Closed. This question needs debugging details. It is not currently accepting answers.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Closed 4 years ago.
Improve this question
I was trying the problem Hash Tables: Ice Cream Parlor on Hackerrank. It is a simple question, but here is the bizzare situation i got to. How does the change of data structure matter?
Case 1:
void whatFlavors(vector<int> cost, int money) {
int ans1,ans2;
vector<int> arr(100,0); //notice this
for(int i=0;i<cost.size();i++){
if(arr[money-cost[i]]!=0){
ans1=i+1;ans2=arr[abs(money-cost[i])];
if(ans1>ans2){
cout<<ans2<<" "<<ans1<<endl;
}else{
cout<<ans2<<" "<<ans1<<endl;
}
break;
}
else{
arr[cost[i]]=i+1;
}
}
}
And output is:
Case 2:
code:
void whatFlavors(vector<int> cost, int money) {
int arr[100]={0}; //notice this
int ans1,ans2;
for(int i=0;i<cost.size();i++){
if(arr[money-cost[i]]!=0){
ans1=i+1;ans2=arr[abs(money-cost[i])];
if(ans1>ans2){
cout<<ans2<<" "<<ans1<<endl;
}else{
cout<<ans2<<" "<<ans1<<endl;
}
break;
}
else{
arr[cost[i]]=i+1;
}
}
}
output:
Let's just notice this part of your code:
if(arr[money-cost[i]]!=0){
ans1=i+1;ans2=arr[abs(money-cost[i])];
This means that your expect money-cost[i] to be negative for some values of i. So you end up reading locations that are outside your array (vector or array) which will lead to undefined behavior in both cases.
Related
Closed. This question needs debugging details. It is not currently accepting answers.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Closed 2 years ago.
Improve this question
Is there any difference in these type of calculating ans in dynamic problems ? Because in case of Code 3 I am getting segmentation fault while Code 1 and Code 2 are passing
int dp[8005][2];//dp is initialized as -1 in main
int a=4;
int b=6;
int tar=90;
int solve(int pos,int ba)
{
if(pos<0||pos>8000||ba>=2)
return 1e6+1;
if(pos==tar)
return 0;
//Code 1
int &ans=dp[pos][ba];
if(ans!=-1)
return dp[pos][ba];
ans=1+solve(pos+a,0);
ans=min(ans,1+solve(pos-b,ba+1));
return ans;
//Code 2
if(dp[pos][ba]!=-1)
return dp[pos][ba];
dp[pos][ba]=1+solve(pos+a,0);
dp[pos][ba]=min(dp[pos][ba],1+solve(pos-b,ba+1));
return dp[pos][ba];
//Code 3
if(dp[pos][ba]!=-1)
return dp[pos][ba];
int ans=1+solve(pos+a,0);
ans=min(ans,1+solve(pos-b,ba+1));
return dp[pos][ba]=ans;
}
Code 3 doesn't update dp until after all the calculations are done.
The first two go
Recurse
Update dp
Recurse
Update dp
Return new value
Note that 1 and 2 are equivalent, since ans is the same as dp[pos][ba].
The third is
Recurse
Recurse
Update dp and return the new value
Closed. This question needs debugging details. It is not currently accepting answers.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Closed 3 years ago.
Improve this question
Problem: https://www.hackerrank.com/challenges/migratory-birds/problem
Test Case Link: https://hr-testcases-us-east-1.s3.amazonaws.com/33294/input04.txt?AWSAccessKeyId=AKIAJ4WZFDFQTZRGO3QA&Expires=1573162301&Signature=MgpSHa3lxX%2FwYwumjzAmF8uviZE%3D&response-content-type=text%2Fplain
for this test case i cant get any output why ?
Thanks in advance
#include<iostream>
using namespace std;
int main()
{
long long n,i=0,num=0,mx=0,r=0;
cin>>n;
long long arr[6]{0};
for(int i=0;i<n;i++)
{
cin>>num;
arr[num]++;
}
for(int i=1;i<6;i++)
{
if(arr[i]>mx)
{
mx=arr[i];
r=i;
}
}
cout<<r;
}
This looks problematic:
long long arr[6]{0};
for(int i=0;i<n;i++)
{
cin>>num;
arr[num]++;
}
If the input value read into num is greater than 5 (or less than zero), undefined behavior will exist when a value is written into an invalid memory location offset from arr.
There is nothing wrong with you code i got accepted all test cases. But the solution can be return in more optimized way
Closed. This question needs debugging details. It is not currently accepting answers.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Closed 8 years ago.
Improve this question
I want to reverse the code as it complete first time. For example pin 1-pin 2-pin 3-pin 4 (it's complete) now it should run as pin 4-pin 3-pin 2-pin 1.
I wrote this code but it's not working in reverse order. Please guide me in this way.
#include<htc.h>
__CONFIG(1,OSCSDIS & HSPLL);
__CONFIG(2,BORDIS & PWRTDIS &WDTDIS);
__CONFIG(3,CCP2RC1);
__CONFIG(4,LVPDIS & STVREN);
__CONFIG(5,UNPROTECT);
__CONFIG(6,WRTEN);
__CONFIG(7,TRU);
define _XTAL_FREQ 40000000
void delay_sec(unsigned char seconds) // This function provides delay in terms of seconds
{
unsigned char i,j;
for(i=0;i<seconds;i++)
for(j=0;j<100;j++)
__delay_ms(10);
}
void led_display(char a)
{
switch(a)
{
case 0: PORTB=0x01;PORTD=0x08; break;
case 1: PORTB=0x02;PORTD=0x04; break;
case 2: PORTB=0x04;PORTD=0x02; break;
case 3: PORTB=0x08;PORTD=0x01; break;
}
}
void main()
{
TRISB=0x00; TRISD=0x00; char a,b;
while(1) {
led_display(a);
a++;
delay_sec(1);
if(a==4) {
a--;
}
}
}
For doing that, you have to remember in which order you are running (reverse or not). So you will have a variable indicating if the order is reverse or not, and change it when you reach the extremities of your counter (0 and 3)
And you can optimize the code with using 1 and -1 for the variable which remembers order and adding it to a.
Your code will seem like this in your main :
int reverse = 1,
char a = 0;
while(1)
{
led_display(a);
delay_sec(1);
if(a==3)
{
reverse=-1;
}
if(a==0)
{
reverse=1;
}
a+=reverse;
}
Regards.
Closed. This question is opinion-based. It is not currently accepting answers.
Want to improve this question? Update the question so it can be answered with facts and citations by editing this post.
Closed 8 years ago.
Improve this question
This is the part that brakes, zod1 zod2 zodA are vectors. In vectors I store int. Cant tell why it crashes cause it gives no error
if(zod1[u]+zod2[y]==zodA[t] || (zod1[u]+zod2[y])%10==zodA[t])
Questions:
1 Is if statement over complicated
2 Can I use vectors like that for solving math in if statement (sorry can't explain better)
3 If 2 question is true how should I solve it
Explanations:
variable names are in my own language (sorry if it looks weird)
Values are zero for all (ex zod1[u]=0)
added whole function (variables going to the function are passed correctly and I know I use some unnecessary thing)
void calc(vector<char> zodis1, vector<char> zodis2, vector<char> zodisAts,int zo1,int zo2,int zoA)
{int i,keliamas=0;
int k =0;
vector<int> zod1(0);
vector<int> zod2(0);
vector<int> zodA(0);
for(i=0;i<zodis1.size();i++)
{
zod1.push_back(0);
}
for(i=0;i<zodis2.size();i++)
{
zod2.push_back(0);
}
for(i=0;i<zodisAts.size();i++)
{
zod2.push_back(0);
}
int u=zodis1.size()-1;
int y=zodis2.size()-1;
int t=zodisAts.size()-1;
if(zod1[u]+zod2[y]==zodA[t] || (zod1[u]+zod2[y])%10==zodA[t])
{//if((zod1[u]+zod2[y])/10==1)
{
keliamas=1;
}
if(u==0||y==0||t!=0)
{
if(keliamas==1)
{
}
}
u--;
y--;
t--;
}
else
{if(zod1[u]!=9)
zod1[u]=zod1[u]+1;
else
{ if(u!=zodis1.size()-1)
u++;
else
{
cout<<"something wrong man";
}
}
}
}
EDIT:
Error is here:
for(i=0;i<zodisAts.size();i++)
{
zod2.push_back(0); // should be zodA
}
Is if statement over complicated
No, I think it is very simple with just one 'or'. However, if zodA[t] is always less than 10, then your if condition can be written as:
if ( (zod1[u]+zod2[y])%10==zodA[t] )
Can I use vectors like that for solving math in if statement (sorry
can't explain better)
Yes you can.
If 2 question is true how should I solve it
If it compiles but then crashes, then probably you are accessing out of bounds indices. Check that your indices are less than the vector sizes.
Closed. This question needs details or clarity. It is not currently accepting answers.
Want to improve this question? Add details and clarify the problem by editing this post.
Closed 8 years ago.
Improve this question
Can someone help me with my code ? I'm trying to use different algorithm but it returns way to big numbers, when i use algorithm between /* */ it works perfect, anyone can see whats wrong with my new code ? (same on java works)
int* czynnikiPierwsze(int n)throw (string){
if(n<0){
string wyjatek1="Nie mozna rozlozyc ujemnej liczby";
throw wyjatek1;
}
int b=0;
while(n>2){
n=n/tab[n-2];
b++;
}
dzielniki=new int[b]();
int j=0;
while(n>2){
dzielniki[j]=tab[n-2];
n=n/tab[n-2];
j++;
}
/* int a=n;
int*dzielniki=new int[30]();
for(int j=0;j<n+a;j++){
while(n>2){
dzielniki[j]=tab[n-2];
n=n/tab[n-2];
break;
}
}*/
return dzielniki;
}
There's no chance your second while(n>2) loop runs even once, since the first loop only exited when that same condition was no longer true.