Left Rotation

Problem Statement :

A left rotation operation on an array of size n shifts each of the array's elements 1 unit to the left. Given an integer, d, rotate the array that many steps left and return the result.


After 2 rotations, arr'=[3,4,5,1,2].

Function Description:

Complete the rotateLeft function in the editor below.

rotateLeft has the following parameters:
   1. int d: the amount to rotate by
   2. int arr[n]: the array to rotate

   1. int[n]: the rotated array

Input Format:

The first line contains two space-separated integers that denote n, the number of integers, and d, the number of left rotations to perform.
The second line contains n space-separated integers that describe arr[].

   1. 1<=n<=10^5 
   2. 1<=d<=n 
   3. 1<=a[i]<=10^6

Solution :


                            Solution in C :

In C:

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {

    int N;
    int d;
    scanf("%d %d", &N, &d);  
    int A[N];
    for (int i = 0; i < N; i++)
        scanf("%d", &A[i]);    
    for (int i = 0; i < N; i++)
        printf("%d ", A[(i + d) % N]);    
    return 0;

In C++:

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

int main() {
    int N, d, i;
    cin >> N >> d;
    int start = N - d;
    int *arr = new int[N];
    for (i=0; i<N; ++i) {
        if (start == N) start = 0;
        cin >> arr[start++];
    for (i=0; i<N; ++i) cout << arr[i] << " ";
    return 0;

In Java:

import java.io.*;
import java.util.*;

public class Solution {

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int N = scan.nextInt();
        int n = scan.nextInt();
        int[] A = new int[N];
        for (int i=0; i<N; ++i) {
            A[i] = scan.nextInt();
        for (int i=0; i<n; ++i) {
        for (int a : A) {
            System.out.print(a+" ");
    private static void rotateArray(int[] A) {
        int t = A[0];
        for (int i=0; i<A.length-1; ++i) {
            A[i] = A[i+1];
        A[A.length-1] = t;

In Python 3:

n, d = map(int, input().split())
arr = [int(x) for x in input().split()]

for i in range(d):
for x in arr[d:]:
    print(x, end=' ')

View More Similar Problems

Tree: Inorder Traversal

In this challenge, you are required to implement inorder traversal of a tree. Complete the inorder function in your editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's inorder traversal as a single line of space-separated values. Input Format Our hidden tester code passes the root node of a binary tree to your $inOrder* func

View Solution →

Tree: Height of a Binary Tree

The height of a binary tree is the number of edges between the tree's root and its furthest leaf. For example, the following binary tree is of height : image Function Description Complete the getHeight or height function in the editor. It must return the height of a binary tree as an integer. getHeight or height has the following parameter(s): root: a reference to the root of a binary

View Solution →

Tree : Top View

Given a pointer to the root of a binary tree, print the top view of the binary tree. The tree as seen from the top the nodes, is called the top view of the tree. For example : 1 \ 2 \ 5 / \ 3 6 \ 4 Top View : 1 -> 2 -> 5 -> 6 Complete the function topView and print the resulting values on a single line separated by space.

View Solution →

Tree: Level Order Traversal

Given a pointer to the root of a binary tree, you need to print the level order traversal of this tree. In level-order traversal, nodes are visited level by level from left to right. Complete the function levelOrder and print the values in a single line separated by a space. For example: 1 \ 2 \ 5 / \ 3 6 \ 4 F

View Solution →

Binary Search Tree : Insertion

You are given a pointer to the root of a binary search tree and values to be inserted into the tree. Insert the values into their appropriate position in the binary search tree and return the root of the updated binary tree. You just have to complete the function. Input Format You are given a function, Node * insert (Node * root ,int data) { } Constraints No. of nodes in the tree <

View Solution →

Tree: Huffman Decoding

Huffman coding assigns variable length codewords to fixed length input characters based on their frequencies. More frequent characters are assigned shorter codewords and less frequent characters are assigned longer codewords. All edges along the path to a character contain a code digit. If they are on the left side of the tree, they will be a 0 (zero). If on the right, they'll be a 1 (one). Only t

View Solution →