Shift the elements to the right (#922)

Algorithm Analysis

Let m be an input array of integers. First, we'll output the last element of the array m[n – 1]. Then, we will sequentially output all the numbers from m[0] to m[n – 2].

Consider an implementation in which the elements of the array will actually be rearranged. For this, it is necessary to move the element from the (i – 1)-th cell to the i-th cell of the array, 1 ≤ i ≤ n – 1. Separately, the element from the (n – 1)-th cell should be moved to the 0-th cell.

Algorithm Implementation

Declare an integer array m.

int m[101];

Read the input numbers into array m.

scanf("%d",&n);
for(i = 0; i < n; i++)
    scanf("%d",&m[i]);

First, we'll output the last number of the array m[n – 1] – it will be the first after the cyclic shift of elements by one position to the right.

printf("%d",m[n-1]);

Then, we'll output all the numbers of the array, starting from the zeroth m[0] to the penultimate m[n – 2].

for(i = 0; i < n - 1; i++)
    printf(" %d",m[i]);
printf("\n");

Implementation – with rearrangement of elements

Declare an integer array m.

int m[101];

Read the input numbers into array m.

scanf("%d",&n);
for(i = 0; i < n; i++)
    scanf("%d",&m[i]);

Remember the last element m[n – 1] in the variable temp.

temp = m[n-1];

Move the element m[i – 1] to m[i] (1 ≤ i ≤ n – 1).

for(i = n - 1; i > 0; i--)
    m[i] = m[i - 1];

Assign the element that was at the end of the array to the beginning of the array.

m[0] = temp;

Output the resulting array.

for(i = 0; i < n; i++)
    printf("%d ",m[i]);
printf("\n");

Implementation – STL rotate

Declare an integer vector.

vector<int> v;

Read the input data.

scanf("%d", &n);
v.resize(n);
for (i = 0; i < n; i++)
    scanf("%d", &v[i]);

Perform a shift of the array elements so that the last element becomes the first.

rotate(v.begin(), v.begin() + n - 1, v.end());

Output the resulting array.

for (i = 0; i < n; i++)
    printf("%d ", v[i]);
printf("\n");

Implementation – dynamic array

#include <stdio.h>

int i, n, a;
int *m;

int main(void)
{
    scanf("%d",&n);
    m = new int[n];

    for(i = 0; i < n; i++)
        scanf("%d",&m[i]);

    printf("%d",m[n-1]);
    for(i = 0; i < n - 1; i++)
        printf(" %d",m[i]);
    printf("\n");

    delete[] m;
    return 0;
}

Java Implementation

import java.util.*;

public class Main
{
    public static void main(String[] args)
    {
        Scanner con = new Scanner(System.in);
        int n = con.nextInt();
        int m[] = new int[n];
        for(int i = 0; i < n; i++)
            m[i] = con.nextInt();

        System.out.print(m[n-1]);
        for(int i = 0; i < n - 1; i++)
            System.out.print(" " + m[i]);
        System.out.println();
        con.close();
    }
}

Python Implementation

n = int(input())
v = list(map(int, input().split()))

v = v[-1:] + v[:-1]

for num in v:
    print(num, end=" ")
print()