve a time complexity of O(n^2). Space complexity is O(1) if in-place modifications are used.

What Interviewers Usually Probe

Can the candidate optimize the removal process by efficiently handling the two-element removal?
Does the candidate consider both removal strategies and invariant tracking?
How does the candidate handle edge cases like multiple equal elements in nums1 and nums2?

Common Pitfalls or Variants

Common pitfalls

Failing to consider all possibilities of element removal and assuming the answer from a single scenario.
Not correctly updating the value of x after adjusting for the removed elements.
Misunderstanding the conditions under which nums1 is transformed into nums2, leading to incorrect assumptions about element removal.

Follow-up variants

Variation in array size or element values, adjusting complexity for larger or smaller arrays.
Changes to the number of elements removed from nums1, modifying the number of potential scenarios to consider.
Introducing additional constraints that may change the solution's approach, such as allowing for negative values in nums1 and nums2.

FAQ

What is the main pattern in the "Find the Integer Added to Array II" problem?

The main pattern is two-pointer scanning with invariant tracking to identify the integer added to nums1 after removing two elements.

How can I solve "Find the Integer Added to Array II" more efficiently?

Focus on efficiently removing two elements from nums1 while tracking the transformation to match nums2, minimizing unnecessary recalculations.

What if nums1 and nums2 contain repeated elements?

Even with repeated elements, the approach of removing two elements and adjusting by a constant integer x remains valid, as long as invariant tracking is maintained.

How does GhostInterview help with this problem?

GhostInterview helps by guiding you through optimization strategies and invariant tracking, ensuring a thorough exploration of all removal possibilities.

Can the "Find the Integer Added to Array II" problem be solved in linear time?

While linear time might not be achievable for all cases, using efficient two-pointer scanning can significantly reduce unnecessary work, bringing the solution closer to O(n).

terminal

Solution

Solution 1: Sorting + Enumeration + Two Pointers

First, we sort the arrays $nums1$ and $nums2$. Since we need to remove two elements from $nums1$, we only need to consider the first three elements of $nums1$, denoted as $a_1, a_2, a_3$. We can enumerate the first element $b_1$ of $nums2$, then we can get $x = b_1 - a_i$, where $i \in \{1, 2, 3\}$. Then we can use the two pointers method to determine whether there exists an integer $x$ that makes $nums1$ and $nums2$ equal, and take the smallest $x$ that satisfies the condition.

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class Solution:
    def minimumAddedInteger(self, nums1: List[int], nums2: List[int]) -> int:
        def f(x: int) -> bool:
            i = j = cnt = 0
            while i < len(nums1) and j < len(nums2):
                if nums2[j] - nums1[i] != x:
                    cnt += 1
                else:
                    j += 1
                i += 1
            return cnt <= 2

        nums1.sort()
        nums2.sort()
        ans = inf
        for i in range(3):
            x = nums2[0] - nums1[i]
            if f(x):
                ans = min(ans, x)
        return ans

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