Plus One Problem

Overview

In this particular LeetCode problem, Plus One, I would have to approach it with a crafty approach. The problem states

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.

Here is an example for a better understanding.

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].

Let's Dive in.

Since we are targeting the end of the array. My first approach was to reverse loop through the array.

var plusOne = function(digits) {
    
     // reverse loop 
    for(let i = digits.length - 1; i >= 0; i--){
}
    
};

Since we can't have double-digit numbers I targeted 9. If the current digit does not equal the value or type of 9 increments by one

var plusOne = function(digits) {
    
     // reverse loop 
    for(let i = digits.length - 1; i >= 0; i--){
        // add constraints 
        if(digits[i] !== 9){
            digits[i] ++
        }
        
    }
    
};

Along with incrementing the number, I decided to return the digit.

var plusOne = function(digits) {
    
     // reverse loop 
    for(let i = digits.length - 1; i >= 0; i--){
        // add constraints 
        if(digits[i] !== 9){
            digits[i] ++
            return digits
        }
        
    }
    
};

I created a SafeNet for 0. If the current number = 0 and if the index is at the current position of 0. Take the array add 1 to it and return the digits.

var plusOne = function(digits) {
    
     // reverse loop 
    for(let i = digits.length - 1; i >= 0; i--){
        // add constraints 
        if(digits[i] !== 9){
            digits[i] ++
            return digits
        }
        digits[i] = 0
        if(i === 0){
            digits.unshift(1)
            return digits
        }
    }
    
};

Thanks for reading.