Plus One LeetCode Solution

Problem – Plus One

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.

Plus One LeetCode Solution in Java

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;
    }
    
    int[] newNumber = new int [n+1];
    newNumber[0] = 1;
    
    return newNumber;
}

Plus One LeetCode Solution in C++

void plusone(vector<int> &digits)
{
	int n = digits.size();
	for (int i = n - 1; i >= 0; --i)
	{
		if (digits[i] == 9)
		{
			digits[i] = 0;
		}
		else
		{
			digits[i]++;
			return;
		}
	}
		digits[0] =1;
		digits.push_back(0);
		
}

Plus One LeetCode Solution in Python

def plusOne(digits):
    num = 0
    for i in range(len(digits)):
    	num += digits[i] * pow(10, (len(digits)-1-i))
    return [int(i) for i in str(num+1)]

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