Try this i will work

*it

Worked bro thanks alot

👍

Hi

I don't know c program

Anybody want to teach me

Why the height of a balanced binary tree O(logn), also explain me please what is the meaning of height being O(logn).

Anyone?

Anyone?

Just think about how the tree is built. If it's a binary tree every node has up to two children nodes. This mean that, assuming it is balanced, at each level you have 2M nodes, where M is the number of nodes of the previous level. Therefore if you have a total of N nodes, you can go as deep as log(N)

Just think about how the tree is built. If it's a binary tree every node has up to two children nodes. This mean that, assuming it is balanced, at each level you have 2M nodes, where M is the number of nodes of the previous level. Therefore if you have a total of N nodes, you can go as deep as log(N)

👍🏽👍🏽

Kisi ke pass DSA ke handwritten notes hai to please send me

Just think about how the tree is built. If it's a binary tree every node has up to two children nodes. This mean that, assuming it is balanced, at each level you have 2M nodes, where M is the number of nodes of the previous level. Therefore if you have a total of N nodes, you can go as deep as log(N)

What you are saying is it not true for any binary tree?

For a binary tree calculating height takes o(n) time right? Then why for height balanced tree it is o(logn)

What you are saying is it not true for any binary tree?

Not really. A binary tree can be developed just on "one side". A balanced one has to be developed equally both sides

Just Google some visual representation and you'll understand what I'm saying

Not really. A binary tree can be developed just on "one side". A balanced one has to be developed equally both sides

But a balanced tree can be like this too. That it have only one child at some nodes

Just Google some visual representation and you'll understand what I'm saying

I've searched but for the height of height balanced binary tree I have not found anything that can help my doubt

But a balanced tree can be like this too. That it have only one child at some nodes

Yes. In that case it is O(log(N) + 1) ~ O(log(N)) Did you get it? It is true that the deepest level can be uncompleted, but it does not really affect time complexity

So the height can be calculated in o(logn) time or does it mean the height is log n

So the height can be calculated in o(logn) time or does it mean the height is log n

The height is O(log(N)), therefore it takes logarithmic time to traverse it till bottom

But calculating the height will still take O(number of nodes) time right?

Because to calculate height, I make a recursive function which at every node ask the height of its left and right sub trees and returns it's height as max of both heights + 1

So to traverse each node I'm making n number of calls. n being the number of nodes

It takes O(log(N))

The number of times you function gets called is equal to the number of levels of the tree, right? Every time, starting from the root node, you check if it has a left or right children. If so, you call the same function on one of the two, right? Therefore you are calling that function for every tree level

The number of times you function gets called is equal to the number of levels of the tree, right? Every time, starting from the root node, you check if it has a left or right children. If so, you call the same function on one of the two, right? Therefore you are calling that function for every tree level

Ok I got it. The height function that I write for a simple binary tree will be different from the height function for a height balanced binary tree right

Ok I got it. The height function that I write for a simple binary tree will be different from the height function for a height balanced binary tree right

Yes

Thanks brother

Thanks brother

No problem

👍

#include <iostream> using namespace std; int main() { long num; cin>>num; cout<<endl; // Reading input from STDIN unsigned long long a,b,sum=0; long i; while(i<num) { cin>>a>>b; cout<<endl; long long j; while(j<=a) { if(j%b==0) { if((j%b)%b==0) { sum+=(j/b)/b; } else{ sum+=j/b; } } else { sum+=j; } } cout<<sum<<endl; } }

#include <iostream> using namespace std; int main() { long num; cin>>num; cout<<endl; // Reading input from STDIN unsigned long long a,b,sum=0; long i; while(i<num) { cin>>a>>b; cout<<endl; long long j; while(j<=a) { if(j%b==0) { if((j%b)%b==0) { sum+=(j/b)/b; } else{ sum+=j/b; } } else { sum+=j; } } cout<<sum<<endl; } }

Can anyone help me with the time complexity of this? When I am compiling it on hacker earth it is showing time limit exceeded error

#include <iostream> using namespace std; int main() { long num; cin>>num; cout<<endl; // Reading input from STDIN unsigned long long a,b,sum=0; long i; while(i<num) { cin>>a>>b; cout<<endl; long long j; while(j<=a) { if(j%b==0) { if((j%b)%b==0) { sum+=(j/b)/b; } else{ sum+=j/b; } } else { sum+=j; } } cout<<sum<<endl; } }

You're not initialising I and j

If j and i are 0 at start so the complexity will be O(a*num). What are the constraints

Can anyone send me resources of recursion,it seems to be difficult to understand in c++

/get cbook /get cppbookguide

I tried many times i can't figure it out

I'd suggest you: 1 - Provide an explanation about what your code should do 2 - State the expected output and the output or error you got 3 - Send a minimal working example of your faulty code in a well-formatted fashion

Anybody want to teach me

/get cbook /get cppbookguide

Can anyone help me with the time complexity of this? When I am compiling it on hacker earth it is showing time limit exceeded error

I'd suggest you: 1 - Provide an explanation about what your code should do 2 - State the expected output and the output or error you got 3 - Send a minimal working example of your faulty code in a well-formatted fashion

@OxFFFFFFFF What about the following message in the rules? 1 - Provide an explanation about what your code should do 2 - State the expected output and the output or error you got 3 - Send a minimal working example of your faulty code in a well-formatted fashion

#include <iostream> using namespace std; int main() { long num; cin>>num; cout<<endl; // Reading input from STDIN unsigned long long a,b,sum=0; long i; while(i<num) { cin>>a>>b; cout<<endl; long long j; while(j<=a) { if(j%b==0) { if((j%b)%b==0) { sum+=(j/b)/b; } else{ sum+=j/b; } } else { sum+=j; } } cout<<sum<<endl; } }

Hi @Mystery_man0007 please read the rules. Use pastebin or similar service.

@OxFFFFFFFF What about the following message in the rules? 1 - Provide an explanation about what your code should do 2 - State the expected output and the output or error you got 3 - Send a minimal working example of your faulty code in a well-formatted fashion

Well, if they don't they're reducing their chances

for (j = x; j <= 4 * x * y; j += y / x)

Well, if they don't they're reducing their chances

What about adding* Fair enough :)

#include <iostream> using namespace std; int main() { long num; cin>>num; cout<<endl; // Reading input from STDIN unsigned long long a,b,sum=0; long i; while(i<num) { cin>>a>>b; cout<<endl; long long j; while(j<=a) { if(j%b==0) { if((j%b)%b==0) { sum+=(j/b)/b; } else{ sum+=j/b; } } else { sum+=j; } } cout<<sum<<endl; } }

O(n^2) time complixty

Hello can anyone assist Describe how parameters are passed between a function call and a function in programming. For each way of passing parameter write a program segment to show how it is implemented.

Hello can anyone assist Describe how parameters are passed between a function call and a function in programming. For each way of passing parameter write a program segment to show how it is implemented.

We r serious autist not ur servants Kevin.. You may find ans to that in a YouTube vid...go on

Hello can anyone assist Describe how parameters are passed between a function call and a function in programming. For each way of passing parameter write a program segment to show how it is implemented.

/report

/report

Hm?

Hm?

Does it break those two rules? * Not checking your problem in google (or any other search engine) first will get you a warn. * Asking to solve an assignment/test/whatever without trying it first yourself will get you a warn or will get you BANNED. We won’t write a code for you, but we can push you forward and suggest something.

Hello guys for my needs I subclassed QProgressBar and created my own the way Im using it is: create an empty QFrame where I want to put my progress bar in it after that add my progress bar to it in my code with setParent, its ok, but problem is in this way the progress bar widget isn't center, so how can I center it? (without using layout of cource)

hello, whats the difference between int const value =5; const int value1 = 5;

hello, whats the difference between int const value =5; const int value1 = 5;

Same

hello, whats the difference between int const value =5; const int value1 = 5;

no difference, it is only different if they were pointers

no difference, it is only different if they were pointers

Yea

i read from txt pairs of numbers how can i check it only 2 numbers in each line and not 3 I want show the line is the problem for example:

the pairs file:

3 25 35 14 42 30 60 70

my code: scanf("%d", &numOfPairs); arrNum = (int *)malloc(sizeof(int) * numOfPairs * numbersInPair); for (int i = 0; i < numOfPairs * numbersInPair; i += 2) { int num1, num2; if (scanf("%d %d \n", &num1, &num2) == 2) { arrNum[i] = num1; arrNum[i + 1] = num2; } else { printf("illegal input at line %d \n", i); } }

console: illegal input at line 4

my code: scanf("%d", &numOfPairs); arrNum = (int *)malloc(sizeof(int) * numOfPairs * numbersInPair); for (int i = 0; i < numOfPairs * numbersInPair; i += 2) { int num1, num2; if (scanf("%d %d \n", &num1, &num2) == 2) { arrNum[i] = num1; arrNum[i + 1] = num2; } else { printf("illegal input at line %d \n", i); } }

That's not the correct way at all. Use std::getline to read whole line then parse it, for example by std::stringstream. If you need C approach, consider fgets to read and sscanf to parse.

/get

i read from txt pairs of numbers how can i check it only 2 numbers in each line and not 3 I want show the line is the problem for example:

Isn't it easier to count the number of blank spaces on each line? If there should be only one pair of numbers per line, then there should be exactly one horizontal blank space per line.

/get cbook /get cppbookguide

Can u send me guidebook of cpp also ,m currently working on cpp😅

Print this pattern in c++ (no of rows =4) ABCD BCDE CDEF DEFG

printf("ABCD\nBCDE\nCDEF\nDEFG");

printf("ABCD\nBCDE\nCDEF\nDEFG");

😂😂😂😂😂

Help me to write Java code Que.) Reverse a String and print it on console "Java Skills"

We r serious autist not ur servants Kevin.. You may find ans to that in a YouTube vid...go on

/report

/report

Hello can anyone assist Describe how parameters are passed between a function call and a function in programming. For each way of passing parameter write a program segment to show how it is implemented.

Your homework?

Heyyy

That's not the correct way at all. Use std::getline to read whole line then parse it, for example by std::stringstream. If you need C approach, consider fgets to read and sscanf to parse.

how change it?

int i, c; FILE *fp; char str[100]; fp = fopen("./pairs.txt", "r"); if (dup2(fileno(fp), STDIN_FILENO < 0)) { printf("Error opening file"); return -1; } if (feof(fp)) printf("Error reading file"); scanf("%d", &numOfPairs); arrNum = (int *)malloc(sizeof(int) * numOfPairs * numbersInPair); for (int i = 0; i < numOfPairs * numbersInPair; i += 2) { int num1, num2; scanf("%d %d", &num1, &num2); arrNum[i] = num1; arrNum[i + 1] = num2; }

If I have hash that stores the frequency of the key...then how do I create a vector which is sorted based on the frequency

int i, c; FILE *fp; char str[100]; fp = fopen("./pairs.txt", "r"); if (dup2(fileno(fp), STDIN_FILENO < 0)) { printf("Error opening file"); return -1; } if (feof(fp)) printf("Error reading file"); scanf("%d", &numOfPairs); arrNum = (int *)malloc(sizeof(int) * numOfPairs * numbersInPair); for (int i = 0; i < numOfPairs * numbersInPair; i += 2) { int num1, num2; scanf("%d %d", &num1, &num2); arrNum[i] = num1; arrNum[i + 1] = num2; }

Don't read integers, read whole line the way I told you. Google how to use those functions.

If I have hash that stores the frequency of the key...then how do I create a vector which is sorted based on the frequency

Iterate over that hash table adding values to the vector. Btw, you can't restore hashed key.

Don't read integers, read whole line the way I told you. Google how to use those functions.

Not working for me.

{ int index, i, numOfPairs; int c; FILE *file; file = fopen("./pairs.txt", "r"); if ((c = getc(file)) == EOF) { perror("Error opening file"); //or return 1; fclose(file); } while ((c = getc(file)) != EOF) putchar(c); { if (pid == ROOT) fscanf(file, "%d", &numOfPairs); allNumbers = (int *)malloc(sizeof(int) * numOfPairs * numbersInPair); //need multiply in 2 numbers for each pair while (!feof(file) && numOfPairs > 0) { int x, y, arrIndex = 0; numOfPairs--; fscanf(file, "%d %d", &x, &y); allNumbers[arrIndex] = x; printf("The X : %d\n", x); allNumbers[arrIndex + 1] = y; printf("THE Y : %d\n", y); arrIndex + 2; } fclose(file); }

it get array garbage

i wrote a code for this, "Construct Binary Tree from Preorder and Inorder Traversal". for sample test cases it runs fine but when i submit the code it gives a run time error and i can't understand the reason why