
The task at hand involves taking the digits of an integer represented as an array in left-to-right order, known as its array-form, and adding another integer k to it. The result of this addition must then be expressed in the same array-form. This problem challenges us to handle the operation directly on an array of digits, reflecting the same approach used in basic arithmetic.
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1 <= num.length <= 1040 <= num[i] <= 9num does not contain any leading zeros except for the zero itself.1 <= k <= 104The main trick in solving this problem is handling the carry in arithmetic correctly when adding the integer k to the number represented by array num. Here's how we can intuitively tackle this:
k digits of the array to an integer and start the addition from there.k to the current digit.k exceeds 9, handle the carry over.k with the carry value (sum divided by 10).sum % 10 to the resultant array to get the correct digit in place.k is still greater than 0), continue resolving by adding digits to the result.Given these steps, the process will simulate the traditional way we perform addition on paper, carrying over whenever a sum reaches two digits. Let's examine this method with the examples provided:
Example 1:
num = [1,2,0,0] and k = 34.k to each digit and resolve the carry.[1,2,3,4] which is indeed 1200 + 34 = 1234.Example 2:
num = [2,7,4] and k = 181.274 + 181 = 455.[4,5,5].Example 3:
num = [2,1,5] and k = 806.215 + 806 = 1021.[1,0,2,1].This approach leverages basic digit-by-digit addition and emphasizes understanding how carry figures in decimal addition. Thus, it aligns well with the constraints and ensures an accurate representation of sum in the array form.
The provided Java solution demonstrates the process of adding a single integer to the array-form representation of another integer. The method, named arrayFormSum, takes two arguments: an integer array numArray representing a number (where each element is a digit), and an integer addNum to be added to this number.
Follow these steps to comprehend the core approach implemented in the code:
len to hold the length of the numArray.carry to the value of addNum as the sum starts with this value.ArrayList named result to store the result of the sum in reverse order.while loop to iterate through the numArray from right to left. The loop continues as long as either there are more digits to process in numArray or there is a nonzero carry. Reduce the index ix by 1 at the start of each iteration.ix is valid (i.e., greater than or equal to 0), add the digit at numArray[ix] to carry.carry (i.e., carry % 10) to the result list. Then update carry by dividing it by 10 to drop the least significant digit.carry, reverse the result list to correct the digit order using Collections.reverse(result).result list which now represents the sum in array-form.These operations collectively convert the array-format number and the additional integer into a summed array-format number efficiently while managing carries over increased place values, applicable notably when significant overflow beyond the original numbers' lengths occurs.
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