[Codeforces]- Help needed in Puzzles(337A) problem.

help
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1

Here is my code. I have understood how to solve this using tutorial approach, but I want to know why my code is giving a WA and what changes should I make to get it accepted.

YOU GUYS ARE A CHARM. THANKS IN ADVANCE…

// JUST LIKE ANIMALS !!!!

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

int main(){
int n,m;cin>>n>>m;vector<int> v(n+2);
int i,ans=INT_MAX,mini,maxi;
for(i=0;i<m;i++)cin>>v[i];
for(i=0;(m-n?i<m-n:i<1);i++){
	maxi=*max_element(&v[i],&v[i+n]);
	mini=*min_element(&v[i],&v[i+n]);
	ans=min(maxi-mini,ans);
}
cout<<ans;
return 0;
}

The accepted solution after corrections as suggested by @mukul_chandel.

// JUST LIKE ANIMALS !!!!

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

int main(){
int n,m;cin>>n>>m;vector<int> v(m+2);
int i,ans=1001,mini,maxi;
for(i=0;i<m;i++)cin>>v[i];
sort(&v[0],&v[m]);
for(i=0;i<=m-n;i++){
    maxi=*max_element(&v[i],&v[i+n]);
    mini=*min_element(&v[i],&v[i+n]);
    ans=min(maxi-mini,ans);
}
cout<<ans;
return 0;
}
1 Like
2
  • size of vector should be m.
  • no point of applying this algorithm without sorting. After sorting you can also find min and max without even using max_element() and min_element() which will also give your solution a better time complexity.
  • Here is your corrected solution.
1 Like
3

@mukul_chandel , Thanks for helping me out but your solution is done using the tutorial approach and I wanted to done it using mine. I was just doing a boo-boo which you pointed out. Thanks :slight_smile: :slight_smile: I will be adding my accepted solution too . :slight_smile:

1 Like
4

#include<stdio.h>

void bubbleSort(int numbers[], int array_size)
{
int i, j, temp;

for (i = (array_size - 1); i >= 0; i–)
{
for (j = 1; j <= i; j++)
{
if (numbers[j-1] > numbers[j])
{
temp = numbers[j-1];
numbers[j-1] = numbers[j];
numbers[j] = temp;
}
}
}
}

int main()
{
int a,b,x;
scanf("%d %d",&a,&b);
int ar[b];
for(x=0;x<b;x++)
{
scanf("%d",&ar[x]);
};

 bubbleSort(ar,b);
 
int min=ar[a-1]-ar[0];
for(x=0;x<b-a+1;x++)
{
   if ((ar[a-1+x]-ar[x])<min) min=(ar[a-1+x]-ar[x]);
};

printf("%d",min);


return 0;

}

My solution for the avove problem in c

5

Here is my solution to this problem.