Array Rotation Codechef Solution|Problem Code: COF1MED1

Array Rotation Codechef Solution: Hello coders, today we are going to solve Array Rotation Codechef Solution.

Codechef solutions

Problem

You are given an array A of N integers. You are to fulfill M queries. Each query has one of the following three types:

  • L d : Rotate the array A to the left by d units.
  • R d : Rotate the array A to the right by d units.
  • Q d : Query the element currently at the d-th index in the array, after all earlier rotations have been carried out.

Input:

The first line contains two numbers – N and M respectively.

The next line contains N space separated Integers, denoting the array A.

Each of the following M lines contains a query in the one of the forms described above.

Output:

For each query of type Q output the answer on a separate line.

Constraints

  • 1≤N≤1000001≤N≤100000
  • 1≤M≤1000001≤M≤100000
  • 1≤d≤N1≤d≤N, in all the queries
  • 1≤elementsofA≤10000001≤elementsofA≤1000000
  • The array A and the queries of the type R follow 1-based indexing.

Example

Sample Input 1 

5 5
5 4 3 3 9
Q 1
L 4
Q 5
R 3
Q 2

Sample Output 1 

5
3
3

Array Rotation – CodeChef Solution in JAVA

import java.util.Arrays;
import java.util.Scanner;
import java.util.stream.Collectors;
public class Main {
  static final int MODULUS = 1_000_000_007;
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int N = sc.nextInt();
    int[] A = new int[N];
    for (int i = 0; i < A.length; ++i) {
      A[i] = sc.nextInt();
    }
    int Q = sc.nextInt();
    int[] x = new int[Q];
    for (int i = 0; i < x.length; ++i) {
      x[i] = sc.nextInt();
    }
    System.out.println(solve(A, x));
    sc.close();
  }
  static String solve(int[] A, int[] x) {
    int[] result = new int[x.length];
    result[0] = multiplyMod(Arrays.stream(A).reduce(Main::addMod).getAsInt(), 2);
    for (int i = 1; i < result.length; ++i) {
      result[i] = multiplyMod(result[i - 1], 2);
    }
    return Arrays.stream(result).mapToObj(String::valueOf).collect(Collectors.joining("\n"));
  }
  static int mod(long x) {
    return (int) ((x % MODULUS + MODULUS) % MODULUS);
  }
  static int addMod(int x, int y) {
    return mod(x + y);
  }
  static int multiplyMod(int x, int y) {
    return mod((long) x * y);
  }
}

Array Rotation – CodeChef Solution in CPP

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int mod = 1000000007;
//--------------------------------------------------------
void solve(){
  ll int n;
  cin>>n;
  ll int sum=0;
  while(n--){
    ll int x;
    cin>>x;
    sum=(sum+x+mod)%mod;
  }
  ll int q;
  cin>>q;
  while(q--){
    ll int x;
    cin>>x;
    sum=(2*sum)%mod;
    cout<<sum<<"\n";
  }
 return ;
}
int main(){
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    srand(chrono::high_resolution_clock::now().time_since_epoch().count());
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif
  solve();
} 

Array Rotation – CodeChef Solution in Python

import io, os, time,math
from sys import stdin ,stdout
def take_input():
    return stdin.readline()
def display_arr(arr):
    stdout.write(" ".join(map(str, arr)) + "\n")
def display_num(num):
    stdout.write(str(num)+  "\n")
mod = 10**9+7
n = int(take_input())
arr = list(map(int,take_input().split()))
test = int(take_input())
queries = list(map(int,take_input().split()))
summe = 0
for i in range(n):
    summe += arr[i]
    summe = summe%mod
for _ in range(test):
    summe = (summe+summe)%mod
    print(summe)
        

Disclaimer: The above Problem (Array Rotation) is generated by CodeChef but the solution is provided by  Chase2learn. This tutorial is only for Educational and Learning purpose.

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