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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
Related
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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
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"unordered_set" is not working properly on codechef and giving wrong output on its online ide whereas I am getting right output on geeksforgeeks ide and codeblocks
for Input like 3 2 10 1 100 4 3 i am getting 4 rows as expected in codeblocks and geeksforgeeks because n+m-1 is 4 whereas I am getting only 2 rows in codechef what can be the reason behind and now how it will work on codechef?
#include<stdio.h>
#include<bits/stdc++.h>
#include<unordered_set>
using namespace std;
int main()
{
int n,m,c=0,d,i,j,sum;
int a[10000];
int b[10000];
unordered_set <int> s;
scanf("%d %d",&n,&m);
for(i=0;i<n;i++)
scanf("%d",&a[i]);
for(i=0;i<m;i++)
scanf("%d",&b[i]);
for(i=0;i<n;i++)
{
for(j=0;j<m;j++)
{
sum=a[i]+b[j];
if(s.find(sum)==s.end())
{
s.insert(sum);
printf("%d %d\n",i,j);
c++;
}
if(c>=(n+m-1))
{d=1;break;}
}
if(d==1)
break;
}
}
Your program exhibits undefined behavior, by way of accessing the value of an uninitialized variable d.
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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.
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I wrote this code and the output is not coming as expected. The function returns the negatives recursively. But it is not doing returning what it should be.
size_t recursive_x(stack<int> mystack){
int count=0;
if(!mystack.empty()){
if(mystack.top()<0){
count+=1;
}
mystack.pop();
count+=recursive_x(mystack);
}cout<<count<<endl;
return count;
}
int main(){
stack<int> mystack;
mystack.push(9);
mystack.push(-2);
mystack.push(-2);
mystack.push(1);
mystack.push(-3);
mystack.push(-1);
mystack.push(99);
mystack.push(-1);
mystack.push(1);
mystack.push(1);
recursive_x(mystack);
}
The output should be 5 but it is coming like this:
Output:
0
0
1
2
2
3
4
4
5
5
5
Assuming you do call the function, the output is exactly what I would expect.
The display shows the number of negative numbers on the stack at the point where
cout<<count<<endl;
is invoked.
It is invoked first for the innermost recursion, because you make your recursive call before doing any output for the current recursion.
count+=recursive_x(mystack);
}cout<<count<<endl;
If you just want the output 5, you could print the result returned from your initial call
int finalCount = recursive_x(mystack);
cout<<finalCount<<endl;
and remove the output in recursive_x().
Each recursive call will print the result.
You just need to move the print command to the main:
size_t recursive_x(stack<int> mystack){
int count=0;
if(!mystack.empty()){
if(mystack.top()<0){
count+=1;
}
mystack.pop();
count+=recursive_x(mystack);
}
return count;
}
int main(){
stack<int> mystack;
mystack.push(9);
mystack.push(-2);
mystack.push(-2);
mystack.push(1);
mystack.push(-3);
mystack.push(-1);
mystack.push(99);
mystack.push(-1);
mystack.push(1);
mystack.push(1);
cout<<recursive_x(mystack)<<endl;
}
Output:
5
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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.