Leet Code: Rotate Array C++ Python || JavaScript || Java solution:

Given an array, rotate the array to the right by k steps, where k is non-negative.

Example 1:

Input: nums = [1,2,3,4,5,6,7], k = 3
Output: [5,6,7,1,2,3,4]
Explanation:
rotate 1 steps to the right: [7,1,2,3,4,5,6]
rotate 2 steps to the right: [6,7,1,2,3,4,5]
rotate 3 steps to the right: [5,6,7,1,2,3,4]

Example 2:

Input: nums = [-1,-100,3,99], k = 2
Output: [3,99,-1,-100]
Explanation: 
rotate 1 steps to the right: [99,-1,-100,3]
rotate 2 steps to the right: [3,99,-1,-100]

Constraints:

• 1 <= nums.length <= 105
• -231 <= nums[i] <= 231 - 1
• 0 <= k <= 105

Follow up:

• Try to come up with as many solutions as you can. There are at least three different ways to solve this problem.
• Could you do it in-place with O(1) extra space?

C++ Solution :

//solution BY Nilesh Vishnu Raut ...Codewithnilesh


#include <bits/stdc++.h>
using namespace std;

int main() {
    string arr[]={"nilesh" ,"Raut" , "nilesh" ,"Raut" };
    unordered_map<string,int>m;
    for(int i=0;i<5;i++){
        m[arr[i]]++;
    }
    
    int max=0;
    string win;
    
    for(auto x:m){
        if(x.second>max){
            max=x.second;
            win=x.first;
            
        }
        else if(max==x.second) {
            if(x.first<win){
                win=x.first;
            }
        }
        
    }
    cout<<win;
	return 0;
}

Java Solution :

class Solution {
public void rotate(int[] nums, int k) {
if(k > nums.length){
k=k%nums.length;
}

    int[] result=new int[nums.length];
    int j=0;
    for(int i=k;i<nums.length;i++){
        result[i]=nums[j];
        j++;
    }
    
    for(int i=0;i<k;i++){
        result[i]=nums[j];
        j++;
        
    }
    System.arraycopy( result, 0, nums, 0, nums.length );
}
}

Python Solution :

class Solution(object):
    def rotate(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: None Do not return anything, modify nums in-place instead.
        """
        def reverse(a,l,h):
            if (l>=h):
                return
            temp=a[l]
            a[l]=a[h]
            a[h]=temp
            reverse(a,l+1,h-1)
        if k>len(nums):
            k=k%len(nums)
        reverse(nums,0,len(nums)-1)
        reverse(nums,0,k-1)
        reverse(nums,k,len(nums)-1)