Leetcode: 66 Plus One Explained Step by Step

 

Problem Description – What is "Plus One"?

The Plus One problem asks you to increment a large integer represented as an array of digits by 1. Each element in the array corresponds to a single digit of the number, with the most significant digit at the front. You need to handle cases where digits can result in a carry (e.g., 9 + 1), requiring changes to multiple digits or even expanding the array.

Example 1:

Input: [1, 2, 3]

Output: [1, 2, 4]

Explanation: The array represents the integer 123, and adding 1 results in 124.

Example 2:

Input: [9, 9, 9]

Output: [1, 0, 0, 0]

Explanation: The array represents the integer 999, and adding 1 results in 1000. Notice how the array grows to accommodate the new digit.

Key Challenges

1. Carry Propagation: When the last digit is 9, adding 1 will result in 10, meaning you need to set the last digit to 0 and carry over 1 to the next digit. If all the digits are 9, you need to increase the array size.
2. Handling Edge Cases: If the input is something like [9, 9, 9], the array needs to expand because the result becomes 1000. You’ll have to create a new array to accommodate the extra digit.

Optimal Java Solution for "Plus One"

Here is a simple and efficient Java solution that solves the problem by working backwards through the array, checking for digits that need to carry over, and modifying the array as necessary.

private static int[] getPlusOne(int[] digits) {

for (int i = digits.length-1; i>=0; i--) {

if (digits[i]<9) {

digits[i]++;

return digits;

}

digits[i]=0;

}

int[] newNumber = new int[digits.length+1];

newNumber[0] = 1;

return newNumber;

}

Step-by-Step Breakdown of the Solution

1. Traversing the Array from Right to Left

We begin by iterating over the array starting from the last element. This is because the least significant digit is the rightmost one, and that’s where we begin adding 1.

for (int i = digits.length - 1; i >= 0; i--) {

• Here, digits.length - 1 gives us the index of the last element, and we iterate backwards.

2. Handling the Case Where the Digit is Less Than 9

If the digit is less than 9, we can directly increment it without needing to worry about carrying over.

if (digits[i] < 9) {

    digits[i]++;

    return digits;

} 

• If the current digit is less than 9, we increment it by 1, and since no further changes are needed, we return the updated digits array.

3. Handling the Case Where the Digit is 9

When a digit is 9, adding 1 makes it 10, which means the current digit becomes 0, and we carry over the 1 to the next digit.

digits[i] = 0;

• We set the current digit to 0 and continue the loop to handle the next digit.

4. Creating a New Array if All Digits Are 9

If all digits are 9, the loop completes, and all digits have been set to 0. In this case, we need to create a new array with an additional digit to store the result.

int[] newNumber = new int[digits.length + 1];

newNumber[0] = 1;

• A new array is created with a size of digits.length + 1. The first element is set to 1, and the rest will automatically be 0. This handles cases like 999 + 1 = 1000.

Time Complexity and Space Complexity

• Time Complexity: The time complexity of this solution is O(n), where n is the number of digits in the input array. This is because we potentially have to iterate through all the digits in the worst case (e.g., when all digits are 9).

• Space Complexity: The space complexity is O(n) since we may need to create a new array of size n+1 in the worst case.

Example Walkthrough

Let’s walk through a few examples to see how the solution works.

Example 1: [1, 2, 3]

1. Start with the last digit (3). Since 3 is less than 9, we increment it to 4 and return [1, 2, 4].

Example 2: [9, 9, 9]

1. Start with the last digit (9). Since it's 9, we set it to 0 and move to the next digit.
2. The second-to-last digit is also 9, so we set it to 0 and move to the next.
3. The first digit is also 9, so we set it to 0 and exit the loop.
4. At this point, all digits are 0, so we create a new array of size 4, set the first element to 1, and return [1, 0, 0, 0].

Why This Solution is Optimal

This solution is efficient because it processes the array from right to left, minimizing unnecessary iterations. It handles both normal and edge cases, such as carrying over and growing the array size when necessary. By leveraging simple conditional logic and array manipulation, this approach solves the problem in linear time.

Conclusion

The Plus One problem may appear straightforward, but it teaches important concepts like handling carry operations and array manipulations in programming. By solving this problem using Java, you can strengthen your understanding of basic operations that are often tested in coding interviews.

Understanding how to approach such problems effectively can give you a competitive edge in your coding interviews. Make sure to practice variations of this problem to become more comfortable with similar challenges.

Happy coding!