What is the time complexity of Plus One?

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

What data structures are used to solve Plus One?

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

What are the key insights for solving Plus One?

**Right-to-Left Carry Propagation:** The natural way to add to a number is from the least significant digit. The solution correctly models this by iterating the `digits` array from `n-1` down to `0`, ensuring carries are handled sequentially. **Early Exit Optimization:** If a digit `digits[i]` is found to be less than `9`, we can simply increment it, and crucially, `return the array immediately`. No further carries will propagate left, making this a highly efficient shortcut for most inputs.

Which companies ask LeetCode 66. Plus One in interviews?

This problem is frequently asked at Accenture, Adobe, Agoda, Amazon, Apple and 9 more companies.

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Plus One - LeetCode 66 Solution

Plus One - Complete Solution Guide

Plus One is LeetCode problem 66, 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 large integer represented as an integer array digits , where each digits[i] is the i th digit of the integer. The digits are ordered from most significant to least significant in left-to-right order. The large integer does not contain any leading 0 's. Increment the large integer by one and return the resulting array of digits . Example 1: Input: digits = [1,2,3] Output: [1,2,4] Explanation: The array represents the integer 123. Incrementing by one gives 123 + 1 = 124. Thus, the

Detailed Explanation

This problem asks us to increment a 'large integer' by one. The interesting twist is that this integer isn't given as a standard `int` or `long`, but as an array of its individual digits. For instance, `[1,2,3]` represents the number 123. The digits are ordered from most significant to least significant, left-to-right, and there are no leading zeros, except for the number 0 itself (which would be `[0]`). The core challenge here lies in accurately simulating elementary addition, specifically handling 'carries'. When you add one to a number like 123, it's straightforward: 123 + 1 = 124, so `[1,2,3]` becomes `[1,2,4]`. However, consider `[4,5,9]`. Adding one makes it 460. This means the `9` becomes `0`, and a `1` is carried over to the `5`, making it `6`. This carry-propagation is where the logic needs to be robust. The most intriguing edge case arises when the number consists entirely of nines, like `[9,9]`. Adding one gives us 100. This means `[9,9]` must transform into `[1,0,0]`. Not only do all the nines become zeros due to carry propagation, but the array itself needs to grow in length to accommodate the new most significant digit. This highlights why a simple `int` conversion isn't suitable, as `[9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9]` would quickly overflow standard integer types.

Solution Approach

The provided solution elegantly simulates the manual process of adding one to a number, starting from the rightmost (least significant) digit and moving left. It iterates backward through the `digits` array. The key insight is that if a digit is less than 9, incrementing it by one will never cause a carry to propagate further left. For example, if we have `[1,2,3]` and are at `3`, we increment it to `4`. The result `[1,2,4]` is final, and we can immediately return. This covers the vast majority of cases and allows for an early exit. However, if the current digit `digits[i]` *is* 9, we must set it to 0. This simulates a carry being generated and propagated to the left. The loop then continues to the next digit to the left, effectively 'carrying the one' to it. If the loop completes, it means every digit in the original array was a 9 (e.g., `[9,9]`). In this specific scenario, all original digits would have been turned into zeros (`[0,0]`). We then need to prepend a `1` to this array to form the final result (`[1,0,0]`), representing the number with an increased digit count.

Step-by-Step Algorithm

Step 1: Iterate through the `digits` array from right to left (from the least significant digit to the most significant digit).
Step 2: If the current digit is less than 9, increment it by 1, and return the modified `digits` array.
Step 3: If the current digit is 9, set it to 0 and continue to the next digit to the left (handle carry-over).
Step 4: If the loop completes (all digits were 9), create a new array with size one greater than the original array. Set the first element of the new array to 1, and copy the remaining elements (all 0s) from the original array.
Step 5: Return the new array.

Key Insights

**Right-to-Left Carry Propagation:** The natural way to add to a number is from the least significant digit. The solution correctly models this by iterating the `digits` array from `n-1` down to `0`, ensuring carries are handled sequentially.
**Early Exit Optimization:** If a digit `digits[i]` is found to be less than `9`, we can simply increment it, and crucially, `return the array immediately`. No further carries will propagate left, making this a highly efficient shortcut for most inputs.
**Elegant Handling of All Nines Edge Case:** The solution cleanly manages the scenario where all digits are 9s (e.g., `[9,9,9]`). If the loop finishes, it implies all digits were converted to 0s. The solution then prepends a `1` to this all-zero array (`[1] + digits`), effectively creating `[1,0,0,0]` without complex array resizing logic, which is both readable and efficient.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Math.

Companies

Asked at: Accenture, Adobe, Agoda, Amazon, Apple, Bloomberg, Intuit, Meta, Microsoft, TikTok, Uber, Visa, Yahoo, tcs.

Plus One | LeetCode 66

Easy

Array Math

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

Solution Code

Available in:

class Solution {
    public int[] plusOne(int[] digits) {
        int n = digits.length;
        for (int i = n - 1; i >= 0; i--) {
            if (digits[i] < 9) {
                digits[i]++;
                return digits;
            }
            digits[i] = 0;
        }
        int[] newDigits = new int[n + 1];
        newDigits[0] = 1;
        return newDigits;
    }
}

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Problem Info

Description

Problem 66: Plus One - Detailed explanation and solution approaches.

Topics

Array Math

Companies

AccentureAdobeAgodaAmazonAppleBloombergIntuitMeta

Stats

Difficulty:Easy

Languages:4

Companies:14

Problem #:66

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

5 steps

1
Step 1: Iterate through the `digits` array from right to left (from the least significant digit to the most significant digit).
2
Step 2: If the current digit is less than 9, increment it by 1, and return the modified `digits` array.
3
Step 3: If the current digit is 9, set it to 0 and continue to the next digit to the left (handle carry-over).
4
Step 4: If the loop completes (all digits were 9), create a new array with size one greater than the original array. Set the first element of the new array to 1, and copy the remaining elements (all 0s) from the original array.
5
Step 5: Return the new array.

💡

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 loop iterates at most once over the input array. Creating a new array in the worst-case scenario, its size is linearly proportional to the input size. Therefore, time complexity is O(n) and space complexity O(1) (considering problem constraints).

📖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

Converting to Integer

Convert the digit array to an integer, add one, then convert the result back to an array of digits.

Time: O(n)Space: O(n) - potentially requires extra space for the integer representation and new array.

Trade-offs: Simpler to understand for beginners, but potentially less efficient for extremely large integers due to the overhead of integer conversion.

✨

Key Insights

1**Right-to-Left Carry Propagation:** The natural way to add to a number is from the least significant digit. The solution correctly models this by iterating the `digits` array from `n-1` down to `0`, ensuring carries are handled sequentially.
2**Early Exit Optimization:** If a digit `digits[i]` is found to be less than `9`, we can simply increment it, and crucially, `return the array immediately`. No further carries will propagate left, making this a highly efficient shortcut for most inputs.
3**Elegant Handling of All Nines Edge Case:** The solution cleanly manages the scenario where all digits are 9s (e.g., `[9,9,9]`). If the loop finishes, it implies all digits were converted to 0s. The solution then prepends a `1` to this all-zero array (`[1] + digits`), effectively creating `[1,0,0,0]` without complex array resizing logic, which is both readable and efficient.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: The input array contains only one digit which is 9. The solution correctly handles this case by creating a new array [1, 0].
Edge case 2: The input array contains all 9s (e.g., [9, 9, 9]). The solution correctly handles this case by creating a new array with an extra digit (1, 0, 0, 0...).
Edge case 3: The input array contains a mix of digits and adding 1 will not result in a carry over at the most significant place. The solution correctly handles this and simply updates the digit.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the carry-over properly. Failing to propagate the carry through the array can lead to incorrect results.

✗Common mistake 2: Not allocating enough memory when creating a new array for the case where a carry-over increases the number of digits (resulting in buffer overflow).

✅

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: Incrementing version numbers or counters represented as strings or arrays of digits (e.g., software versioning).

🏦Real-world scenario 2: Handling large numerical calculations where the numbers exceed the limits of standard integer data types, especially in financial systems or scientific computations.

💡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: August 6, 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

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

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

@InterviewProJul 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.

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 "Plus One"?

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 math techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?