ROTATE ARRAY BY K TIMES-THREE APPRAOCH

 Given an array of integers rotate the array by k times

INPUT
7 3
1 2 3 4 5 6 7
6 3
1 2 3 4 5 6
12 4
1 2 3 4 5 6 7 8 9 10 11 12

OTPUT
5 6 7 1 2 3 4
4 5 6 1 2 3 
9 10 11 12 1 2 3 4 5 6 7 8 

Brute force approach:

A simple approach is to use two pointer one start for  i=0 and other pointer start from i=i+k

simply swap the both number and once again iterate form i+k to n and simply assign the value of i=0 to i=k  in an array.

 #include<bits/stdc++.h>
    using namespace std;
     void rotate(vector<int>& nums, int k) {
            
            int i=0;
            k=k%nums.size();
            int j=nums.size()-k;
            vector<int>kk;
            kk=nums;
            int c=0;
            while(i<k&&j<nums.size())
            {
                swap(nums[i],nums[j]);
                i++;
                j++;
                c++;
            }
            for(int i=0;i<nums.size()-c;i++)
            {
                nums[i+k]=kk[i];
            }
        
        }
    
    int main()
    {
        int n,k;
        cin>>n>>k;
      vector<int>v;
        for(int i=0;i<n;i++)
        {
         int x;
         cin>>x;
         v.push_back(x);
        }
        rotate(v,k);
        for(int i=0;i<n;i++)
        {
            cout<<v[i]<<" ";
        }
    }
   

TIME COMPLEXITY-O(N)

SPACE COMPLEXITY-O(N)

How can we Optimise the space?

1.we will first reverse the array 

2. we will reverse the first k element

3.we will reverse the remaining element from i+k to n

#include<iostream>
#include<bits/stdc++.h>
using namespace  std;
 
 
    void rotate(vector<int>& nums, int k) {
        
        int i=0;
        k=k%nums.size();
        int j=nums.size()-k;
        vector<int>kk;
        kk=nums;
        int c=0;
        while(i<k&&j<nums.size())
        {
           
            i++;
            j++;
            c++;
        }
        reverse(nums.begin(),nums.end());
        reverse(nums.begin(),nums.begin()+k);
        reverse(nums.begin()+k,nums.end());
     
      //  reverse(nums.begin()+k+1,nums.end());
    }

int main()
{
    int n,k;
    cin>>n>>k;
  vector<int>v;
    for(int i=0;i<n;i++)
    {
     int x;
     cin>>x;
     v.push_back(x);
    }
    rotate(v,k);
    for(int i=0;i<n;i++)
    {
        cout<<v[i]<<" ";
    }
}

TIME COMPLEXITY:-O(N)

SPACE COMPLEXITY:-O(1)

3.solution using juggling algorithm in the next blog😀😊