What is the time complexity of Find the Integer Added to Array II?

The optimal solution has a time complexity of O(n log n).

What data structures are used to solve Find the Integer Added to Array II?

This problem uses Array, Two Pointers, Sorting, Enumeration. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Find the Integer Added to Array II?

Insight 1: Sorting both arrays makes it easier to check for equality after applying the shift `x`. Sorting allows us to use a two-pointer approach. Insight 2: Since we remove only two elements, we can iterate through the first few elements of the sorted `nums1` and calculate the candidate `x` values based on the first element of `nums2`. This significantly reduces the search space for `x`.

Which companies ask LeetCode 3132. Find the Integer Added to Array II in interviews?

This problem is frequently asked at Mitsogo.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Find the Integer Added to Array II - LeetCode 3132 Solution

Find the Integer Added to Array II - Complete Solution Guide

Find the Integer Added to Array II is LeetCode problem 3132, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given two integer arrays nums1 and nums2 . From nums1 two elements have been removed, and all other elements have been increased (or decreased in the case of negative) by an integer, represented by the variable x . As a result, nums1 becomes equal to nums2 . Two arrays are considered equal when they contain the same integers with the same frequencies. Return the minimum possible integer x that achieves this equivalence. Example 1: Input: nums1 = [4,20,16,12,8], nums2 = [14,18,10] Output:

Detailed Explanation

The problem asks us to find the smallest integer `x` such that after removing two elements from `nums1` and adding `x` to all the remaining elements, the resulting array becomes equal to `nums2`. Two arrays are considered equal if they contain the same integers with the same frequencies. We're given that such an `x` always exists and `nums2` always has length two less than `nums1`.

Solution Approach

The provided solution first sorts both `nums1` and `nums2`. It then iterates through the first three elements of `nums1`, calculating a potential `x` value based on the difference between the first element of `nums2` and the current element of `nums1`. For each candidate `x`, a `check` function verifies if applying `x` to `nums1` (after removing two elements) results in `nums2`. The `check` function uses a two-pointer approach to efficiently compare the elements. The minimum `x` that satisfies the condition is returned.

Step-by-Step Algorithm

Step 1: Sort `nums1` and `nums2` in ascending order using a sorting algorithm (e.g., quicksort or mergesort).
Step 2: Initialize `min_x` to a large value (e.g., infinity or `Integer.MAX_VALUE`).
Step 3: Iterate through the first three elements of `nums1` (index `i` from 0 to 2).
Step 4: Calculate a candidate `x` value as `nums2[0] - nums1[i]`. This is a potential shift value.
Step 5: Call the `check` function with the candidate `x` to determine if applying `x` to `nums1` results in `nums2` after removing two elements.
Step 6: In the `check` function, use two pointers (`p1` for `nums1`, `p2` for `nums2`). If `nums1[p1] + x == nums2[p2]`, increment both pointers. Otherwise, increment `p1` and increment removed_count.
Step 7: If, after the loop, p2 == len(nums2), then the condition of equality can only be satisfied if the number of removed elements from nums1 is exactly 2. Remove elements equals to the sum of the elements removed during the loop with p1 and p2 and the number of elements left in nums1 that need to be removed (len(nums1) - p1)
Step 8: If the `check` function returns `true` (meaning applying `x` works), update `min_x` to the minimum of its current value and the candidate `x`.
Step 9: After iterating through the first three elements, return the `min_x` value.

Key Insights

Insight 1: Sorting both arrays makes it easier to check for equality after applying the shift `x`. Sorting allows us to use a two-pointer approach.
Insight 2: Since we remove only two elements, we can iterate through the first few elements of the sorted `nums1` and calculate the candidate `x` values based on the first element of `nums2`. This significantly reduces the search space for `x`.
Insight 3: Because of the constraints of nums1 length being at least 3, we can derive x from nums2[0] - nums1[i] with i < 3.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(1)

Topics

This problem involves: Array, Two Pointers, Sorting, Enumeration.

Companies

Asked at: Mitsogo.

Find the Integer Added to Array II | LeetCode 3132

Medium

Array Two Pointers Sorting Enumeration

Time: O(n log n)Space: O(1)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int minimumAddedInteger(int[] nums1, int[] nums2) {
        Arrays.sort(nums1);
        Arrays.sort(nums2);

        class Checker {
            public boolean check(int x) {
                int p1 = 0, p2 = 0;
                int removed_count = 0;

                while (p1 < nums1.length && p2 < nums2.length) {
                    if (nums1[p1] + x == nums2[p2]) {
                        p1++;
                        p2++;
                    } else {
                        removed_count++;
                        p1++;
                    }
                }

                if (p2 == nums2.length) {
                    removed_count += nums1.length - p1;
                    return removed_count == 2;
                }

                return false;
            }
        }

        Checker checker = new Checker();

        int min_x = Integer.MAX_VALUE;

        for (int i = 0; i < Math.min(3, nums1.length); i++) {
            int candidate_x = nums2[0] - nums1[i];
            if (checker.check(candidate_x)) {
                min_x = Math.min(min_x, candidate_x);
            }
        }

        return min_x;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 3132: Find the Integer Added to Array II - Detailed explanation and solution approaches.

Topics

Array Two Pointers Sorting Enumeration

Companies

Mitsogo

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3132

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Sort `nums1` and `nums2` in ascending order using a sorting algorithm (e.g., quicksort or mergesort).
2
Step 2: Initialize `min_x` to a large value (e.g., infinity or `Integer.MAX_VALUE`).
3
Step 3: Iterate through the first three elements of `nums1` (index `i` from 0 to 2).
4
Step 4: Calculate a candidate `x` value as `nums2[0] - nums1[i]`. This is a potential shift value.
5
Step 5: Call the `check` function with the candidate `x` to determine if applying `x` to `nums1` results in `nums2` after removing two elements.
6
Step 6: In the `check` function, use two pointers (`p1` for `nums1`, `p2` for `nums2`). If `nums1[p1] + x == nums2[p2]`, increment both pointers. Otherwise, increment `p1` and increment removed_count.
7
Step 7: If, after the loop, p2 == len(nums2), then the condition of equality can only be satisfied if the number of removed elements from nums1 is exactly 2. Remove elements equals to the sum of the elements removed during the loop with p1 and p2 and the number of elements left in nums1 that need to be removed (len(nums1) - p1)
8
Step 8: If the `check` function returns `true` (meaning applying `x` works), update `min_x` to the minimum of its current value and the candidate `x`.
9
Step 9: After iterating through the first three elements, return the `min_x` value.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The dominant operation is sorting, which has a time complexity of O(n log n). The iteration and `check` function contribute O(n) * O(1) to the time complexity, where the 3 iterations is constant time so it doesn't affect overall complexity.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Hashing

Count the frequency of each number in both arrays using hash maps. Then, iterate through possible x values, check if nums1 can become nums2 by adding x and removing two elements. This avoids sorting, but might have a higher constant factor.

Time: O(n*k) where k is the search range for x, potentially up to the range of nums1 elementsSpace: O(n)

Trade-offs: Hashing avoids the initial sorting step, potentially making it faster when n is small. However, its performance depends on the range of possible x values. The current sorting method would be better for large arrays with larger range.

✨

Key Insights

1Insight 1: Sorting both arrays makes it easier to check for equality after applying the shift `x`. Sorting allows us to use a two-pointer approach.
2Insight 2: Since we remove only two elements, we can iterate through the first few elements of the sorted `nums1` and calculate the candidate `x` values based on the first element of `nums2`. This significantly reduces the search space for `x`.
3Insight 3: Because of the constraints of nums1 length being at least 3, we can derive x from nums2[0] - nums1[i] with i < 3.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: `nums1` and `nums2` contain duplicate elements. The sorting and two-pointer approach correctly handle this case because they compare elements based on their values, regardless of their original indices.
Edge case 2: `x` is negative. The solution correctly calculates and handles negative values of `x`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the arrays first. Without sorting, it's much harder to efficiently compare the elements and determine if the arrays are equal after applying the shift.

✗Common mistake 2: Incorrectly implementing the `check` function. It's important to accurately track the number of removed elements and ensure that exactly two elements are removed from `nums1`.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Data reconciliation. Imagine two databases that should contain the same data. If some data is missing or has been transformed by a constant offset, this problem can be used to find the offset value and identify the missing data.

🏦Real-world scenario 2: Signal processing. If a signal has been shifted by a constant value and some data points are missing due to noise or errors, this algorithm can be adapted to find the shift and reconstruct the original signal.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 12, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

When is the two-pointer technique most effective?

Two pointers work best on sorted arrays or when searching for pairs/triplets that satisfy a condition. Common patterns include: moving pointers inward from both ends (like in two-sum variants), fast/slow pointers for cycle detection, or left/right pointers for partitioning.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Find the Integer Added to Array II"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether two pointers techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?