数组中不同的相邻元素

给定一个数组，找出是否有可能通过交换两个相邻数组元素来获得具有不同相邻元素的数组。
例子：

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 .

只有当大多数出现的元素的频率小于或等于数组大小的一半时，即（<= (n+1)/2），才能获得具有不同相邻元素的数组。为了更清楚，考虑不同的例子
第一个例子：a[] = {1, 1, 2, 3, 1}
我们可以通过以下方式获得数组 {1, 2, 1, 3, 1}
交换数组 a 中的（第二个和第三个）元素。
这里 1 出现最多，其频率为
3.所以 3 <= ((5+1)/2) 。
因此，这是可能的。
下面是这种方法的实现。

C++

// C++ program to check if we can make
// neighbors distinct.
#include 
using namespace std;
 
void distinctAdjacentElement(int a[], int n)
{
    // map used to count the frequency
    // of each element occurring in the
    // array
    map 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;
}

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 m = new Dictionary();
 
    // 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

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 m = new HashMap();
 
// 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

Javascript

输出：

NO