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:
- Build a Priority_queue or max_heap, pq that stores elements and their frequencies.
…… Priority_queue or max_heap is built on the basis of the frequency of elements. - 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 }
- 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.
- 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 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 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 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# 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. |
<script> // Javascript 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 function rearrangeArray(arr, N)
{ // Store frequencies of all elements
// of the array
var mp = new Map();
var visited = new Map();
for ( var i = 0; i < N; i++)
{
if (mp.has(arr[i]))
mp.set(arr[i], mp.get(arr[i])+1)
else
mp.set(arr[i], 1)
}
var pq = [];
// Adding high freq elements
// in descending order
for ( var i = 0; i < N ; i++)
{
var val = arr[i];
if (mp.get(val) > 0 && visited[val] != 1)
{
pq.push([mp.get(val), val]);
}
visited[val] = 1;
}
pq.sort();
// 'result[]' that will store resultant value
var result = Array(N).fill(0);
// Work as the previous visited element
// initial previous element will be ( '-1' and
// it's frequency wiint also be '-1' )
var prev = [-1, -1];
var l = 0;
// Traverse queue
while (pq.length != 0)
{
// Pop top element from queue and add it
// to result
var k = pq[pq.length-1];
pq.pop();
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.push(prev);
}
pq.sort();
// Make current element as the previous
// decrease frequency by 'one'
k[0]--;
prev = k;
l++;
}
for ( var it of result)
{
if (it == 0)
{
// If found 0, No valid result
// array possible
document.write( "Not valid Array" + "<br>" );
return ;
}
}
for ( var it of result)
{
document.write(it + " " );
}
} // Driver Code var A = [1, 1, 1, 1, 2, 2, 3, 3];
// Size of the array var N = A.length;
rearrangeArray(A, N); // This code is contributed by rutvik_56. </script> |
1 3 1 2 1 3 2 1
Time Complexity: O(N*logN)
Auxiliary Space: O(N)