Add to Array-Form of Integer - Leetcode 989 - Python

Solving Add to Array-Form of Integer - Leetcode 989, today's daily leetcode problem.

0:00 - Read the problem
1:02 - Drawing Explanation
9:00 - Coding Explanation

leetcode 989

#neetcode #leetcode #python

[titajohn]

I was stuck on this one and couldn't understand the solutions either. This video showed me the way. Thank you so much

[litheshshetty]

Complexity is not O(number of digits in K), you reversed array so its O(n), n is size of num.

[AhmedMohamed-eszo]

what's tool's name that you use to draw on the screen?

[MeMe-twxb]

Wouldnt reversing be an O(n) operation every time?

[techwithkiran]

By using pointers we don't need to reverse twice,

carry = 0
res = []
i = len(num)-1
for n in range(max(len(num), len(str(k)))):
digit1 = num[i] if i>= 0 else 0
digit2 = k%10 if k else 0
sum = digit1+digit2+carry
carry = sum//10
sum = sum%10 if sum>=10 else sum
res.append(sum)
i -= 1
k = k//10 if k else 0

if carry:
res.append(carry)
return res[::-1]

[vidhya]

Why reverse the number though, oh its for cases when k>numbers in array

[user.-mnix]

class Solution {
public List<Integer> addToArrayForm(int[] num, int k) {
List<Integer> list = new ArrayList<>();
int carry = 0;

for (int i = num.length - 1; i >= 0 || k > 0 || carry > 0; i--) {
int x = (i >= 0) ? num[i] : 0;
int sum = x + k % 10 + carry;
list.add(sum % 10);
carry = sum / 10;
k /= 10;
}

Collections.reverse(list);
return list;
}
}

[souvikmukherjee]

This is not supposed to be EASY level :(

[democreation]

270 ms code beats more than 95% submissions ?