How do you solve Plus One on LeetCode?

Plus One (LeetCode #66) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Direct manipulation of the array avoids unnecessary conversions.

What is the brute force solution for Plus One?

The brute force approach for Plus One is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves converting the array of digits into a single integer, incrementing it by one, and then converting it back into an array of digits. This is straightforward but inefficient for large integers. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Plus One?

Plus One uses the following concepts: Array, Math. The recommended patterns to study are: Array, Math.

What are the interview tips for Plus One?

Think about edge cases, like all digits being 9. Explain your thought process clearly as you code. Consider both time and space complexity when discussing your solution.

What are common mistakes when solving Plus One?

Converting the entire array to an integer without considering overflow. Not handling the carry properly when all digits are 9.

#66

Plus One

Easy

ArrayMathArrayMath

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal solution directly manipulates the array without converting it to an integer. It handles the carry from the least significant digit to the most significant efficiently, which is crucial for large numbers.

⚙️

Algorithm

4 steps

1Step 1: Start from the last digit of the array.
2Step 2: Increment the last digit by one.
3Step 3: If the result is 10, set the current digit to 0 and move to the next digit to the left, adding the carry.
4Step 4: If there's still a carry after processing all digits, insert 1 at the beginning of the array.

solution.py7 lines

1def plus_one(digits):
2    for i in range(len(digits) - 1, -1, -1):
3        if digits[i] < 9:
4            digits[i] += 1
5            return digits
6        digits[i] = 0
7    return [1] + digits

ℹ

Complexity note: The time complexity is O(n) because we may need to traverse the entire array in the worst case (e.g., when all digits are 9). The space complexity is O(1) since we are modifying the input array in place.

1Direct manipulation of the array avoids unnecessary conversions.
2Handling carry is crucial for correct results.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.