Program to add one to a number that is shown as a digit list in Python

Suppose we have an array called nums, containing decimal digits of a number. For example, [2, 5, 6] represents 256. We have to add 1 to this number and return the list in the same format as before.

So, if the input is like nums = [2, 6, 9], then the output will be [2, 7, 0] (269 + 1 = 270).

Algorithm Steps

To solve this, we will follow these steps −

• i := size of nums - 1

• while i >= 0, do

  • if nums[i] + 1

    • nums[i] := nums[i] + 1

    • come out from loop

  • otherwise,

    • nums[i] := 0

    • i := i - 1

• if i

  • insert 1 into nums at position 0

• return nums

Example

Let us see the following implementation to get better understanding ?

def solve(nums):
    i = len(nums) - 1
    while i >= 0:
        if nums[i] + 1 <= 9:
            nums[i] = nums[i] + 1
            break
        else:
            nums[i] = 0
            i -= 1
    if i < 0:
        nums.insert(0, 1)
    return nums

nums = [2, 6, 9]
print(solve(nums))

The output of the above code is ?

[2, 7, 0]

How It Works

The algorithm starts from the rightmost digit and adds 1. If the digit becomes greater than 9, it sets the digit to 0 and carries over to the next position. This process continues until we find a digit that doesn't overflow or we reach the beginning of the array.

Edge Case Example

Let's see what happens when all digits are 9 ?

def solve(nums):
    i = len(nums) - 1
    while i >= 0:
        if nums[i] + 1 <= 9:
            nums[i] = nums[i] + 1
            break
        else:
            nums[i] = 0
            i -= 1
    if i < 0:
        nums.insert(0, 1)
    return nums

nums = [9, 9, 9]
print(solve(nums))

The output of the above code is ?

[1, 0, 0, 0]

Conclusion

This algorithm efficiently handles digit addition with carry-over by processing digits from right to left. It correctly handles all cases including when all digits are 9, requiring an additional digit at the front.