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Distinct adjacent elements in an array
  • Difficulty Level : Medium
  • Last Updated : 07 Jan, 2020

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++






// 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 occurring 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;
}


Java




// 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 occurring 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


Python3




# Python program to check if we can make
# neighbors distinct.
def distantAdjacentElement(a, n):
  
    # dict used to count the frequency
    # of each element occurring 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


C#




// 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 occurring 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



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.

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