通过从一对中删除绝对最大差为K的较大元素，将数组简化为单个元素

给定一个由N个整数组成的数组arr [] 和正整数K ，任务是检查是否可以通过重复删除绝对差最大为K的一对中存在的两个元素中较大的一个，来将给定的数组简化为单个元素。如果可以将数组简化为单个元素，则打印“是” 。否则，打印“否” 。

例子：

Input: arr[] = {2, 1, 1, 3}, K = 1
Output: Yes
Explanation:
Operation 1: Select the pair {arr[0], arr[3]} ( = (2, 3), as | 3 – 2 | ≤ 1. Now, remove 3 from the array. The array modifies to {2, 1, 1}.
Operation 2: Select the pair {arr[0], arr[1]} ( = (2, 1), as | 2 – 1 | ≤ 1. Now, remove 2 from the array. The array modifies to {1, 1}.
Operation 3: Remove 1 from the array. The array modifies to {1}. 
Therefore, the last remaining array element is 1.

Input: arr[] = {1, 4, 3, 6}, K = 1
Output: No

方法：可以使用贪婪方法解决给定的问题。想法是在所有可能的移动中删除具有最大值的元素。请按照给定的步骤解决问题：

• 以降序对给定数组arr []进行排序。
• 在索引[0，N – 2]的范围内遍历数组arr [ ] 。如果arr [i]和arr [i + 1]的绝对值大于K ，则打印“否”并退出循环。
• 如果循环完全执行，则打印“是” 。

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 an array can be
// reduced to single element by removing
// maximum element among any chosen pairs
void canReduceArray(int arr[], int N, int K)
{
    // Sort the array in descending order
    sort(arr, arr + N, greater());
 
    // Traverse the array
    for (int i = 0; i < N - 1; i++) {
 
        // If the absolute difference
        // of 2 consecutive array
        // elements is greater than K
        if (arr[i] - arr[i + 1] > K) {
 
            cout << "No";
            return;
        }
    }
 
    // If the array can be reduced
    // to a single element
    cout << "Yes";
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 1, 1, 3 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int K = 1;
 
    // Function Call to check
    // if an array can be reduced
    // to a single element
    canReduceArray(arr, N, K);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG
{
 
// Function to check if an array can be
// reduced to single element by removing
// maximum element among any chosen pairs
static void canReduceArray(int arr[], int N, int K)
{
   
    // Sort the array in descending order
    Arrays.sort(arr);
    int b[] = new int[N];
        int j = N;
        for (int i = 0; i < N; i++) {
            b[j - 1] = arr[i];
            j = j - 1;
        }
    // Traverse the array
    for (int i = 0; i < N - 1; i++) {
 
        // If the absolute difference
        // of 2 consecutive array
        // elements is greater than K
        if (arr[i] - arr[i + 1] > K) {
 
            System.out.print("No");
            return;
        }
    }
 
    // If the array can be reduced
    // to a single element
    System.out.print("Yes");
}
 
// Driven Code
public static void main(String[] args)
{
    int arr[] = { 2, 1, 1, 3 };
    int N = arr.length;
    int K = 1;
 
    // Function Call to check
    // if an array can be reduced
    // to a single element
    canReduceArray(arr, N, K);
}
}
 
// This code is contributed by splevel62

Python3

# Python3 program for the above approach
 
# Function to check if an array can be
# reduced to single element by removing
# maximum element among any chosen pairs
def canReduceArray(arr, N, K):
     
    # Sort the array in descending order
    arr = sorted(arr)
 
    # Traverse the array
    for i in range(N - 1):
         
        # If the absolute difference
        # of 2 consecutive array
        # elements is greater than K
        if (arr[i] - arr[i + 1] > K):
            print ("No")
            return
 
    # If the array can be reduced
    # to a single element
    print ("Yes")
 
# Driver Code
if __name__ == '__main__':
     
    arr = [2, 1, 1, 3]
    N = len(arr)
    K = 1
 
    # Function Call to check
    # if an array can be reduced
    # to a single element
    canReduceArray(arr, N, K)
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to check if an array can be
// reduced to single element by removing
// maximum element among any chosen pairs
static void canReduceArray(int[] arr, int N,
                           int K)
{
     
    // Sort the array in descending order
    Array.Sort(arr);
    int[] b = new int[N];
    int j = N;
     
    for(int i = 0; i < N; i++)
    {
        b[j - 1] = arr[i];
        j = j - 1;
    }
     
    // Traverse the array
    for(int i = 0; i < N - 1; i++)
    {
         
        // If the absolute difference
        // of 2 consecutive array
        // elements is greater than K
        if (arr[i] - arr[i + 1] > K)
        {
            Console.WriteLine("No");
            return;
        }
    }
 
    // If the array can be reduced
    // to a single element
    Console.WriteLine("Yes");
}
 
// Driver Code
public static void Main(String []args)
{
    int[] arr = { 2, 1, 1, 3 };
    int N = arr.Length;
    int K = 1;
 
    // Function Call to check
    // if an array can be reduced
    // to a single element
    canReduceArray(arr, N, K);
}
}
 
// This code is contributed by souravghosh0416

输出：

Yes

时间复杂度： O(N * log N)
辅助空间： O(1)