Count Number of Peaks CodeChef Solution – Code Drive Solution

About Count Number of Peaks CodeChef Solution

• This is a coding contest based on algorithms, data structures, and problem-solving.
• Organizer: The contest is hosted by NIT Trichy.
• Prizes: NA
• Registrations for prizes: NA

Problem: Count Number of Peaks CodeChef Solution

You are given an integer NN where N≤10N≤10. Consider any array AA with length NN where each element can either be 00, 11, or 22, we define f(A)f(A) as the number of extrema in AA. You need to find the sum of f(A)f(A) over all possible arrays AA.

Note:

In an array AA, we consider AiAi as an extremum if it is strictly greater than both of its neighbors (i.e. Ai>Ai−1Ai>Ai−1 and Ai>Ai+1Ai>Ai+1), or if it is strictly smaller than both of its neighbors (i.e. Ai<Ai−1Ai<Ai−1 and Ai<Ai+1Ai<Ai+1). Note that first and last elements are not counted as extremum.

Extrema is the plural of extremum.

• The first line of the input contains an integer TT – the number of test cases. The test cases then follow.
• The only line of each test case contains one integer NN

Output Format

For each test case, output on a single the sum of f(A)f(A) over all possible arrays AA.

Constraints

• 1≤T≤101≤T≤10
• 1≤N≤101≤N≤10

Sample Input 1 

3
1
3
5

Sample Output 1 

0
10
270

Explanation

• Test case 11:
  • A=[0]A=[0]. Then f(A)=0f(A)=0.
  • A=[1]A=[1]. Then f(A)=0f(A)=0.
  • A=[2]A=[2]. Then f(A)=0f(A)=0.

Therefore the answer is 00.

• Test case 22: There are 1010 arrays AA with f(A)=1f(A)=1:
  • A=[0,1,0]A=[0,1,0].
  • A=[0,2,0]A=[0,2,0].
  • A=[0,2,1]A=[0,2,1].
  • A=[1,2,0]A=[1,2,0].
  • A=[1,2,1]A=[1,2,1].
  • A=[2,1,2]A=[2,1,2].
  • A=[2,0,2]A=[2,0,2].
  • A=[2,0,1]A=[2,0,1].
  • A=[1,0,2]A=[1,0,2].
  • A=[1,0,1]A=[1,0,1].

The remaining AA’s has f(A)=0f(A)=0. Therefore the answer is 1⋅10=101⋅10=10.

Solution: Count Number of Peaks CodeChef Solution

Count Number of Peaks CodeChef Solution Using C++

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

#define ff              first
#define ss              second
#define int             long long
#define pb              push_back
#define mp              make_pair
#define mt              make_tuple
#define pii             pair<int,int>
#define vi              vector<int>
#define mii             map<int,int>
#define umii            unordered_map<int,int>
#define pqb             priority_queue<int>
#define pqs             priority_queue<int,vi,greater<int> >
#define setbits(x)      __builtin_popcountll(x)
#define mod             1000000007
#define inf             1e18
#define ps(x,y)         fixed<<setprecision(y)<<x
#define mk(arr,n,type)  type *arr=new type[n];
#define range(a,b)      substr(a,b-a+1)
#define endl            "\n"
#define FIO             ios_base::sync_with_stdio(false);cin.tie(NULL);


int32_t main(){
    FIO;
    int t;cin>>t;
    while(t--){
        int n;cin>>n;
        int ans=0;
        if(n==3){
            cout<<10<<endl;

        }
        else if(n>3){
            ans=pow(3,(n-3))*8*(n-2);
            ans=ans+ans/4;
            cout<<ans<<endl;
        }
        else{
            cout<<0<<endl;
        }
    }
}

Count Number of Peaks CodeChef Solution Using Python

# cook your dish here
for _ in range(int(input())):
    n = int(input())
    s = 0
    if(n==3): print(10)
    elif(n>3):
        s = pow(3,(n-3))*8*(n-2)
        s+= s//4
        print(s)
    else:
        print(0)

Contest Details:

• This is an External Rated Contest.
• Duration: 3 Hours
• Start time: 30th December 2021, 20:00 hrs IST
• End time: 30th December 2021, 23:00 hrs IST
  You may check your timezone here.
• This contest is rated only for Division 2 and Division 3 users. Division 1 users can participate unofficially in this contest.

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