检查是否可以为 N 个对象着色，以便对于第 i 个对象，使用完全 arr[i] 不同的颜色

给定一个由N个正整数组成的数组arr[] ，任务是检查是否可以为N个对象着色，使得对于数组的第 i^个元素，恰好存在用于为所有对象着色的arr[i]不同颜色除了第i^个对象。

例子：

Input: arr[] = {1, 2, 2}
Output: Yes
Explanation: 
One of the possible ways to color is:  {“Red”, “blue”, “blue”}.

1. For arr[0](=1), there is exactly 1 distinct color, which is “blue”.
2. For arr[1](=2), there are exactly 2 distinct colors, which are “blue” and “red”.
3. For arr[2](=3), there are exactly 2 distinct colors, which are “blue” and “red”.

Input: arr[] = {4, 3, 4, 3, 4}
Output: No

方法：可以根据以下观察解决问题：

1. For 2 objects, there are N-2 objects common while calculating the number of distinct colors. Therefore, there can be a difference of at most 1 between the maximum and minimum element of the array arr[].
2. Now there are two cases:
   1. If the maximum and minimum elements are equal, then the answer is “Yes”,  only if the element is N – 1 or the element is less than or equal to N/2, because every color can be used more than once.
   2. If the difference between the maximum and minimum element is 1, then the number of distinct colors in the N object must be equal to the maximum element, because the minimum element is 1 less than the maximum element.
      • Now, assuming the frequency of the minimum element as X and the frequency of the maximum element as Y, then the answer is “Yes” if and only if X+1 ≤ A ≤ X+ Y/2 (observation-based).

请按照以下步骤解决问题：

• 首先按升序对数组进行排序。
• 如果arr[N-1]和arr[0]的差大于1，则打印“ No ”。
• 否则，如果arr[N-1]等于arr[0] ，则检查以下内容：
  • 如果arr[N-1] = N-1或2*arr[N-1] <= N ，则打印“是”。
  • 否则，打印“否”。
• 否则，计算 min 和 max 元素的频率并将它们存储在变量中，比如X和Y ，然后执行以下操作：
  • 如果arr[N-1]大于X并且arr[N-1]小于或等于X+Y/2 ，则打印“ Yes ”。
  • 否则，打印“否”。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to check if coloring is
// possible with give conditions
string checkValid(int arr[], int N)
{
 
    // Sort the vector
    sort(arr, arr + N);
 
    // Coloring not possible in case of
    // maximum - minimum element > 1
    if (arr[N - 1] - arr[0] > 1)
        return "No";
 
    // case 1
    else if (arr[N - 1] == arr[0]) {
 
        // If h is equal to N-1 or
        // N is greater than 2*h
        if (arr[N - 1] == N - 1
            || 2 * arr[N - 1] <= N)
            return "Yes";
        else
            return "No";
    }
    // Case 2
    else {
        // Stores frequency of minimum element
        int x = 0;
 
        for (int i = 0; i < N; i++) {
 
            // Frequency  of minimum element
            if (arr[i] == arr[0])
                x++;
        }
 
        // Stores frequency of maximum element
        int y = N - x;
 
        // Condition for case 2
        if ((arr[N - 1] >= x + 1)
            and (arr[N - 1] <= x + y / 2))
            return "Yes";
        else
            return "No";
    }
}
 
// Driver Code
int main()
{
    // GivenInput
    int arr[] = { 1, 2, 2 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    cout << checkValid(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
     
// Function to check if coloring is
// possible with give conditions
static String checkValid(int arr[], int N)
{
     
    // Sort the vector
    Arrays.sort(arr);
 
    // Coloring not possible in case of
    // maximum - minimum element > 1
    if (arr[N - 1] - arr[0] > 1)
        return "No";
 
    // Case 1
    else if (arr[N - 1] == arr[0])
    {
         
        // If h is equal to N-1 or
        // N is greater than 2*h
        if (arr[N - 1] == N - 1
            || 2 * arr[N - 1] <= N)
            return "Yes";
        else
            return "No";
    }
     
    // Case 2
    else
    {
         
        // Stores frequency of minimum element
        int x = 0;
 
        for(int i = 0; i < N; i++)
        {
             
            // Frequency  of minimum element
            if (arr[i] == arr[0])
                x++;
        }
 
        // Stores frequency of maximum element
        int y = N - x;
 
        // Condition for case 2
        if ((arr[N - 1] >= x + 1) &&
            (arr[N - 1] <= x + y / 2))
            return "Yes";
        else
            return "No";
    }
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given Input
    int[] arr = { 1, 2, 2 };
    int N = arr.length;
 
    // Function Call
    System.out.print(checkValid(arr, N));
}   
}
 
// This code is contributed by sanjoy_62

Python3

# Python3 program for the above approach
 
# Function to check if coloring is
# possible with give conditions
def checkValid(arr, N):
 
    # Sort the vector
    arr = sorted(arr)
 
    # Coloring not possible in case of
    # maximum - minimum element > 1
    if (arr[N - 1] - arr[0] > 1):
        return "No"
 
    # case 1
    elif (arr[N - 1] == arr[0]):
 
        # If h is equal to N-1 or
        # N is greater than 2*h
        if (arr[N - 1] == N - 1 or
        2 * arr[N - 1] <= N):
            return "Yes"
        else:
            return "No"
    # Case 2
    else:
         
        # Stores frequency of minimum element
        x = 0
 
        for i in range(N):
 
            # Frequency of minimum element
            if (arr[i] == arr[0]):
                x += 1
 
        # Stores frequency of maximum element
        y = N - x
 
        # Condition for case 2
        if ((arr[N - 1] >= x + 1) and
            (arr[N - 1] <= x + y // 2)):
            return "Yes"
        else:
            return "No"
 
# Driver Code
if __name__ == '__main__':
     
    # Given Input
    arr = [ 1, 2, 2 ]
    N = len(arr)
 
    # Function Call
    print (checkValid(arr, N))
     
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to check if coloring is
// possible with give conditions
static string checkValid(int []arr, int N)
{
 
    // Sort the vector
    Array.Sort(arr);
 
    // Coloring not possible in case of
    // maximum - minimum element > 1
    if (arr[N - 1] - arr[0] > 1)
        return "No";
 
    // case 1
    else if (arr[N - 1] == arr[0]) {
 
        // If h is equal to N-1 or
        // N is greater than 2*h
        if (arr[N - 1] == N - 1
            || 2 * arr[N - 1] <= N)
            return "Yes";
        else
            return "No";
    }
    // Case 2
    else {
        // Stores frequency of minimum element
        int x = 0;
 
        for (int i = 0; i < N; i++) {
 
            // Frequency  of minimum element
            if (arr[i] == arr[0])
                x++;
        }
 
        // Stores frequency of maximum element
        int y = N - x;
 
        // Condition for case 2
        if ((arr[N - 1] >= x + 1)
            && (arr[N - 1] <= x + y / 2))
            return "Yes";
        else
            return "No";
    }
}
 
// Driver Code
public static void Main()
{
    // GivenInput
    int []arr = { 1, 2, 2 };
    int N = arr.Length;
 
    // Function Call
    Console.Write(checkValid(arr, N));
 
}
}
 
// This code is contributed by SURENDRA_GANGWAR.

Javascript

 
// Javascript program for the above approach
 
// Function to check if coloring is
// possible with give conditions
function checkValid(arr, N)
{
     
    // Sort the vector
    arr.sort((a, b) => a - b);
 
    // Coloring not possible in case of
    // maximum - minimum element > 1
    if (arr[N - 1] - arr[0] > 1)
        return "No";
 
    // Case 1
    else if (arr[N - 1] == arr[0])
    {
         
        // If h is equal to N-1 or
        // N is greater than 2*h
        if (arr[N - 1] == N - 1 ||
        2 * arr[N - 1] <= N)
            return "Yes";
        else
            return "No";
    }
     
    // Case 2
    else
    {
         
        // Stores frequency of minimum element
        let x = 0;
 
        for(let i = 0; i < N; i++)
        {
             
            // Frequency  of minimum element
            if (arr[i] == arr[0])
                x++;
        }
 
        // Stores frequency of maximum element
        let y = N - x;
 
        // Condition for case 2
        if ((arr[N - 1] >= x + 1) &&
            (arr[N - 1] <= x + y / 2))
            return "Yes";
        else
            return "No";
    }
}
 
// Driver Code
 
// GivenInput
let arr = [ 1, 2, 2 ];
let N = arr.length;
 
// Function Call
document.write(checkValid(arr, N));
 
// This code is contributed by gfgking
 

输出

Yes

时间复杂度： O(N*log(N))
辅助空间： O(1)