Why is it wrong, even though it is correct?

1

CodeChef: Practical coding for everyone my solution
CodeChef: Practical coding for everyone question

2

@ssjgz can you pls help//

3

Not sure where to start with this one:

[simon@simon-laptop][15:48:59]
[~/tmp/DONTSYNC/hackerrank/otherpeoples]>./compile-latest-cpp.sh 
Compiling shoyam-AVERAGE.cpp
+ g++ -std=c++14 shoyam-AVERAGE.cpp -O3 -g3 -Wall -Wextra -Wconversion -DONLINE_JUDGE -D_GLIBCXX_DEBUG -fsanitize=undefined -ftrapv
+ set +x
Successful
[simon@simon-laptop][15:49:03]
[~/tmp/DONTSYNC/hackerrank/otherpeoples]>cat failed_test_case.txt
15
28
28
23
15
16
53
53
19
4
30
31
28
22
52
20
[simon@simon-laptop][15:49:05]
[~/tmp/DONTSYNC/hackerrank/otherpeoples]>cat failed_test_case.txt | ./a.out 
shoyam-AVERAGE.cpp:8:26: runtime error: signed integer overflow: 13 * 479001600 cannot be represented in type 'int'
shoyam-AVERAGE.cpp:23:30: runtime error: signed integer overflow: 1932053504 * 2 cannot be represented in type 'int'
shoyam-AVERAGE.cpp:8:26: runtime error: signed integer overflow: 13 * 479001600 cannot be represented in type 'int'
shoyam-AVERAGE.cpp:8:26: runtime error: signed integer overflow: 14 * 1932053504 cannot be represented in type 'int'
shoyam-AVERAGE.cpp:8:26: runtime error: signed integer overflow: 15 * 1278945280 cannot be represented in type 'int'
shoyam-AVERAGE.cpp:25:14: runtime error: variable length array bound evaluates to non-positive value -4
21

First things first - there’s no need to use factorials in this :slight_smile:

1 Like
4

I think factorial is neccessary for finding the no of combination of t numbers where any 2 are added.

5 
#include <iostream>
using namespace std;

int fac(int n){
    if(n==1){ // Check if n == 0 too.
        return 1;
    }else{
        return(n*fac(n-1));
    }
}

int s=0;
int main() {
	int n;
	cin>>n;
	
	float r[n];
	
	for(int i=0 ; i<n ;i++){
	    cin>>r[i];    
	}
	
	int t= (fac(n))/(fac(n-2)*2); // Because n-2 can be -1
	
	float a[t];
	for(int i=0 ; i<n ;i++){
	    for(int j=i+1;j<n;j++){
	        a[s]= (r[i]+r[j])/2;
	        if(s==t-1){
	            break;
	        }else{
	            s++;
	        }
	    }
	}
	
	int s=0;
	for(int i=0 ; i<n ;i++){
	    for(int j=0;j<n;j++){
	        if(a[i]==r[j]){
	            s++;
	        }
	    }
	}
	
	
	cout<<s<<endl;
	
	return 0;
}

This is your code. I added two comments, look at them. And as said by @ssjgz , no need to calculate factorials. You can use two pointers and binary search with an overall time complexity of O(N\log{N}).

int t = (fac(n))/fac(n-2)*2);
// Can be re-written as
int t = (n * (n - 1)) / 2;
1 Like
6

Thnx @ssjgz @suman_18733097 for the help

2 Likes
7

Try submitting it in C++17