Leetcode Biweekly Contest 94 | 2513 : Minimize the Maximum of Two Arrays Solution | Newton School

[@KaranMashru]

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Code :
class Solution {
public:
long gcd(long a, long b)
{
if((a%b)==0){
return b;
}else{
return gcd(b, a%b);
}
}

void bs(long l, long h, int divisor1, int divisor2, long lcm, long * ans, int c1, int c2){
if(l

[@aseembaranwal]

WHat is the reason to choose 10^10 as the upper limit for the search space?

[@securelyinsaycure]

Very helpful video thank you so much!!!

[@rahulkhandelwal7034]

Bhai atcoder ke solution bhi post kiya kro , kaafi helpful h tmhre video

[@Rajat_maurya]

app bahut accha padhate ho kash mai 3rd year m apka channel dekha hota

[@LogicArena01]

Thanks for that explanation...❤

[@maddycoder1294]

[@affanmokarram121]

oh..bhai...mza agya....I like the explanation which is beginner friendly. Thanks

[@SB_09124]

Very nice explanation ...#concept clearr👍👍

[@ManishkumarRajakCS--]

Are you doing #6comp30days challenge by arsh goyal??

[@socify4410]

bhaii sab kya question tha and kya explaination hai

[@aryamankumar645]

Good job

[@DeepSingh-zs2oi]

very good explanation

[@factholicdunyia2049]

Very well explained.. thanku sir

[@ManishkumarRajakCS--]

Are you doing 6comp30day challenge by arsh goyal??

[@factholicdunyia2049]

@@ManishkumarRajakCS-- yes

[@manaskumarpanda8961]

@Karan Mashru
#include
class Solution {
public:
int find_gcd(int a , int b){
if(b == 0)return a ;
else return find_gcd(b , a % b) ;
}
public:
bool check(int divisor1 , int divisor2 , int uniqueCnt1 , int uniqueCnt2 , int mid , int lcm){

int temp1 = mid - mid/divisor1 , temp2 = mid - mid/divisor2 ;

int temp3 = mid - (mid/divisor1 + mid/divisor2 - (mid / lcm)) ;
int temp4 = mid - (mid / lcm) ;
if((temp1 >= uniqueCnt1) && (temp2 >= uniqueCnt2) && ((temp1 + temp2 - temp3) >= (uniqueCnt1 + uniqueCnt2)))return true ;
else return false ;

}
public:
int minimizeSet(int divisor1, int divisor2, int uniqueCnt1, int uniqueCnt2) {
int lo = 1 , hi = INT_MAX , ans = 0 ;
int lcm = ((long long)divisor1 * (long long)divisor2) / find_gcd(divisor1 , divisor2) ;

while(lo

[@ankitaryan5153]

just use long it will give you correct answer