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Reverse The Array

Easy

0/40

Average time to solve is 15m

2141 upvotes

Asked in companies

Problem statement

Given an array/list 'ARR' of integers and a position ‘M’. You have to reverse the array after that position.

Example:

We have an array ARR = {1, 2, 3, 4, 5, 6} and M = 3 , considering 0 
based indexing so the subarray {5, 6} will be reversed and our 
output array will be {1, 2, 3, 4, 6, 5}.

Detailed explanation ( Input/output format, Notes, Images )

Input format:

The very first line of input contains an integer ‘T’ denoting the number of test cases. 

The first line of every test case contains one integer ‘N’ where ‘N’ denotes the number of elements and an integer ‘M’ which denotes after which position the array has to be reversed.

The second line of every test case contains ‘N’ space-separated integers which denote the elements of input of array/list.

Output format:

For each test case, return the required array.

Output for each test case is printed on a separate line.

Note:

You do not need to print anything, it has already been taken care of. Just implement the given function.

Consider 0-based indexing of the array.

Constraints:

1 <= T <= 10
0 <= M <= N <= 5*10^4
-10^9 <= ARR[i] <= 10^9

Time Limit: 1 sec

Sample Input 1:

2
6 3
1 2 3 4 5 6
5 2
10 9 8 7 6

Sample Output 1:

1 2 3 4 6 5
10 9 8 6 7

Explanation 1:

For the first test case, 
Considering 0-based indexing we have M = 3 so the 
subarray[M+1 … N-1] has to be reversed.
Therefore the required output will be {1, 2, 3, 4, 6, 5}.

For the second test case, 
Considering 0-based indexing we have M = 2 so the 
subarray[M+1 … N-1] has to be reversed.
Therefore the required output will be {10, 9, 8, 6, 7}.

Sample Input 2:

2
7 3
1 4 5 6 6 7 7 
9 3
10 4 5 2 3 6 1 3 6

Sample Output 2:

 1 4 5 6 7 7 6
 10 4 5 2 6 3 1 6 3

Hints:

1. Try to think by creating another array
2. Try to think which elements are beign swapped.

Approaches (2)

Approach 1

Approach 2

Reverse The Array

Approach:

Here we will use another array for reversing the elements.
The basic idea is just to store the elements up to the Mth position in the newly created array in the same way as they are given in the original array. Now after the Mth position store the elements in the reverse order.

Algorithm:

Firstly we will create another array say ‘BRR’ of the same size as the original array.
We will take one variable P = 0.
Now run a for loop in the forward direction from i=0 to i=M and fill the array ‘BRR’ in the same way as the original array i.e BRR[P++] = ARR[i]. Here ‘i’ will be incremented by one in each iteration.
Now run a second for loop in the reverse direction from j=N-1 to j=k+1 and fill the array ‘BRR’.Now the elements will be filled in the reversed order i.e BRR[p++] = ARR[j]. Here ‘j’ will be decremented by one in each iteration.
Finally copy back the array ‘BRR’ to array ‘ARR’ which the required array we want.

Time Complexity

O(N), where ‘N’ is the number of elements in the array.

Traversing over (M) elements will take O(M) time. Again traversing the array in the reverse direction over (N-M-1) elements will take (N-M-1) time. Thus total time complexity will be O(N).

Space Complexity

O(N)

Another array is created for reversing the elements.

(75% EXP penalty)

(100% EXP penalty)

Reverse The Array

Samar Singla

Interview problems

best code in cpp , faster than 100%

void reverseArray(vector<int> &arr , int m) {

int s = m+1;

int e = arr.size() - 1;

int a;

while(s<=e){

a = arr[s];

arr[s]=arr[e];

arr[e]=a;

s++;

e--;

}

1 view

0 replies

0 upvotes

Noothan Kp

Interview problems

simple way in c/c++

void reverseArray(vector<int> &arr , int m) {

int s = m+1,e=arr.size()-1;

while(s<=e)

{

swap(arr[s],arr[e]);

s++;e--;

}

6 views

0 replies

0 upvotes

Deep Chhatbar

Interview problems

(TWO POINTER)16ms run time better than 99% solution most basic example

void reverseArray(vector<int> &arr , int m) { 
     int start  = m+1;
     int end = arr.size();
     int mid = (start + end)/2;   
     while( start < mid){
          swap(arr[start],arr[end-1]);
          start++;
          end--;
     }   	
}

26 views

0 replies

1 upvote

ABHISHEK CODER

Interview problems

void reverseArray(vector<int> &arr , int m) { int n = sizeof(arr)/sizeof(arr[0]); int b[n]; for(int i =0;i<n;i++){ if(i<=m){b[i]=arr[i];} else { b[n+m-i] = arr[i]; } } for(int j=0;j<n;j++){ cout << b[j]; } }

what is wrong in my approach

11 views

0 replies

0 upvotes

Abhinav.exe

Interview problems

Use built-in STL function

#include<algorithm>

void reverseArray(vector<int> &arr , int m) {

// Write your code here

reverse(arr.begin()+m+1,arr.end());

}

40 views

1 reply

0 upvotes

Akash Singh

Interview problems

java code

import java.util.* ;

import java.io.*;

import java.util.ArrayList;

public class Solution

{

public static void reverseArray(ArrayList<Integer> arr, int m)

{

// Write your code here.

int left = m+1;

int right = arr.size()-1;

while(left<right){

int temp =arr.get(left);

arr.set(left,arr.get(right));

arr.set(right,temp);

left++;

right--;

}

80 views

0 replies

0 upvotes

Ankita Paul

Interview problems

Using two pointer concept

void reverseArray(vector<int> &arr , int m) {
    // Write your code here 
    int n=arr.size();
    int i=m+1,j=n-1;
    while(i<=j){
        swap(arr[i],arr[j]);
        i++;
        j--;
    }
}

90 views

0 replies

2 upvotes

Utsav Raj Singh

Interview problems

Reverse an array using two pointer approach

void reverseArray(vector<int> &arr , int m) {
    // Write your code here       
    int start =	m+1;
    int end=arr.size()-1;
    while(start<end){
        int temp = arr[start];
        arr[start]=arr[end];
        arr[end]=temp;
        start++;
        end--;
    }
}

45 views

0 replies

1 upvote

Pothanaboina Lavanya

Interview problems

Reverse The Array

import java.util.* ;

import java.io.*;

import java.util.ArrayList;

public class Solution

{

public static void reverseArray(ArrayList<Integer> arr, int m)

{

// Write your code here.

int start=m+1;

int end=arr.size()-1;

while(start<end){

int temp=arr.get(start);

arr.set(start,arr.get(end));

arr.set(end,temp);

start++;

end--;

}

58 views

0 replies

1 upvote

srinivas lingampelli

Interview problems

easy💯 || Simple 🧠 || java || O(n)

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

public class Solution 
{

   
    public static void reverseArray(ArrayList<Integer> arr, int m)
    {
        // Write your code here.


        int left=m+1;
        int right=arr.size()-1;

        while(left<right)
        {
            int temp=arr.get(left);
            arr.set(left,arr.get(right));
            arr.set(right,temp);
            left++;
            right--;

        }
    }
}

52 views

0 replies

0 upvotes

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