通过从任何递增对中重复删除一个元素，将数组减少为单个元素

给定一个由[1, N]范围内的排列组成的数组arr[] ，任务是检查给定的数组是否可以通过执行以下操作减少为单个非零元素：

Select indices i and j such that i < j and arr[i] < arr[j] and convert one of the two elements to 0.

如果可以将数组减少为单个元素，则在每个操作中打印“是” ，然后是所选索引以及被替换元素的索引。否则，打印“否” 。

例子：

Input: arr[] = {2, 4, 6, 1, 3, 5} 
Output: 
Yes 
0 1 1 
0 2 2 
3 4 3 
0 4 4 
0 5 5 
Explanation: 
In the first operation choose the elements 2 and 4 at indices 0 and 1. Convert the element 4 at index 1 to 0, arr[] = {2, 0, 6, 1, 3, 5} 
In the second operation choose the elements 2 and 6 at indices 0 and 2. Convert the element 6 at index 2 to 0, arr[] = {2, 0, 0, 1, 3, 5} 
In the third operation choose the elements 1 and 3 at indices 3 and 4. Convert the element 1 at index 3 to 0, arr[] = {2, 0, 0, 0, 3, 5} 
In the forth operation choose the elements 2 and 3 at indices 0 and 4. Convert the element 3 at index 4 to 0, arr[] = {2, 0, 0, 0, 0, 5} 
In the fifth operation choose the elements 2 and 5 at indices 0 and 5. Convert the element 5 at index 5 to 0, arr[] = {2, 0, 0, 0, 0, 0} 
So, the array is reduced to a single positive element in 5 operations.

Input: arr[] = {2, 3, 1} 
Output: No 
Explanation: 
There is not set of operations in which the given array can be converted to a single element.

方法：想法是首先将所有元素从索引[1, N – 2] 转换为0 ，然后在最后一步中消除第 0 个或第(N – 1)^个元素之一以获得单例数组。以下是解决问题的步骤：

1. 选择一组可以执行给定操作的有效索引。
2. 现在要选择要转换为0 的元素，请检查以下条件：
   • 如果数组的第 0 个索引是这些索引的一部分，则将另一个索引处的元素转换为0 。
   • 如果第 0 个索引不是所选索引的一部分，则将两个元素中较小的一个替换为0 。
3. 直到单个正元件保持在阵列中继续此过程。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to print the order of
// indices of converted numbers
void order_of_removal(int a[], int n)
{
    // Stores the values of indices
    // with numbers > first index
    queue greater_indices;
 
    int first = a[0];
 
    // Stores the index of the closest
    // consecutive number to index 0
    int previous_index = 0;
 
    // Push the indices into the queue
    for (int i = 1; i < n; i++) {
        if (a[i] > first)
            greater_indices.push(i);
    }
 
    // Traverse the queue
    while (greater_indices.size() > 0) {
 
        // Stores the index of the
        // closest number > arr[0]
        int index = greater_indices.front();
 
        greater_indices.pop();
 
        int to_remove = index;
 
        // Remove elements present in
        // indices [1, to_remove - 1]
        while (--to_remove > previous_index) {
 
            cout << to_remove << " "
                 << index << " "
                 << to_remove << endl;
        }
 
        cout << 0 << " " << index << " "
             << index << endl;
 
        // Update the previous_index
        // to index
        previous_index = index;
    }
}
 
// Function to check if array arr[] can
// be reduced to single element or not
void canReduced(int arr[], int N)
{
    // If array element can't be
    // reduced to single element
    if (arr[0] > arr[N - 1]) {
        cout << "No" << endl;
    }
 
    // Otherwise find the order
    else {
        cout << "Yes" << endl;
        order_of_removal(arr, N);
    }
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 1, 2, 3, 4 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    canReduced(arr, N);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
     
// Function to print the order of
// indices of converted numbers
public static void order_of_removal(int[] a, int n)
{
     
    // Stores the values of indices
    // with numbers > first index
    Queue greater_indices = new LinkedList<>();
   
    int first = a[0];
   
    // Stores the index of the closest
    // consecutive number to index 0
    int previous_index = 0;
   
    // Push the indices into the queue
    for(int i = 1; i < n; i++)
    {
        if (a[i] > first)
            greater_indices.add(i);
    }
   
    // Traverse the queue
    while (greater_indices.size() > 0)
    {
         
        // Stores the index of the
        // closest number > arr[0]
        int index = greater_indices.peek();
   
        greater_indices.remove();
   
        int to_remove = index;
   
        // Remove elements present in
        // indices [1, to_remove - 1]
        while (--to_remove > previous_index)
        {
            System.out.println(to_remove + " " +
                                   index + " " +
                                   to_remove);
        }
   
        System.out.println(0 + " " + index +
                               " " + index);
   
        // Update the previous_index
        // to index
        previous_index = index;
    }
}
   
// Function to check if array arr[] can
// be reduced to single element or not
public static void canReduced(int[] arr, int N)
{
     
    // If array element can't be
    // reduced to single element
    if (arr[0] > arr[N - 1])
    {
        System.out.println("No");
    }
   
    // Otherwise find the order
    else
    {
        System.out.println("Yes");
        order_of_removal(arr, N);
    }
} 
 
// Driver Code
public static void main(String[] args)
{
     
    // Given array arr[]
    int arr[] = { 1, 2, 3, 4 };
   
    int N = arr.length;
   
    // Function call
    canReduced(arr, N);
}
}
 
// This code is contributed divyeshrabadiya07

Python3

# Python3 program for the above approach
 
# Function to print the order of
# indices of converted numbers
def order_of_removal(a, n) :
 
    # Stores the values of indices
    # with numbers > first index
    greater_indices = []
 
    first = a[0]
 
    # Stores the index of the closest
    # consecutive number to index 0
    previous_index = 0
 
    # Push the indices into the queue
    for i in range(1, n) :
        if (a[i] > first) :
            greater_indices.append(i)
 
    # Traverse the queue
    while (len(greater_indices) > 0) :
 
        # Stores the index of the
        # closest number > arr[0]
        index = greater_indices[0];
 
        greater_indices.pop(0)
 
        to_remove = index
 
        # Remove elements present in
        # indices [1, to_remove - 1]
        to_remove =- 1
        while (to_remove > previous_index) :
 
            print(to_remove, index, to_remove)
 
        print(0, index, index)
 
        # Update the previous_index
        # to index
        previous_index = index
 
# Function to check if array arr[] can
# be reduced to single element or not
def canReduced(arr, N) :
 
    # If array element can't be
    # reduced to single element
    if (arr[0] > arr[N - 1]) :
        print("No")
 
    # Otherwise find the order
    else :
        print("Yes")
        order_of_removal(arr, N)
 
# Given array arr[]
arr = [ 1, 2, 3, 4 ]
 
N = len(arr)
 
# Function Call
canReduced(arr, N)
 
# This code is contributed by divyesh072019

C#

// C# program for
// the above approach
using System;
using System.Collections.Generic;
class GFG{
     
// Function to print the order of
// indices of converted numbers
public static void order_of_removal(int[] a,
                                    int n)
{
  // Stores the values of indices
  // with numbers > first index
  Queue greater_indices = new Queue();
 
  int first = a[0];
 
  // Stores the index of the closest
  // consecutive number to index 0
  int previous_index = 0;
 
  // Push the indices into the queue
  for(int i = 1; i < n; i++)
  {
    if (a[i] > first)
      greater_indices.Enqueue(i);
  }
 
  // Traverse the queue
  while (greater_indices.Count > 0)
  {
    // Stores the index of the
    // closest number > arr[0]
    int index = greater_indices.Peek();
 
    greater_indices.Dequeue();
 
    int to_remove = index;
 
    // Remove elements present in
    // indices [1, to_remove - 1]
    while (--to_remove > previous_index)
    {
      Console.WriteLine(to_remove + " " +
                        index + " " + to_remove);
    }
 
    Console.WriteLine(0 + " " + index +
                      " " + index);
 
    // Update the previous_index
    // to index
    previous_index = index;
  }
}
 
// Function to check if array []arr can
// be reduced to single element or not
public static void canReduced(int[] arr,
                              int N)
{
  // If array element can't be
  // reduced to single element
  if (arr[0] > arr[N - 1])
  {
    Console.WriteLine("No");
  }
 
  // Otherwise find the order
  else
  {
    Console.WriteLine("Yes");
    order_of_removal(arr, N);
  }
} 
 
// Driver Code
public static void Main(String[] args)
{
  // Given array []arr
  int []arr = {1, 2, 3, 4};
 
  int N = arr.Length;
 
  // Function call
  canReduced(arr, N);
}
}
 
// This code is contributed by Rajput-Ji

Javascript

      // JavaScript program for
      // the above approach
      // Function to print the order of
      // indices of converted numbers
      function order_of_removal(a, n) {
        // Stores the values of indices
        // with numbers > first index
        var greater_indices = [];
 
        var first = a[0];
 
        // Stores the index of the closest
        // consecutive number to index 0
        var previous_index = 0;
 
        // Push the indices into the queue
        for (var i = 1; i < n; i++) {
          if (a[i] > first) greater_indices.push(i);
        }
 
        // Traverse the queue
        while (greater_indices.length > 0) {
          // Stores the index of the
          // closest number > arr[0]
          var index = greater_indices[0];
 
          greater_indices.shift();
 
          var to_remove = index;
 
          // Remove elements present in
          // indices [1, to_remove - 1]
          to_remove -= 1;
          while (to_remove > previous_index) {
            document.write(to_remove + " " + index + " " + to_remove + "<br>");
          }
 
          document.write(0 + " " + index + " " + index + "<br>");
 
          // Update the previous_index
          // to index
          previous_index = index;
        }
      }
 
      // Function to check if array []arr can
      // be reduced to single element or not
      function canReduced(arr, N) {
        // If array element can't be
        // reduced to single element
        if (arr[0] > arr[N - 1]) {
          document.write("No" + "<br>");
        }
 
        // Otherwise find the order
        else {
          document.write("Yes" + "<br>");
          order_of_removal(arr, N);
        }
      }
 
      // Driver Code
      // Given array []arr
      var arr = [1, 2, 3, 4];
 
      var N = arr.length;
 
      // Function call
      canReduced(arr, N);
       
      // This code is contributed by rdtank.
    

输出：

Yes
0 1 1
0 2 2
0 3 3

时间复杂度： O(N)
辅助空间： O(N)

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