Roman to Integer | LeetCode 13 | Coding Interview Tutorial

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Roman to Integer solution: LeetCode 13

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Roman to Integer | LeetCode 13 | Coding Interview
#romantointeger #leetcode #algorithms #terriblewhiteboard #codinginterview

Click the time stamp to jump to different parts of the video.
00:00 Title
00:06 Problem readout
02:22 Whiteboard solution
06:38 Coding solution
15:36 Result and outro

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Комментарии

If there are any videos you'd like me to make or if you have any ideas on how to optimize this solution, let me know!

TerribleWhiteboard

You legit have some of the best explanations and problem break downs. Drawing out the problem and solving it side by side with the code is an excellent idea.

alvarocastro-cid

Dude. Such a great job with the explanation! Thanks so much! Also, see you haven't posted in about 2 years, hope you're doing well.

CameraMeetsWorld

thank you for making these! Still helpful 2 years later! Your explanations are super helpful!

chips-n-salsa

Dude, this was so helpful with the visuals on the left side breaking down the code! liked and subbed

TheCaioFreitas

Best explanation so far, seen other videos from different channel but only video explaining in so much depth.

shashibhushansingh

Why aren't you uploading anymore? Your explanation makes the logic crystal clear.

rv-bz

Great explanation! Breaking down the problem and turning your logic into code step by step was a great way to tackle the problem. I believe it is a great approach :)

hunorvadasz-perhat

thanks for the video.Good explanation :)

AyushiSharmaDSA

Thank you so much, your videos are awesome

TolgayToklar

Very easy to follow explanation! Thanks!

logandallalio

Amazing. Thanks.
var romanToInt = function (s) {
const list = {
I: 1,
V: 5,
X: 10,
L: 50,
C: 100,
D: 500,
M: 1000,
};

let total = 0;

for (let i = 0; i < s.length; i++) {
let currentInt = list[s.charAt(i)]; //Look at the first roman numeral pass it through the map and get the value
let nextInt = list[s.charAt(i + 1)];

if (nextInt) {
//Checks for presence of a nextInt (edge case)
if (currentInt >= nextInt) {
//If currentInt is greater than or equal to nextInt
total += currentInt; //Add the current integer to the total
} else {
total += nextInt - currentInt; //If currentInt is not greater than or equal to nextInt we subtract and add it to total
i++; //Since we have accounted for both values we are incrementing to check next set
}
} else {
total += currentInt;
}
}
return total;
};

DmitriyMalayevProfile

Beautiful, I was following the other approach which was not elegant. Yours is better.
total += dicto[l1[i]]
total -= 2*dicto[l1[i-1]]

historyvspresent

my solution was indentical to this in cpp but the problem its slow, 40 ms to be exact, and is faster than 5.93% of cpp programs..
any ideas on why that's the case ? is there a faster implementation algorithm ??

seal

Thanks for the great explanation. But when i try this in python it gives an error for nextInt :- string index out of range
Can you please help?

mortysmith

Why are you not uploading videos? Please make more youtube videos

shivashankar

I think your spoken explanation makes more sense: the number 4 is written as IV. It's not that you do subtraction, it's that the written symbol IV is 4. The rule doesn't have anything to do with "the number on the left is smaller than the number on the right" - the rule is that IV is the way you write 4, and CM is 900. It's just a lookup table. (Having said that, I think the problem is so simple that anyone who does anything sensible is going to be fine in an interview.)

JamesMooreSeattle

A cleaner alternative that avoids O(n)^2 is

const romanToInt = (s) => {
let map = {
I: 1,
V: 5,
X: 10,
L: 50,
C: 100,
D: 500,
M: 1000
}
let total = 0
for (let i = 0; i < s.length; i++) {
let current = map[s[i]]
let next = map[s[i + 1]]
if (next && current >= next) {
total += current
} else if (next && current < next) {
total += next - current
i++
} else {
total += current
}
}
return total
};

hideinbush

better than nick white's 🤭 explanations

WhatIsThisAllAbout

Channel name should be wonderful whiteboard not a terrible whiteboard.

anandparmar

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