Divide Two Integers | Leetcode - 29

Detailed Explanation for Divide Two Integers Leetcode 29 problem.
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[@whys2016]

There is a bug in the code. Taking the absolute value is dangerous because abs(Int.MIN_VALUE) causes overflow, and you will end up with Int.MIN_VALUE

[@codeman0017]

there is an edge case to handle it my friend

[@gcziprusz]

I struggled to understand even after watching this the intuition behind doubling of the divisor.

[@deepaksaroha522]

yes, explanation is not there, just algorithm..

[@zr60]

This guy literally just copy pasted the solution and read the code out loud LOL. He probably doesn't understand the intuition behind it either.

[@bmuralikrishna8054]

As per my understanding, the intuition behind this problem is, If we can exactly remove the dividable value from the given value then, finally we can reach the remaining value which cannot be dividable. We can do this by subtracting the divisor from the numerator but it is not efficient. Instead, we can do doubling the denominator and can reach the edge quickly. Once we reach the edge or cross the edge, just subtract the value from the numerator and the properly dividable value will be left out and the number of times that value can be dividable is 1

[@rohanjuneja1778]

yes thats correct . @@bmuralikrishna8054

[@MrLeyt1125]

(divisor

[@aaditya3937]

If we divide 15/2 in your way i.e, doubling the divisor then I'am getting the soution as: 3+2+1(6) which actually is wrong(ans must be 7)

[@DivyanshSaraswat-fl9iz]

04:14 What's the result variable showing here?

[@himanshuchhikara4918]

how time complexity is logN^2 can you please explain little bit

[@gouravkrroy3625]

i dont understand some lines!..1.if (dividend==1

[@anubhavnegi4230]

1. yeah first statement is wrong 2. if both are -ve then it becomes false == false so condition becomes true hence it is correct

[@hangto0401]

Sorry but I don't understand time complexity why time complexity is O(logN^2). Could you please explain that? :(

[@rm0ndo]

Outer loop will run for log n times (dividend is continuously reduced by 2^count) and inner loop also runs in log n time (divisor keeps doubling). So total time complexity is O(log n * log n) = O(log n^2)

[@shubhamgarge7001]

i doesn't understand overflow condition

[@bcinerd]

it shows error interger overflow and runtime error

[@radoonhadoon]

im getting overflow at dividend - (divisor

[@AshishKumar-tb6go]

getting the same have u sort that out?

[@radoonhadoon]

@@AshishKumar-tb6go it's working in java but not in c++. I cannot understand why it isn't working in c++, if you find out why , please reply

[@marvellouschandan]

​@@radoonhadoon @Ashish Kumar We need to handle the case when our variables going to overflow, for that you have convert all variables to long data type. Below solution might help class Solution { public: int divide(int dividend, int divisor) { if (dividend == 1 result is negative. bool sign = (dividend >=0) == (divisor >=0) ? true : false; long divd = abs(dividend); long divs = abs(divisor); long result=0; while (divd >= divs) { long count=0; // We are doubling each time while (divd >= divs

[@mohitmundotia8950]

@@marvellouschandan Thanks got it

[@divya-ul9vi]

use long long instead of int for c++.

[@manupathria1073]

Wrote same code in c++. Getting runtime error on Dividend= INT_MAX DIVISOR=1 please help to understand the base case of overflow.

[@abhishek-singh31]

Divisor = -1

[@anubhavnegi4230]

Put it dividend = INT_MIN divisor = -1

[@satyamgupta6030]

what a great solution just one correction . tc will be O(log N) because inner loop will run for O(logN) time for the first time and when inner control come to inner loop 2nd time with the new dividend value it will always execute for 1 time therefore the tc - will be O(LogN) only? Please correct me if I'm wrong. btw thanks for the great solution .

[@ashutoshdas3010]

isn't abs a mod operator? @AlgorithmsMadeWrong

[@21sivansrawat]

1 condition missing in sign assignment that if both dividend and divisor are negative then also sign will be positive public int divide(int dividend, int divisor) { if(dividend == 1= 0) || (dividend

[@your_daddy_mf]

NAIVE THIS IS ALSO HANDLED BY SAME LINE

[@anubhavnegi4230]

hey your first line of code seems wrong , isnt dividend an int so max value supplied to it would 1

[@pironobcoding]

you have explained really well, but I can't understand your code its really complex

[@pranaynagpure2804]

Time complexity is wrong it should be O(log n *log n ) that is log^2 n because both loops exist log n times

[@GHOST-mg7uu]

int divid(long long dividend, long long divisor) { // Calculate sign of divisor i.e., // sign will be negative only iff // either one of them is negative // otherwise it will be positive int sign = ((dividend < 0) ^ (divisor < 0)) ? -1 : 1; // remove sign of operands dividend = abs(dividend); divisor = abs(divisor); // Initialize the quotient long long quotient = 0, temp = 0; // test down from the highest bit and // accumulate the tentative value for // valid bit for (int i = 31; i >= 0; --i) { if (temp + (divisor

[@MrLeyt1125]

You cannot use long long, read descrpition: Note: Assume we are dealing with an environment that could only store integers within the 32-bit signed integer range

[@anirudhrathore6601]

ok but can you explain how you can get absolute of -2147483648 and then solving it for divisor 2 will be hard because doubling 1073741824 will return negative number making it an infinite loop.

[@himanshukandwal8710]

why (1

[@osamaayman3765]

it is actually Integer.MIN_VALUE because it is written in the 2's complement from: the first bit from the left is taken with a negative sign and all other bits are positive. So, 1

[@adrikagupta5573]

What is the intuition behind this?

[@manojboganadham9071]

yes i too can not understand that

[@kondapanenitinusripriyanka7719]

Understood??

[@eeshanchanpura3356]

its giving wrong result for input (1,-1)

[@AliveGamerNo1]

class Solution { public int divide(int dividend, int divisor) { // lets go if(dividend==Integer.MIN_VALUE && divisor==-1) return Integer.MAX_VALUE; // if(dividend==Integer.MIN_VALUE && divisor==1) return Integer.MIN_VALUE; int temp=0; int ans=0; int sign = (dividend

[@Soh965]

That was awesome. Such an amazing solution... How did people come up with these, dang

[@AlgorithmsMadeEasy]

Thanks!!

[@reethikavanaparthy6984]

Does Math.abs work for INT_MIN??

[@MDFAIZANSARI]

The algo didn't work for 10 / 2

[@rajkaran9720]

Hi As per your explanation the below code is working for 10 ,3 but not for 16,3, can you help me what i'm missing here public int divide2(int dividend, int divisor) { int result = 0; int temp = divisor; while(dividend >= divisor) { int count = 0; while(dividend-temp >=0 ) { count++; temp1; result = result + count; } return result; }

[@khatariinsaan5284]

Code is wrong..Here is the Right Code if(dividend == 1=0) ? true : false; dividend = Math.abs(dividend); divisor = Math.abs(divisor); int result=0; while(dividend - divisor >= 0 ){ int count=0; while(dividend - (divisor

[@gaganbedi4332]

I watched it 3 times, but did't understood what he did in inner while loop.

[@shivanshupandey3637]

try to take few more testcases....not satisfying....

[@ashishkumarjayswal]

bhai tujhe hame samjhana hai ,ye nahi btana ki tujhe aata hai

[@somnathmaityik]

op🤣🤣🤣🤣

[@vamshi3307]

🤣🤣🤣

[@Man_of_Culture.]

🤣

[@vishichaurasiya4427]

🤣🤣🤣

[@suzy1107]

The solution couldn't pass certain tests unfortunately..

[@suzy1107]

nvm if I write a >= b it won't work. If I write a - b = 0, it works..

[@iitjee258]

nice explanation

[@TheShrivivek]

class Solution { public int divide(int dividend, int divisor) { int sign = 1; boolean add = false; if (dividend == divisor) return 1; if (divisor == Integer.MIN_VALUE) return 0; if (dividend == Integer.MIN_VALUE && divisor == -1) return Integer.MAX_VALUE; if (divisor == 1 ) return dividend; else if (divisor == -1) return -dividend; sign = (dividend >=0) == (divisor>=0) ? 1 : -1; divisor = Math.abs(divisor); if (dividend == Integer.MIN_VALUE) { dividend = Integer.MIN_VALUE + divisor; add = true; } dividend = Math.abs(dividend); int ans = 0; while (dividend >= divisor){ dividend -= divisor; ans++; } if (add) return sign*(ans+1); return sign*ans; } } simple approach java solution

[@aashishshah1187]

Great explanation 🔥🔥 , i suffer entire day to solve this question finally i will have sound sleep 🙂

[@AlgorithmsMadeEasy]

Glad you liked it !! Keep growing !!

[@zr60]

you solve it but you don't understand it what's the point?

[@nitianvlog]

time waste kra diye optimal appraoch bataye hi nahi

[@amolmishra8621]

Awesome explanation!

[@AlgorithmsMadeEasy]

Thanks @Amol.

[@yt-spikegaming7394]

I was guys who liked this video at 666th count XD

[@hymnish_you]

Nice Explanation

[@AlgorithmsMadeEasy]

Thanks

[@rajeshbammidy180]

If you do Math.abs(Integer.MIN_VALUE) it will give you Integer.MIN_VALUE only so this would fail for the some testcases

[@gcziprusz]

There is a bug on line `boolean sign = (dividend>=0) == (divisor>=0) ? true: false;` This example should yield a positive result -2/-1

[@mohdanas458]

what's wrong in it ? it is showing the positive result as LHS is equal to RHS condition true answer would be true

[@suyashrahatekar4964]

The thing I hate about all solns on youtube are they dont explain the mathematical logic behind this. They present it like an algorithm that we have studied in 1st grade itself. Gave a dislike to all of em but fuck youtube

[@ShreyaSingh-gm1pk]

It's too complex

[@shivaarukonda2640]

poorly explained

[@ItsYourCandyBro]

Bro 4

[@AlgorithmsMadeEasy]

4

[@hardikp5192]

error in this case -2147483648 -1