What is the time complexity of Find the Array Concatenation Value?

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

What data structures are used to solve Find the Array Concatenation Value?

This problem uses Array, Two Pointers, Simulation. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Find the Array Concatenation Value?

Using two pointers (left and right) to efficiently access the first and last elements of the array, reducing the need for array resizing or copying. Recognizing that the problem can be solved using a two-pointer approach with O(1) space complexity, significantly improving efficiency compared to approaches that modify the array directly.

Which companies ask LeetCode 2562. Find the Array Concatenation Value in interviews?

This problem is frequently asked at IBM.

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Find the Array Concatenation Value - LeetCode 2562 Solution

Find the Array Concatenation Value - Complete Solution Guide

Find the Array Concatenation Value is LeetCode problem 2562, a Easy 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 a 0-indexed integer array nums . The concatenation of two numbers is the number formed by concatenating their numerals. For example, the concatenation of 15 , 49 is 1549 . The concatenation value of nums is initially equal to 0 . Perform this operation until nums becomes empty: If nums has a size greater than one, add the value of the concatenation of the first and the last element to the concatenation value of nums , and remove those two elements from nums . For example, if the nu

Detailed Explanation

The problem asks you to calculate the 'concatenation value' of an integer array. The concatenation value is computed iteratively. In each iteration, you take the first and last elements of the array, concatenate them as strings (treating them as strings), convert the resulting string back to an integer, and add this integer to the running total (concatenation value). Then, you remove the first and last elements from the array. This process continues until the array is empty or contains only one element. If the array has only one element left, that element is added to the concatenation value.

Solution Approach

The provided solutions employ a two-pointer approach. Two pointers, `left` and `right`, are initialized to point to the beginning and end of the array, respectively. The algorithm iteratively concatenates the elements pointed to by `left` and `right`, adds the result to the `concatenationValue`, and then moves `left` one step to the right and `right` one step to the left. This continues until `left` and `right` cross each other (meaning the array is exhausted). The final `concatenationValue` is the answer. If the array has only one element left, that element is directly added to the total.

Step-by-Step Algorithm

Initialize `concatenationValue` to 0, `left` to 0, and `right` to `nums.length - 1`.
Enter a `while` loop that continues as long as `left` is less than or equal to `right`.
Inside the loop, check if `left` and `right` are equal (only one element remains). If so, add `nums[left]` to `concatenationValue` and break out of the loop.
If `left` and `right` are not equal, concatenate `nums[left]` and `nums[right]` as strings, convert the resulting string to an integer, and add it to `concatenationValue`.
Increment `left` and decrement `right` to move the pointers towards the center.
After the loop finishes, return `concatenationValue`.

Key Insights

Using two pointers (left and right) to efficiently access the first and last elements of the array, reducing the need for array resizing or copying.
Recognizing that the problem can be solved using a two-pointer approach with O(1) space complexity, significantly improving efficiency compared to approaches that modify the array directly.
Handling the edge case where the array has only one element or is empty.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Two Pointers, Simulation.

Companies

Asked at: IBM.

Find the Array Concatenation Value | LeetCode 2562

Easy

Array Two Pointers Simulation

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

Solution Code

Available in:

class Solution {
    public long findTheArrayConcVal(int[] nums) {
        long concatenationValue = 0;
        int left = 0;
        int right = nums.length - 1;

        while (left <= right) {
            if (left == right) {
                concatenationValue += nums[left];
                break;
            } else {
                String concatenated = String.valueOf(nums[left]) + String.valueOf(nums[right]);
                concatenationValue += Long.parseLong(concatenated);
                left++;
                right--;
            }
        }
        return concatenationValue;
    }
}

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Problem Info

Description

Problem 2562: Find the Array Concatenation Value - Detailed explanation and solution approaches.

Topics

Array Two Pointers Simulation

Companies

IBM

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:2562

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

6 steps

1
Initialize `concatenationValue` to 0, `left` to 0, and `right` to `nums.length - 1`.
2
Enter a `while` loop that continues as long as `left` is less than or equal to `right`.
3
Inside the loop, check if `left` and `right` are equal (only one element remains). If so, add `nums[left]` to `concatenationValue` and break out of the loop.
4
If `left` and `right` are not equal, concatenate `nums[left]` and `nums[right]` as strings, convert the resulting string to an integer, and add it to `concatenationValue`.
5
Increment `left` and decrement `right` to move the pointers towards the center.
6
After the loop finishes, return `concatenationValue`.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The algorithm iterates through the array once using two pointers. The number of iterations is proportional to the length of the array (n). No additional data structures are used beyond a few integer variables, resulting in constant space 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

Using a deque

Instead of two pointers, use a deque data structure. Concatenate and remove from the front and back of the deque.

Time: O(n)Space: O(n)

Trade-offs: This approach is less efficient in space complexity than the two-pointer method. It's only beneficial if other deque operations are already needed.

✨

Key Insights

1Using two pointers (left and right) to efficiently access the first and last elements of the array, reducing the need for array resizing or copying.
2Recognizing that the problem can be solved using a two-pointer approach with O(1) space complexity, significantly improving efficiency compared to approaches that modify the array directly.
3Handling the edge case where the array has only one element or is empty.

📚

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.

Edge Cases

Empty array: The `while` loop condition `left <= right` handles the empty array case gracefully because the loop won't execute. The function will correctly return 0.
Single-element array: The `if (left == right)` condition directly handles this. The single element is added to the `concatenationValue`.

⚠️

Common Mistakes to Avoid

✗Incorrect handling of edge cases: Forgetting to handle the case where the input array has only one element or is empty can lead to errors.

✗Inefficient string concatenation: Repeatedly creating strings within the loop can lead to performance issues. Using a more efficient string manipulation technique (like in the C solution with direct arithmetic) can help.

✅

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?

💼Data processing: This type of concatenation could be used in processing identification numbers, where combining parts of a code is meaningful.

🏦Number puzzle solving: Similar logic can be applied to number puzzles or games where you need to combine number parts strategically.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ 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: July 13, 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.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

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 Array Concatenation Value"?

For this easy-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.

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