Plus One

 66. Plus One

Easy

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.

Code:

class Solution: def plusOne(self, digits: List[int]) -> List[int]: a='' l=[] for i in digits: a+=str(i) a=str(int(a)+1) for i in a: l.append(int(i)) return l

#instead of second for loop : return [int(temp[i]) for i in range(len(temp))]

Other C++:

Intuition

If we thought that all elements of the vector is a number we need to increase it by 1. The input can be 100 digits so we must handle that through digits.

Approach

First we increment the first digit (last element) by 1, if it becomes 10 we make it 0 ans add 1 to the second digit.. until the last digit (first element), if it becoms 10 we make it 1 and push_back a leading zero.

Complexity

• Time complexity:
  O(n)

Code

class Solution {
public:
    vector<int> plusOne(vector<int>& v) {
        int n = v.size();
        for(int i = n-1; i >= 0; i--){
            if(i == n-1)
                v[i]++;
            if(v[i] == 10){
                v[i] = 0;
                if(i != 0){
                    v[i-1]++;
                }
                else{
                    v.push_back(0);
                    v[i] = 1;
                }
            }
        }
        return v;
    }
};