[Array] Plus One - Easy

The Array Plus One problem is a coding challenge we often see. The goal is to add one to a number that is shown as an array of digits. Each part of the array shows one digit of the number. We start from the most important digit. To solve this, we go through the array from the last digit. We add one and take care of any carry that comes from this.

In this article, we will look at different parts of the Array Plus One problem. We will understand the problem, write solutions in Java, Python, and C++, and deal with edge cases. We will also talk about ways to make space usage better, common errors to avoid, and things to think about for good performance. The following headers will help us with our discussion:

Understanding the Array Plus One Problem
Java Solution for Array Plus One
Python Implementation for Array Plus One
C++ Approach to Array Plus One
Handling Carry in Array Plus One
Optimizing Space Complexity in Array Plus One
Testing Edge Cases in Array Plus One
Performance Considerations for Array Plus One
Common Mistakes in Array Plus One Implementation
Frequently Asked Questions

Java Solution for Array Plus One

To solve the “Array Plus One” problem in Java, we want to add one to the number that an array of digits represents. Each digit is in an array. Our goal is to give back a new array that shows the result.

Here is a simple way to do it:

public 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; // Set current digit to 0
        }
        
        // If all digits were 9, we need a new array
        int[] result = new int[n + 1];
        result[0] = 1; // Set the first digit to 1
        return result;
    }
}

Explanation:

We start from the last digit of the array and try to add one to it.
If the digit is less than 9, we just add one and return the array.
If the digit is 9, we set it to 0 and move to the next digit.
If we finish the loop, it means we had a carry for all digits. For example, from 999 to 1000. We create a new array that is size n + 1 and set the first element to 1.

This way helps us to manage the carry correctly and keeps the original array safe. Use this Java solution to solve the Array Plus One problem easily.

If you want to see more about array problems, you can check out Array Two Sum.

Python Implementation for Array Plus One

To solve the Array Plus One problem in Python, we will go through the array from the last element. Our goal is to add one to the last digit and deal with any carry from this addition. Here is a simple implementation:

def plus_one(digits):
    n = len(digits)
    for i in range(n - 1, -1, -1):
        if digits[i] < 9:
            digits[i] += 1
            return digits
        digits[i] = 0  # Set current digit to 0 if it becomes 10
        
    # If all digits were 9, we need to add a new leading 1
    return [1] + digits

Explanation:

We go through the list from the end to the start.
If the current digit is less than 9, we just add one and return the updated array.
If the digit is 9, we change it to 0 and move to the next digit.
If we finish the loop and all digits were 9, we need to add a new leading 1 to the list. This gives us the correct output.

Example Usage:

result = plus_one([1, 2, 3])  # Output: [1, 2, 4]
result = plus_one([9, 9, 9])  # Output: [1, 0, 0, 0]

This implementation works well to manage the carry and changes the input array directly. It uses little extra space. If you want to learn more about array manipulation, you can check the Array Rotate Array article.

C++ Approach to Array Plus One

In the C++ way to solve the Array Plus One problem, we want to add one to the number that the array of digits shows. We need to take care of any carry that comes from the addition. This is really important when we have trailing nines. Here is a simple code:

#include <vector>
using namespace std;

class Solution {
public:
    vector<int> plusOne(vector<int>& digits) {
        int n = digits.size();
        for (int i = n - 1; i >= 0; --i) {
            if (digits[i] < 9) {
                digits[i]++;
                return digits;
            }
            digits[i] = 0; // Change current digit to 0 and move to next digit
        }
        // If we finish the loop, it means we have a carry that needs a new digit
        digits.insert(digits.begin(), 1);
        return digits;
    }
};

Explanation of the Code:

Loop Through the Array: We go from the last number to the first.
Increment Logic: If the current digit is less than 9, we just add one to it and return the array.
Set to Zero: If the digit is 9, we change it to 0 and keep going.
Handle Carry: If all digits are 9, we put a 1 at the start of the array.

Performance:

Time Complexity: O(n), where n is how many digits we have in the array.
Space Complexity: O(1) if we do not count the output array or O(n) if we count the output array.

This code works well for adding one. It also deals with carry situations while keeping good performance. For more examples and solutions, we can check Array Two Sum or Array Maximum Subarray.

Handling Carry in Array Plus One

In the Array Plus One problem, we need to handle the carry. This is important for updating the array that shows a non-negative integer. The challenge is to add one to the integer and think about any carry values that may come up during the addition.

Steps to Handle Carry:

Start from the End: We begin at the last digit of the array. This is because we add one to the least significant digit first.
Add One: We add one to the current digit.
Check for Carry: If the digit is less than 10, we just update the array and we are done. If it is 10, we set the current position to 0 and carry over 1 to the next digit.
Continue: We move to the next digit on the left and repeat the steps until we finish all digits or there is no carry left.
Handle Overflow: If we finish all digits and still have a carry, we add a new digit at the start of the array.

Example Implementation in Java:

public int[] plusOne(int[] digits) {
    for (int i = digits.length - 1; i >= 0; i--) {
        if (digits[i] < 9) {
            digits[i]++;
            return digits;
        }
        digits[i] = 0; // Set current digit to 0 and carry over
    }
    int[] result = new int[digits.length + 1];
    result[0] = 1; // New digit for the carry
    return result;
}

Example Implementation in Python:

def plus_one(digits):
    for i in range(len(digits) - 1, -1, -1):
        if digits[i] < 9:
            digits[i] += 1
            return digits
        digits[i] = 0
    return [1] + digits

Example Implementation in C++:

vector<int> plusOne(vector<int>& digits) {
    for (int i = digits.size() - 1; i >= 0; --i) {
        if (digits[i] < 9) {
            digits[i]++;
            return digits;
        }
        digits[i] = 0;
    }
    digits.insert(digits.begin(), 1); // Insert 1 at the start
    return digits;
}

These examples show how we can handle the carry well. They give the right output for different situations.

Optimizing Space Complexity in Array Plus One

We want to optimize space complexity in the Array Plus One problem. The goal is to do the operation in-place. This means we only use a small amount of extra space. The problem is about adding one to a number that is shown as an array of digits. Each item in the array is a single digit.

In-Place Solution

The in-place solution changes the original array. We do not need to make another array. The algorithm begins from the last digit and goes to the first digit. Here is how we do it:

Start from the last digit and add one.
If the digit is 10, set it to 0 and carry over 1 to the next digit.
Keep doing this until there are no carries left or we have checked all digits.
If there is still a carry after checking all digits, for example from 999 to 1000, we make a new array that is one size bigger. We put 1 at the start and then all zeros.

Java Implementation

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; // carry over
    }
    // Create a new array if needed
    int[] result = new int[n + 1];
    result[0] = 1; // set the leading 1
    return result;
}

Python Implementation

def plus_one(digits):
    n = len(digits)
    for i in range(n - 1, -1, -1):
        if digits[i] < 9:
            digits[i] += 1
            return digits
        digits[i] = 0  # carry over
    return [1] + digits  # handle carry for 999... case

C++ Implementation

vector<int> plusOne(vector<int>& digits) {
    int n = digits.size();
    for (int i = n - 1; i >= 0; i--) {
        if (digits[i] < 9) {
            digits[i]++;
            return digits;
        }
        digits[i] = 0; // carry over
    }
    digits.insert(digits.begin(), 1); // handle carry
    return digits;
}

Key Points

The algorithm runs in O(n) time complexity. Here n is the number of digits.
Space complexity is O(1) if we do not count the output array. We only use a small amount of space for variables.
Changing the input array in-place means we do not need extra storage. This makes our solution better for space.

This in-place method is important for using space wisely in the Array Plus One problem. We can handle even big inputs without using too much memory. For similar problems, we can look at Array Two Sum or Array Maximum Subarray.

Testing Edge Cases in Array Plus One

When we work on the Array Plus One problem, it is very important to test different edge cases. This helps us make sure our solution is strong and can handle all possible situations. Here are some important edge cases to think about:

Single Digit Array:
- Input: [9]
- Expected Output: [1, 0]
- Explanation: The carry from 9 makes an extra digit.
Multiple Digits Without Carry:
- Input: [1, 2, 3]
- Expected Output: [1, 2, 4]
- Explanation: We just increase the last digit with no carry.
Multiple Digits With Carry:
- Input: [1, 2, 9]
- Expected Output: [1, 3, 0]
- Explanation: The last digit goes over, so we need a carry.
All Nines:
- Input: [9, 9, 9]
- Expected Output: [1, 0, 0, 0]
- Explanation: All digits are nine, so we get a full carry.
Leading Zeros:
- Input: [0, 0, 1]
- Expected Output: [0, 0, 2]
- Explanation: Leading zeros do not change the result.
Empty Array:
- Input: []
- Expected Output: [1]
- Explanation: An empty input means the number 0, and adding one gives us 1.
Very Large Number:
- Input: [9, 9, 9, 9, 9, 9, 9, 9]
- Expected Output: [1, 0, 0, 0, 0, 0, 0, 0, 0]
- Explanation: We test how the code works with big inputs and many carries.

Sample Test Code

Here is a simple code for testing these edge cases in Python:

def plus_one(digits):
    n = len(digits)
    for i in range(n - 1, -1, -1):
        if digits[i] < 9:
            digits[i] += 1
            return digits
        digits[i] = 0
    return [1] + digits

# Edge case tests
edge_cases = [
    ([9], [1, 0]),
    ([1, 2, 3], [1, 2, 4]),
    ([1, 2, 9], [1, 3, 0]),
    ([9, 9, 9], [1, 0, 0, 0]),
    ([0, 0, 1], [0, 0, 2]),
    ([], [1]),
    ([9, 9, 9, 9, 9, 9, 9, 9], [1, 0, 0, 0, 0, 0, 0, 0, 0]),
]

for input_case, expected in edge_cases:
    assert plus_one(input_case) == expected

By adding these edge cases in our tests, we help make sure our solution for the Array Plus One problem is reliable and fast. For more related problems, we can check articles like Array Two Sum or Array Remove Duplicates.

Performance Considerations for Array Plus One

When we work on the “Array Plus One” problem, we should think about performance. This is important to make sure our solution runs well, especially with big inputs. Here are the main points that affect performance:

Time Complexity:
- The algorithm usually runs in O(n) time, where n is how long the input array is. We need to go through the array to manage carry propagation.
- In the worst case, like with an array full of 9s, we must process every element. This means we check all the elements in the array.
Space Complexity:
- The space complexity is O(1) if we change the array directly. We only need a few variables for carry and index.
- If we create a new array for the bigger number, the space complexity will be O(n).
Handling Large Numbers:
- If the input array is a big number, like [9, 9, 9], we need to check for overflow. We do this by looking for carry after the most important digit.
- For example, if we still have a carry after the last digit, we need a new array with extra space for the result.
Edge Cases:
- We should think about edge cases like an empty array or an array with just one number.
- If we do not handle these cases well, performance can drop. This can cause extra work or errors.
Optimizing Iterations:
- We should avoid going through the array too much, especially if the carry is zero. We can stop early when the addition does not give us a carry.
- For example, if the last operation has no carry, we do not need to check the other elements.
Code Implementation:
- A good implementation can make performance much better. Here is an example in Python:

def plus_one(digits):
    n = len(digits)
    for i in range(n - 1, -1, -1):
        if digits[i] < 9:
            digits[i] += 1
            return digits
        digits[i] = 0
    return [1] + digits

Benchmarking:
- We should check how long the code takes with different input sizes. This helps us find slow parts.
- We can use tools to watch memory use and execution time so that our solution stays efficient.

By following these performance points, we can make sure our “Array Plus One” solution is both efficient and easy to scale. For more related array problems, we can read this Array Two Sum article.

Common Mistakes in Array Plus One Implementation

When we work on the “Array Plus One” problem, we can make some common mistakes. These mistakes can give wrong results or slow code. Here are some of the usual problems:

Incorrect Carry Handling:

If we do not handle carry well, we can get wrong results when the last digit overflows.
We must add the carry to the next digit the right way.

// Java example of handling carry
int[] plusOne(int[] digits) {
    for (int i = digits.length - 1; i >= 0; i--) {
        if (digits[i] < 9) {
            digits[i]++;
            return digits;
        }
        digits[i] = 0; // reset current digit
    }
    // handle case where all digits are 9
    int[] result = new int[digits.length + 1];
    result[0] = 1;
    return result;
}

Modifying the Input Array:
- Changing the input array directly can cause problems in other functions. These functions might need the original array.
Assuming Fixed Size for Output:
- We must think about cases where the new array is bigger than the input. For example, turning [9] into [1, 0].
Off-by-One Errors:
- Mistakes in loops can make us skip or process digits wrong.
Inefficient Space Usage:
- Using extra data structures can make space use worse. We should try to use fewer extra arrays.
Ignoring Edge Cases:
- We must test for cases like empty arrays or arrays with all 9s. We need to handle these edge cases.
Inconsistent Return Types:
- We should keep the return type the same. Returning void when we expect an array can cause errors.
Not Considering Leading Zeros:
- If needed, we must handle leading zeros when adding to the array.
Failure to Validate Input:
- We need to check the input array. For example, we should make sure it only has digits. This can prevent runtime errors.
Performance Overheads:
- Using slow algorithms, like nested loops for simple tasks, can hurt performance a lot.

By knowing these common mistakes, we can do the “Array Plus One” algorithm better and faster. For more problems and solutions with arrays, check out the Array Contains Duplicate article.

Frequently Asked Questions

1. What is the Array Plus One problem?

The Array Plus One problem is about adding one to a non-negative number that is shown as an array of digits. Each digit is in order, with the biggest digit at the front. The task is to add one to this number and give the answer as a new array. We also need to deal with any carries that happen during the addition.

2. How do you handle carry in the Array Plus One implementation?

When we make the Array Plus One solution, we must handle the carry from the last digit well. We start from the last digit in the array. We add one to this digit and see if it is more than 9. If it is, we change that digit to 0 and carry over 1 to the next digit. We keep doing this until there are no more carries or we have checked all digits.

3. What is the time complexity of the Array Plus One algorithm?

The time complexity of the Array Plus One algorithm is O(n). Here n is the number of digits in the input array. This happens because, in the worst case, we may need to check every digit to fix carries. This is especially true if the array is all 9s.

4. How can I optimize space complexity in the Array Plus One solution?

To make space complexity better in the Array Plus One problem, we can change the original array if we can. This way, we don’t need to make new arrays for the result. But if we need a new digit (like changing [9] to [1, 0]), we may need to make a new array that is n+1 size to fit the new number.

5. Are there common mistakes to avoid in the Array Plus One implementation?

Some common mistakes in the Array Plus One implementation are not handling the carry correctly, forgetting about cases when all digits are 9, and changing the original array when we should not. We should always test edge cases to make sure our solution works for inputs like [9, 9, 9] and [0]. For more tips on working with arrays, see the article on Array Remove Duplicates from Sorted Array.