Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Form a number using bit rotations of N having same frequency of each digit

  • Last Updated : 24 Dec, 2021

Given a number N, the task is to convert it into a number in which all distinct digits have the same frequency, by rotating its bits either clockwise or anti-clockwise. If it is not possible to convert the number print -1, otherwise print the number

Examples: 

Input: N = 331
Output: 421
Explanation: 
Binary representation of 331: 101001011
After an anti-clockwise rotation – 421: 110100101
421 is a steady number

Input: N = 58993
Output: 51097

 

Approach: The idea is to check all the possible scenarios, after rotating the bits of the number. Follow the below steps to solve this problem:

  1. Use a map to keep track of the frequencies of the digits.
  2. Use a set to check if all the frequencies are the same or not.
  3. If the number has the same frequencies for all digits, then print it as the answer
  4. Else rotate the binary representation of the number and check again.
  5. If after all the rotations, no number can be found having the same frequencies of all digits, then print -1.

Below is the representation of the above approach.

C++




// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to rotate
// the number by one place
int rotate(int N, int bits)
{
 
    // Getting the least significant bit
    int ls = N & 1;
 
    // Rotating the number
    return (N >> 1) | (int(pow(2, (bits - 1))) * ls);
}
 
// Function to check
// if a number is steady or not
bool checkStead(int N)
{
 
    // Map for the frequency of the digits
    map<char, int> mp;
    for (auto i : to_string(N))
        mp[i]++;
 
    // Checking if all the
    // frequencies are same or not
    set<int> se;
    for (auto it = mp.begin(); it != mp.end(); ++it)
        se.insert(it->second);
    if (se.size() == 1)
        return true;
    return false;
}
 
void solve(int N)
{
 
    // Getting the number of bits needed
    int bits = int(log2(N)) + 1;
    bool flag = true;
 
    // Checking all the possible numbers
    for (int i = 1; i < bits; ++i) {
        if (checkStead(N)) {
            cout << N << "\n";
            flag = false;
            break;
        }
        N = rotate(N, bits);
    }
    if (flag)
        cout << -1;
}
 
// Driver Code
int main()
{
 
    int N = 331;
    solve(N);
 
    return 0;
}
 
    // This code is contributed by rakeshsahni

Java




// Java code for the above approach
import java.util.HashMap;
import java.util.HashSet;
 
class GFG
{
   
    // Function to rotate
    // the number by one place
    public static int rotate(int N, int bits) {
 
        // Getting the least significant bit
        int ls = N & 1;
 
        // Rotating the number
        return (N >> 1) | (int) (Math.pow(2, (bits - 1))) * ls;
    }
 
    // Function to check
    // if a number is steady or not
public static boolean checkStead(int N)
{
 
    // Map for the frequency of the digits
    HashMap<String, Integer> mp = new HashMap<String, Integer>();
    for (String i : Integer.toString(N).split("")){
        if(mp.containsKey(i)){
            mp.put(i, mp.get(i) + 1);
        }else{
            mp.put(i, 1);
        }
    }
 
    // Checking if all the
    // frequencies are same or not
    HashSet<Integer> se = new HashSet<Integer>();
    for (String it : mp.keySet())
        se.add(mp.get(it));
    if (se.size() == 1)
        return true;
    return false;
}
 
public static void solve(int N)
{
 
    // Getting the number of bits needed
    int bits = (int)(Math.log(N) / Math.log(2)) + 1;
    boolean flag = true;
 
    // Checking all the possible numbers
    for (int i = 1; i < bits; ++i) {
        if (checkStead(N)) {
            System.out.println(N);
            flag = false;
            break;
        }
        N = rotate(N, bits);
    }
    if (flag)
        System.out.println(-1);
}
 
    // Driver Code
    public static void main(String args[]) {
 
        int N = 331;
        solve(N);
    }
}
 
// This code is contributed by Saurabh Jaiswal

Python3




# Python code for the above approach
import math
 
def solve(N):
 
    # Getting the number of bits needed
    bits = int(math.log(N, 2))+1
    flag = True
 
    # Checking all the possible numbers
    for i in range(1, bits):
        if checkStead(N):
            print(N)
            flag = False
            break
        N = rotate(N, bits)
 
    if flag:
        print(-1)
 
# Function to rotate
# the number by one place
def rotate(N, bits):
   
    # Getting the least significant bit
    ls = N & 1
     
    # Rotating the number
    return (N >> 1) | (2**(bits-1)*ls)
 
# Function to check
# if a number is steady or not
def checkStead(N):
   
    # Map for the frequency of the digits
    mp = {}
    for i in str(N):
        if i in mp:
            mp[i] += 1
        else:
            mp[i] = 1
 
    # Checking if all the
    # frequencies are same or not
    if len(set(mp.values())) == 1:
        return True
    return False
 
 
# Driver Code
N = 331
solve(N)

C#




// C# code for the above approach
using System;
using System.Collections.Generic;
 
class GFG {
 
  // Function to rotate
  // the number by one place
  public static int rotate(int N, int bits)
  {
 
    // Getting the least significant bit
    int ls = N & 1;
 
    // Rotating the number
    return (N >> 1)
      | (int)(Math.Pow(2, (bits - 1))) * ls;
  }
 
  // Function to check
  // if a number is steady or not
  public static bool checkStead(int N)
  {
 
    // Map for the frequency of the digits
    Dictionary<char, int> mp
      = new Dictionary<char, int>();
    foreach(char i in N.ToString())
    {
      if (mp.ContainsKey(i)) {
        mp[i]++;
      }
      else {
        mp[i] = 1;
      }
    }
 
    // Checking if all the
    // frequencies are same or not
    HashSet<int> se = new HashSet<int>();
    foreach(KeyValuePair<char, int> it in mp)
      se.Add(it.Value);
    if (se.Count == 1)
      return true;
    return false;
  }
 
  public static void solve(int N)
  {
 
    // Getting the number of bits needed
    int bits = (int)(Math.Log(N) / Math.Log(2)) + 1;
    bool flag = true;
 
    // Checking all the possible numbers
    for (int i = 1; i < bits; ++i) {
      if (checkStead(N)) {
        Console.WriteLine(N);
        flag = false;
        break;
      }
      N = rotate(N, bits);
    }
    if (flag)
      Console.WriteLine(-1);
  }
 
  // Driver Code
  public static void Main()
  {
 
    int N = 331;
    solve(N);
  }
}
 
// This code is contributed by ukasp.

Javascript




<script>
 
// JavaScript code for the above approach
 
// Function to check if a number
// is steady or not
function checkStead(N)
{
     
    // Map for the frequency of the digits
    let mp = new Map();
    let s = [];
 
    while (N != 0)
    {
        s.push(N % 10);
        N = Math.floor(N / 10)
    }
 
    for(let i = 0; i < s.length; i++)
    {
        if (mp.has(s[i]))
        {
            mp.set(s[i], mp.get(s[i]) + 1)
        }
        else
        {
            mp.set(s[i], 1);
        }
    }
 
    // Checking if all the
    // frequencies are same or not
    let st = new Set([...mp.values()]);
 
    if (st.size == 1)
        return true;
    else
        return false
}
 
// Function to rotate
// the number by one place
function rotate(N, bits)
{
     
    // Getting the least significant bit
    let ls = N & 1;
 
    // Rotating the number
    return ((N >> 1) | (Math.pow(2, bits - 1) * ls))
}
 
function solve(N)
{
     
    // Getting the number of bits needed
    let bits = Math.floor(Math.log2(N)) + 1
    let flag = true
 
    // Checking all the possible numbers
    for(let i = 1; i < bits; i++)
    {
        if (checkStead(N) == 1)
        {
            document.write(N)
            flag = false
            break
        }
        N = rotate(N, bits)
    }
     
    if (flag)
        document.write(-1)
}
 
// Driver Code
N = 331
 
solve(N)
 
// This code is contributed by Potta Lokesh
 
</script>
Output: 
421

 

Time Complexity: O(log2N)
Auxiliary Space: O(log10N)


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!