Find maximum sum possible equal sum of three stacks | Greedy | DSA Sheet | Amazon 🔥
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- Опубликовано: 12 сен 2024
- #greedy #competitiveprogramming #coding #dsa
Hey, Guys in this video I have explained with code how we can solve the problem 'Find maximum sum possible equal sum of three stacks'.
CODE-- ideone.com/B22D4L
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Superb explanation
You were supposed to do in O(1)space
Okay so , finally after 4 days (exactly at 4 th day ) I had completed this greedy playlist ... Thank u ❤️✨
bro how can we calculate sum given that we can only pop an element from
top of the stack?
can we do this by a priority queue
thnk bro bcz of this i completed greedy with all questions
Ye series krne k liye thank you so much puri complete Krna please ❤️
Bhaiya coding mai khud se seekkh raha hoo aur sawal ho hi nahi rahe hai.. mai kya karu aap ko coding karte dekhta hoo to siochta ho kaise kar lete ho aap..please kuch bta do aap kaise seekh ho aap..
practice karo bhai ho jayega aasan lagega
@@justanaverageguy4739 bro ye code jo bhai ne karaya ye time limit exceeded dikha Raha hai. maine bhi yehi logic likha tha but thoda alag uska bhi time limit exceeded aaraha tha. Fir ye dekhke kiya to bhi same.
Bhaiya please ek course banao interview preparation ka . .. bdw super as usual 🙏
Huffman Encoding(Greedy) pe bhi video bana do plz
Bro can you upload your codes somewhere like in github so that we can refer them before few days of interview??
Bro please make a video on which algorithm is used in which cases 🙏🙏
Just like you explain dp significance in the problem while solving
OK
Please make a video about how much competitive programming should we do to clear coding rounds?
ok
Plz sir tell about any online internship which is madantory for my 5th sem
The rate might be the factor so tried with this way ->
#include
using namespace std;
bool check(int arr[]){
for(int i=1;i pr2.first;
}
int main(void){
int tcs{0};
cin>>tcs;
while(tcs--){
int n1,n2,n3;
cin>>n1>>n2>>n3;
int no;
pair arr[3];
vector v(3);
int sum[3] = {0};
for(int i=0;i>no;sum[0]+=no;v[0].push(no);}
for(int i=0;i>no;sum[1]+=no;v[1].push(no);}
for(int i=0;i>no;sum[2]+=no;v[2].push(no);}
for(int i=0;i