Queries on Left and Right Circular shift on array

Given an array A of N integers. There are three type of type of commands:

  • 1 x : Right Circular Shift the array x times. If an array is a[0], a[1], …., a[n – 1], then after one right circular shift the array will become a[n – 1], a[0], a[1], …., a[n – 2].
  • 2 y : Left Circular Shift the array y times. If an array is a[0], a[1], …., a[n – 1], then after one right circular shift the array will become a[1], …., a[n – 2], a[n – 1], a[0].
  • 3 l r : Print the sum of all integers in the subarray a[l…r] (l and r inclusive).

Given Q queries, the task is execute each query.

Examples:



Input : n = 5, arr[] = { 1, 2, 3, 4, 5 }
        query 1 = { 1, 3 }
        query 2 = { 3, 0, 2 }
        query 3 = { 2, 1 }
        query 4 = { 3, 1, 4 }
Output : 12
         11
Initial array arr[] = { 1, 2, 3, 4, 5 }
After query 1, arr[] = { 3, 4, 5, 1, 2 }.
After query 2, sum from index 0 to index 
               2 is 12, so output 12.
After query 3, arr[] = { 4, 5, 1, 2, 3 }.
After query 4, sum from index 1 to index 
               4 is 11, so output 11.

Method 1 : (Brute Force)Implement three function, rotateR(arr, k) which will right rotate array arr by k times, rotateL(arr, k) which will rotate array arr by k times, sum(arr, l, r) which will output sum of array arr from index l to index r. On the input of value 1, 2, 3 call the appropriate function.

Method 2 : (Efficient Approach)Initially, there are no rotations and we have many queries asking for sum of integers present in a range od indexes.
We can evaluate the prefix sum of all elements in the array, prefixsum[i] will denote the sum of all the integers upto ith index.
Now, if we want to find sum of elements between two indexes i.e l and r, we compute it in constant time by just calculating prefixsum[r] – prefixsum[l – 1] .
Now for rotations, if we are rotating the array for every query, that will be highly inefficient.
We just need to track the net rotation. If the tracked number is negative, it means left rotation has domainated else right rotation has dominated. When we are tracking the net rotations, we need to do mod n. As after every n rotation, array will return to its original state.
We need to observe it in such a way that every time we rotate the array, only its indexes are changing.
If we need to answer any query of third type and we have l and r. We need to find what l and r were in the original order. We can easily find it out by adding the net rotations to the index and taking mod n.
Every command can be executed in O(1) time.

Below is C++ implemenatation of this approach:

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// CPP Program to solve queries on Left and Right 
// Circular shift on array
#include <bits/stdc++.h>
using namespace std;
  
// Function to solve query of type 1 x.
void querytype1(int* toRotate, int times, int n)
{
    // Decreasing the absolute rotation
    (*toRotate) = ((*toRotate) - times) % n;
}
  
// Function to solve query of type 2 y.
void querytype2(int* toRotate, int times, int n)
{
    // Increasing the absolute rotation.
    (*toRotate) = ((*toRotate) + times) % n;
}
  
// Function to solve queries of type 3 l r.
void querytype3(int toRotate, int l, int r, 
                      int preSum[], int n)
{
    // Finding absolute l and r.
    l = (l + toRotate + n) % n;
    r = (r + toRotate + n) % n;
  
    // if l is before r.
    if (l <= r) 
        cout << (preSum[r + 1] - preSum[l]) << endl;    
  
    // If r is before l.
    else 
        cout << (preSum[n] + preSum[r + 1] - preSum[l])
             << endl;    
}
  
// Wrapper Function solve all queries.
void wrapper(int a[], int n)
{
    int preSum[n + 1];
    preSum[0] = 0;
  
    // Finding Prefix sum
    for (int i = 1; i <= n; i++)
        preSum[i] = preSum[i - 1] + a[i - 1];
  
    int toRotate = 0;
  
    // Solving each query
    querytype1(&toRotate, 3, n);
    querytype3(toRotate, 0, 2, preSum, n);
    querytype2(&toRotate, 1, n);
    querytype3(toRotate, 1, 4, preSum, n);
}
  
// Driver Program
int main()
{
    int a[] = { 1, 2, 3, 4, 5 };
    int n = sizeof(a) / sizeof(a[0]);
    wrapper(a, n);
    return 0;
}

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Output:

12
11


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