A Game of Numbers

 Problem:-

You are given an array A of N integers. Now, two functions 

F(X)

 and 

G(X)

 are defined:

F(X)

 : This is the smallest number Z such that 

X<ZN

 and 

A[X]<A[Z]

G(X)

 : This is the smallest number Z such that 

X<ZN

 and 

A[X]>A[Z]

Now, you need to find for each index i of this array 

G(F(i))

, where 

1iN

 . If such a number does not exist, for a particular index i, output 1 as its answer. If such a number does exist, output 

A[G(F(i))]

Input :

The first line contains a single integer N denoting the size of array A. Each of the next N lines contains a single integer, where the integer on the 

ith

 line denotes 

A[i]

.

Output :

Print N space separated integers on a single line, where the 

ith

 integer denotes 

A[G(F(i))]

 or 1, if 

G(F(i))

 does not exist.

Constraints:

1N30000

0A[i]1018

Sample Input
8
3
7
1
7
8
4
5
2
Sample Output
1 4 4 4 -1 2 -1 -1
Time Limit: 1
Memory Limit: 256
Source Limit:
Explanation

Next Greater     Next Smaller
3 –> 7                 7 –>1
7 –> 8                 8 –>4
1 –> 7                 7 –> 4
7 –> 8                 8 –> 4
8 –> -1                -1 –> -1
4 –> 5                 5 –> 2
5 –> -1                -1 –> -1
2 –> -1                -1 –> -1


Code:-

#include<stdio.h>
int main()
{
int n,i,flag=0,j,k;
long long int a[30000];
scanf(“%d”,&n);
for(i=0;i<n;i++)
{
scanf(“%lld”,&a[i]);
}
for(i=0;i<n;i++)
{
flag=0;
for( j=i;j<n;j++)
{
if(a[i]<a[j])
break;
}
if(j==n){
printf(“-1 “);
flag=1;
}
else{
for(k=j;k<n;k++)
if(a[j]>a[k])
break;
}
if(k==n&&flag==0)
printf(“-1 “);
else
if(flag==0)
printf(“%lld “,a[k]);
}
return 0;
}





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