Distinct adjacent elements in an array

Given an array, find whether it is possible to obtain an array having distinct neighbouring elements by swapping two neighbouring array elements.

Examples:

Input : 1 1 2
Output : YES
swap 1 (second last element) and 2 (last element), 
to obtain 1 2 1, which has distinct neighbouring 
elements .

Input : 7 7 7 7
Output : NO
We can't swap to obtain distinct elements in 
neighbor .

To obtain an array having distinct neighbouring elements is possible only, when the frequency of most occurring element is less than or equal to half of size of array i.e ( <= (n+1)/2 ). To make it more clear consider different examples



1st Example : a[] = {1, 1, 2, 3, 1}
We can obtain array {1, 2, 1, 3, 1} by
swapping (2nd and 3rd) element and
(4th and 5th) elements from array a.
Here 1 occurs most and its frequency is
3 (6+1)/2.
Hence, it will never possible.

Below is the implementation of this approach .

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to check if we can make
// neighbors distinct.
#include <bits/stdc++.h>
using namespace std;
  
void distinctAdjacentElement(int a[], int n)
{
    // map used to count the frequency
    // of each element occuring in the
    // array
    map<int, int> m;
  
    // In this loop we count the frequency
    // of element through map m .
    for (int i = 0; i < n; ++i)
        m[a[i]]++;
  
    // mx store the frequency of element which
    // occurs most in array .
    int mx = 0;
  
    // In this loop we calculate the maximum
    // frequency and store it in variable mx.
    for (int i = 0; i < n; ++i)
        if (mx < m[a[i]])
            mx = m[a[i]];
  
    // By swapping we can adjust array only
    // when the frequency of the element
    // which occurs most is less than or
    // equal to (n + 1)/2 .
    if (mx > (n + 1) / 2)
        cout << "NO" << endl;
    else
        cout << "YES" << endl;
}
  
// Driver program to test the above function
int main()
{
    int a[] = { 7, 7, 7, 7 };
    int n = sizeof(a) / sizeof(a[0]);
    distinctAdjacentElement(a, n);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to check if we can make
// neighbors distinct.
import java.io.*;
import java.util.HashMap;
import java.util.Map;
class GFG {
      
    static void distinctAdjacentElement(int a[], int n)
    {
        // map used to count the frequency
        // of each element occuring in the
        // array
        HashMap<Integer,Integer> m = new HashMap<Integer,
                                             Integer>();
       
        // In this loop we count the frequency
        // of element through map m .
        for (int i = 0; i < n; ++i){
     
            // checks if map already contains a[i] then 
            // update the previous value by incrementing 
          // by 1
            if(m.containsKey(a[i])){
                int x = m.get(a[i]) + 1;
                m.put(a[i],x); 
            }
            else{
                m.put(a[i],1);
            }
              
        }
       
        // mx store the frequency of element which
        // occurs most in array .
        int mx = 0;
       
        // In this loop we calculate the maximum
        // frequency and store it in variable mx.
        for (int i = 0; i < n; ++i)
            if (mx < m.get(a[i]))
                mx = m.get(a[i]);
       
        // By swapping we can adjust array only
        // when the frequency of the element
        // which occurs most is less than or
        // equal to (n + 1)/2 .
        if (mx > (n + 1) / 2)
            System.out.println("NO");
        else
            System.out.println("YES");
    }
          
    // Driver program to test the above function
    public static void main (String[] args) {
        int a[] = { 7, 7, 7, 7 };
        int n = 4;
        distinctAdjacentElement(a, n);
    }
}
// This code is contributed by Amit Kumar

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python program to check if we can make
# neighbors distinct.
def distantAdjacentElement(a, n):
  
    # dict used to count the frequency
    # of each element occuring in the
    # array
    m = dict()
  
    # In this loop we count the frequency
    # of element through map m
    for i in range(n):
        if a[i] in m:
            m[a[i]] += 1
        else:
            m[a[i]] = 1
  
    # mx store the frequency of element which
    # occurs most in array .
    mx = 0
  
    # In this loop we calculate the maximum
    # frequency and store it in variable mx.
    for i in range(n):
        if mx < m[a[i]]:
            mx = m[a[i]]
  
    # By swapping we can adjust array only
    # when the frequency of the element
    # which occurs most is less than or
    # equal to (n + 1)/2 .
    if mx > (n+1) // 2:
        print("NO")
    else:
        print("YES")
  
  
# Driver Code
if __name__ == "__main__":
    a = [7, 7, 7, 7]
    n = len(a)
    distantAdjacentElement(a, n)
  
# This code is contributed by
# sanjeev2552

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to check if we can make 
// neighbors distinct. 
using System;
using System.Collections.Generic;
  
class GFG {
  
public static void distinctAdjacentElement(int[] a, int n)
{
    // map used to count the frequency 
    // of each element occuring in the 
    // array 
    Dictionary<int, int> m = new Dictionary<int, int>();
  
    // In this loop we count the frequency 
    // of element through map m . 
    for (int i = 0; i < n; ++i)
    {
  
        // checks if map already 
        // contains a[i] then 
        // update the previous
        // value by incrementing 
        // by 1 
        if (m.ContainsKey(a[i]))
        {
            int x = m[a[i]] + 1;
            m[a[i]] = x;
        }
        else
        {
            m[a[i]] = 1;
        }
  
    }
  
    // mx store the frequency
    // of element which 
    // occurs most in array . 
    int mx = 0;
  
    // In this loop we calculate
    // the maximum frequency and
    // store it in variable mx. 
    for (int i = 0; i < n; ++i)
    {
        if (mx < m[a[i]])
        {
            mx = m[a[i]];
        }
    }
  
    // By swapping we can adjust array only 
    // when the frequency of the element 
    // which occurs most is less than or 
    // equal to (n + 1)/2 . 
    if (mx > (n + 1) / 2)
    {
        Console.WriteLine("NO");
    }
    else
    {
        Console.WriteLine("YES");
    }
}
  
    // Main Method
    public static void Main(string[] args)
    {
        int[] a = new int[] {7, 7, 7, 7};
        int n = 4;
        distinctAdjacentElement(a, n);
    }
}
  
// This code is contributed
// by Shrikant13

chevron_right



Output:

NO

This article is contributed by Surya Priy. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



My Personal Notes arrow_drop_up



Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.