Skip to content
Related Articles

Related Articles

Rearrange numbers in an array such that no two adjacent numbers are same
  • Difficulty Level : Medium
  • Last Updated : 23 Feb, 2021

Given an array of integers. The task is to rearrange elements of the array such that no two adjacent elements in the array are same.

Examples:  

Input: arr[] = {1, 1, 1, 2, 2, 2}
Output: {2, 1, 2, 1, 2, 1}

Input: arr[] = {1, 1, 1, 1, 2, 2, 3, 3}
Output: {1, 3, 1, 3, 2, 1, 2, 1}

The idea is to put the highest frequency element first (a greedy approach). We use a priority queue (Or Binary Max Heap) and put all elements and ordered by their frequencies (highest frequency element at the root). We then one by one take the highest frequency element from the heap and add it to result. After we add, we decrease the frequency of the element and temporarily move this element out of priority queue so that it is not picked next time.

We have to follow the step to solve this problem, they are: 

  1. Build a Priority_queue or max_heap, pq that stores elements and their frequencies. 
    …… Priority_queue or max_heap is built on the bases of the frequency of elements.
  2. Create a temporary Key that will be used as the previously visited element (the previous element in the resultant array. Initialize it { num = -1, freq = -1 }
  3. While pq is not empty. 
    • Pop an element and add it to the result.
    • Decrease frequency of the popped element by ‘1’.
    • Push the previous element back into the priority_queue if it’s frequency > ‘0’.
    • Make the current element as the previous element for the next iteration.
  4. If the length of resultant string and original are not equal, print “not possible”. Else print result.

Below is the implementation of the above approach:

C++14




// C++14 program to rearrange numbers in
// an Array such that no two numbers are
// adjacent
#include <bits/stdc++.h>
using namespace std;
 
// Function to rearrange numbers in array such
// that no two adjacent numbers are same
void rearrangeArray(int arr[], int N)
{
     
    // Store frequencies of all elements
    // of the array
    map<int, int>mp, visited;
     
    for(int i = 0; i < N; i++)
    {
        mp[arr[i]]++;
    }
     
    priority_queue<pair<int, int>>pq;
     
    // Adding high freq elements
    // in descending order
    for(int i = 0; i < N ; i++)
    {
        int val = arr[i];
         
        if (mp[val] > 0 and visited[val] != 1)
        {
            pq.push({mp[val], val});
        }
        visited[val] = 1;
    }
     
    // 'result[]' that will store resultant value
    vector<int>result(N);
     
    // Work as the previous visited element
    // initial previous element will be ( '-1' and
    // it's frequency wiint also be '-1' )
    pair<int, int>prev = { -1, -1 };
    int l = 0;
     
    // Traverse queue
    while (pq.size() != 0)
    {
         
        // Pop top element from queue and add it
        // to result
        pair<int,int>k = pq.top();
        pq.pop();
        result[l] = k.second;
         
        // If frequency of previous element is less
        // than zero that means it is useless, we
        // need not to push it
        if (prev.first > 0)
        {
            pq.push(prev);
        }
         
        // Make current element as the previous
        // decrease frequency by 'one'
        k.first--;
        prev = k;
        l++;
    }
     
    for(auto it : result)
    {
        if (it == 0)
        {
             
            // If found 0, No valid result
            // array possible
            cout << "Not valid Array" << endl;
            return;
        }
    }
     
    for(auto it : result)
    {
        cout << it << ", ";
    }
}
 
// Driver Code
int main()
{
    int A[] = { 1, 1, 1, 1, 2, 2, 3, 3 };
     
    // Size of the array
    int N = sizeof(A) / sizeof(A[0]);
     
    rearrangeArray(A, N);
}
 
// This code is contributed by koulick_sadhu

Java




// Java Program to rearrange numbers in an Array
// such that no two numbers are adjacent
 
import java.util.Comparator;
import java.util.PriorityQueue;
 
// Comparator class to sort in descending order
class KeyComparator implements Comparator<Key>
{
 
    // Overriding compare()method of Comparator
    public int compare(Key k1, Key k2)
    {
        if (k1.freq < k2.freq)
            return 1;
        else if (k1.freq > k2.freq)
            return -1;
        return 0;
    }
}
 
// Object of num and its freq
class Key
{
    int freq; // store frequency of character
    int num;
    Key(int freq, int num)
    {
        this.freq = freq;
        this.num = num;
    }
}
 
public class GFG
{
 
    // Function to rearrange numbers in array such
    // that no two adjacent numbers are same
    static void rearrangeArray(int[] arr)
    {
        int n = arr.length;
 
        // Store frequencies of all elements
        // of the array
        int[] count = new int[10000];
        int visited[] = new int[10000];
 
        for (int i = 0; i < n; i++)
            count[arr[i]]++;
 
        // Insert all characters with their frequencies
        // into a priority_queue
        PriorityQueue<Key> pq
            = new PriorityQueue<>(new KeyComparator());
 
        // Adding high freq elements in descending order
        for (int i = 0; i < n; i++)
        {
            int val = arr[i];
 
            if (count[val] > 0 && visited[val] != 1)
                pq.add(new Key(count[val], val));
            visited[val] = 1;
        }
 
        // 'result[]' that will store resultant value
        int result[] = new int[n];
 
        // work as the previous visited element
        // initial previous element will be ( '-1' and
        // it's frequency will also be '-1' )
        Key prev = new Key(-1, -1);
 
        // Traverse queue
        int l = 0;
        while (pq.size() != 0)
        {
            // pop top element from queue and add it
            // to result
            Key k = pq.peek();
            pq.poll();
            result[l] = k.num;
 
            // If frequency of previous element is less
            // than zero that means it is useless, we
            // need not to push it
            if (prev.freq > 0)
                pq.add(prev);
 
            // make current element as the previous
            // decrease frequency by 'one'
            (k.freq)--;
            prev = k;
            l++;
        }
 
        // If length of the resultant array and original
        // array is not same then the array is not valid
        if (l != result.length)
        {
            System.out.println(" Not valid Array ");
        }
        // Otherwise Print the result array
        else
        {
            for (int i : result)
            {
                System.out.print(i + " ");
            }
        }
    }
 
    // Driver Code
    public static void main(String args[])
    {
        int arr[] = new int[] { 1, 1, 1, 1, 2, 2, 3, 3 };
 
        rearrangeArray(arr);
    }
}

Python3




# Python3 program to rearrange numbers in
# an Array such that no two numbers are
# adjacent
 
# Function to rearrange numbers in array such
# that no two adjacent numbers are same
def rearrangeArray(arr, N) :
       
    # Store frequencies of all elements
    # of the array
    mp = {}
    visited = {}    
    for i in range(N) :
        if(arr[i] in mp) :
            mp[arr[i]] += 1
        else :
            mp[arr[i]] = 1
       
    pq = []
       
    # Adding high freq elements
    # in descending order
    for i in range(N) :
        val = arr[i]
        if((val in mp) and ((val not in visited) or (visited[val] != 1))) :
            pq.append([mp[val], val])
        visited[val] = 1   
    pq.sort()
    pq.reverse()
       
    # 'result[]' that will store resultant value
    result = [0]*N
       
    # Work as the previous visited element
    # initial previous element will be ( '-1' and
    # it's frequency wiint also be '-1' )
    prev = [-1, -1]
    l = 0
       
    # Traverse queue
    while (len(pq) != 0) :
           
        # Pop top element from queue and add it
        # to result
        k = pq[0]
        pq.pop(0)
        result[l] = k[1]
           
        # If frequency of previous element is less
        # than zero that means it is useless, we
        # need not to push it
        if (prev[0] > 0) :
         
            pq.append(prev)
            pq.sort()
            pq.reverse()
           
        # Make current element as the previous
        # decrease frequency by 'one'
        prev = [k[0] - 1, k[1]]
        l += 1
         
    for it in result :
        if (it == 0) :
               
            # If found 0, No valid result
            # array possible
            print("Not valid Array")
            return   
    for it in result :
        print(it , end = " ")
         
A = [ 1, 1, 1, 1, 2, 2, 3, 3 ]
       
# Size of the array
N = len(A)
rearrangeArray(A, N)
 
#This code is contributed by divyesh072019.

C#




// C# program to rearrange numbers in
// an Array such that no two numbers are
// adjacent
using System;
using System.Collections.Generic;
class GFG{
     
    // Function to rearrange numbers in array such
    // that no two adjacent numbers are same
    static void rearrangeArray(int[] arr, int N)
    {
          
        // Store frequencies of all elements
        // of the array
        Dictionary<int, int> mp = new Dictionary<int, int>();
        Dictionary<int, int> visited = new Dictionary<int, int>();
          
        for(int i = 0; i < N; i++)
        {
            if(mp.ContainsKey(arr[i]))
            {
                mp[arr[i]] += 1;
            }
            else{
                mp[arr[i]] = 1;
            }
        }
          
        List<Tuple<int, int>> pq = new List<Tuple<int,int>>();
          
        // Adding high freq elements
        // in descending order
        for(int i = 0; i < N ; i++)
        {
            int val = arr[i];
              
            if (mp.ContainsKey(val) && (!visited.ContainsKey(val) || visited[val] != 1))
            {
                pq.Add(new Tuple<int,int>(mp[val], val));
            }
            visited[val] = 1;
        }
         
        pq.Sort();
        pq.Reverse();
          
        // 'result[]' that will store resultant value
        int[] result = new int[N];
          
        // Work as the previous visited element
        // initial previous element will be ( '-1' and
        // it's frequency wiint also be '-1' )
        Tuple<int, int> prev = new Tuple<int,int>( -1, -1 );
        int l = 0;
          
        // Traverse queue
        while (pq.Count != 0)
        {
              
            // Pop top element from queue and add it
            // to result
            Tuple<int,int> k = pq[0];
            pq.RemoveAt(0);
            result[l] = k.Item2;
              
            // If frequency of previous element is less
            // than zero that means it is useless, we
            // need not to push it
            if (prev.Item1 > 0)
            {
                pq.Add(prev);
                pq.Sort();
                pq.Reverse();
            }
              
            // Make current element as the previous
            // decrease frequency by 'one'
            prev = new Tuple<int,int>(k.Item1 - 1, k.Item2);
            l++;
        }
        foreach(int it in result)
        {
            if (it == 0)
            {
                  
                // If found 0, No valid result
                // array possible
                Console.WriteLine("Not valid Array");
                return;
            }
        }
        foreach(int it in result)
        {
            Console.Write(it + " ");
        }
    }
 
  // Driver code
  static void Main()
  {
    int[] A = { 1, 1, 1, 1, 2, 2, 3, 3 };
      
    // Size of the array
    int N = A.Length;
    rearrangeArray(A, N);
  }
}
 
// This code is contributed by divyeshrabadiya07.
Output: 
1 3 1 2 1 3 2 1

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.




My Personal Notes arrow_drop_up
Recommended Articles
Page :