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# 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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