In this HackerEarth Group Photo problem solution, You and your friends want to take group photos. The process of taking photos can be described as follows:

On the photo, each photographed friend occupies a rectangle of pixels: the ith of them occupies the rectangle of width wi pixels and height hi pixels. On the group photo everybody stands in a line, thus the minimum pixel size of the photo including all the photographed friends is W * H, where W is the total sum of all widths and H is the maximum height among the heights of all the photographed friends.
The friends made n photos - the jth (1 <= j <= n) photo had everybody except for the jth friend as he was the photographer.
Print the minimum size of each made photo in pixels.

HackerEarth Group Photo problem solution

HackerEarth Group Photo problem solution.

using namespace std ;
#define pb push_back
#define mp make_pair
#define infile() freopen("sample.txt","r",stdin);
#define output() freopen("sampleout.txt","w",stdout);
#define ll long long
#define sc(t); scanf("%d",&t);
#define scl(t); scanf("%lld",&t);
#define sc2(n,m); scanf("%d%d",&n,&m);
#define scl2(n,m); scanf("%lld%lld",&n,&m);
#define debug(); printf("tushar\n");
#define N 200005
#define mod 1000000007
#define printi(n) printf("%d",n);
vector < ll > v ;
int n ;
int a[N] ;
int w[N],h[N];
int main()
int i , j , t ;
sc(n) ;
multiset < int > s ;
multiset < int > :: iterator it ;
ll sum = 0LL;
scanf("%d %d",&w[i],&h[i]) ;
sum = (ll)(sum + w[i]) ;
s.insert(h[i]) ;
for(i = 1 ; i <= n ; i++ )
ll tmp = (ll)(sum - w[i]) ;
it = s.find(h[i]) ;
s.erase(it) ;
it = s.end() ;
it-- ;
ll ans = (ll)(tmp * (*it)) ;
printf("%lld ",ans);
return 0 ;

Second solution

using namespace std;
int main()
int n;
int w[200005],h[200005],cpy[200005];
int sum=0;
assert(n>=1 && n<=2e5);
for(int i=0;i<n;i++)
assert(w[i]>=1 && w[i]<=10);assert(h[i]>=1 && h[i]<=1e3);
if(n==1){cout<<"0\n";return 0;}
for(int i=0;i<n;i++)
int wgt=sum-w[i];
int hgt=(cpy[n-1]==h[i])?cpy[n-2]:cpy[n-1];
cout<<wgt*hgt<<" ";
return 0;