LC 66 - Plus One

Question

You are given a large integer represented as an integer array digits, where each digits[i] is the ith 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 result should be [1,2,4].

Example 2:

Input: digits = [4,3,2,1]
Output: [4,3,2,2]

Explanation: The array represents the integer 4321. Incrementing by one gives 4321 + 1 = 4322. Thus, the result should be [4,3,2,2].

Example 3:

Input: digits = [9]
Output: [1,0]

Explanation: The array represents the integer 9. Incrementing by one gives 9 + 1 = 10. Thus, the result should be [1,0].

Constraints:

• 1 <= digits.length <= 100
• 0 <= digits[i] <= 9
• digits does not contain any leading 0’s.

Links

Question here and solution here

Solution

concept

Just add one to the least significant digit and handle carry properly.

code

class Solution:
    def plusOne(self, digits: List[int]) -> List[int]:
        q = deque(digits)

        carry = False
        for i in range(len(q) - 1, -1, -1):
            if i == len(q) - 1 or carry:
                q[i] += 1

            if q[i] >= 10:
                q[i] %= 10
                carry = True
            else:
                carry = False

        if carry:
            q.appendleft(1)
        
        return list(q)
        
class NeetSolution:
    def plusOne(self, digits: List[int]) -> List[int]:
        one = 1
        i = 0
        digits.reverse()

        while one:
            if i < len(digits):
                if digits[i] == 9:
                    digits[i] = 0
                else:
                    digits[i] += 1
                    one = 0
            else:
                digits.append(one)
                one = 0
            i += 1

        digits.reverse()
        return digits
        
class NeetSolution:
    def plusOne(self, digits: List[int]) -> List[int]:
        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
        
class NeetSolution:
	"""
	recursion
	"""
    def plusOne(self, digits: List[int]) -> List[int]:
        if not digits:
            return [1]

        if digits[-1] < 9:
            digits[-1] += 1
            return digits
        else:
            return self.plusOne(digits[:-1]) + [0]

Complexity

time: $O(n)$
space: $O(n)$