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Asked: 2024-09-24T19:18:54+05:30 In:

You are given an array of digits representing a non-negative integer. Each element in the array corresponds to a single digit of that integer, where the most significant digit is at the start of the array. Your task is to develop a function that adds one to this integer and returns the resulting number as an array of digits. The solution should account for cases where the addition results in a carry that affects multiple digits. For example, if the input array is [2, 9, 9], your function should return [3, 0, 0] after adding one to the integer represented by the array. Make sure to handle edge cases such as when the number consists entirely of nines.

I have a fun little coding challenge for you! Imagine you have an array of digits that represents a non-negative integer. Each element in this array corresponds to a single digit in that integer, with the most significant digit at the beginning.

Here’s where it gets interesting: suppose you want to add one to this integer, but you need to do it in such a way that your function can handle carries gracefully. For example, if the array you have is [2, 9, 9], the integer represented is 299. Adding one to this gives you 300, which should be returned as the array [3, 0, 0].

What I’m curious about is how you would approach this problem, especially if the input array might consist entirely of nines. Think about it—if your input is [9, 9, 9], after adding one, the resulting number would be 1000, which should turn into the array [1, 0, 0, 0]. That’s a tricky edge case, right?

To give you a clearer picture of what to consider, here are some more examples:
1. Input: [1, 2, 3] should yield [1, 2, 4].
2. Input: [4, 3, 2, 1] should show [4, 3, 2, 2].
3. Input: [0] should just return [1], since you’re adding one to zero.

So now, the challenge is to write a function that can handle these scenarios smoothly. It should account for all cases, from a simple increment to handling multiple carries.

How would you structure your function? Would you loop through the array from the end to the beginning (to catch those carries)? Would you use a separate variable to track any carry-over? What kind of edge cases do you think are crucial to consider? I’m excited to see how you would code this up and what insights you might have on the problem! Let’s brainstorm together!

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    2 Answers

    1. To solve the problem of adding one to an integer represented as an array of digits, the approach involves iterating through the array from the last element (least significant digit) towards the first. This way, we can easily manage any carries that occur during addition. The algorithm starts by initializing a variable, `carry`, to 1, since we want to add one. We loop through the digits in reverse. For each digit, we add the `carry` to the current digit. If this sum reaches 10, we set the current digit to 0 and keep the carry as 1 for the next iteration. If it’s less than 10, we simply update the digit and reset the `carry` to 0, breaking out of the loop since there’s no more carry to handle.

      After iterating through the array, if the `carry` is still 1 (which occurs in cases like when the input is entirely nines, e.g., [9, 9, 9]), we need to insert a new most significant digit at the beginning of the array. This can be easily accomplished by using an array method to unshift a new digit. Finally, if the carry is not triggered, our function would return the modified array. This solution effectively handles various edge cases such as single-digit arrays, arrays filled with nines, or even an array with a single zero.


    2. Coding Challenge: Adding One to an Array of Digits

      Okay, here’s what I think about this problem! So, we have an array of numbers and we want to add one to it. Like, if we have [2, 9, 9], it should turn into [3, 0, 0] after adding one.

      I guess the tricky part is the carry-over, especially when we get to 9s. If we have a number like [9, 9, 9], it needs to change to [1, 0, 0, 0] after we add one. That sounds like a big jump!

      Here’s how I probably would do it:

      1. Start from the end of the array since that’s where we want to add one first.
      2. Use a loop to check each digit. If it’s less than 9, I could just add one and finish up. Yay!
      3. But if it’s a 9, I guess I should set it to 0 and carry over the 1 to the next digit.
      4. I might need to keep doing this until I run out of digits. If I get to the front and still have a carry, that means we need to add another digit at the start.

      Here’s a rough idea of how the code might look like:

            function addOne(digits) {
          for (let i = digits.length - 1; i >= 0; i--) {
              if (digits[i] < 9) {
                  digits[i]++;
                  return digits;
              }
              digits[i] = 0; // Reset current digit to 0 if it's 9
          }
          // If we get here, all the digits were 9
          digits.unshift(1); // Add a new most significant digit
          return digits;
      }

      So, I think this should handle the cases! I’m not super sure but it seems kind of straightforward if I break it down step by step. I guess the edge cases are mostly when we have lots of 9s. I hope this makes sense!


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