需要训练且力量和耐力比其他任何球员都低的球员人数

给定具有三个组件的2D阵列播放器[power、urance、id] 。如果一个球员的力量和耐力比任何其他球员都低，那么他就需要训练。任务是用他们的id找出需要训练的玩家数量。

例子：

Input: {{5, 4, 1}, {6, 3, 2}, {3, 5, 3}}
Output: 0
Explanation: There is no player which has strictly greater power and endurance than any other player.

Input: {{1, 1, 0}, {2, 2, 1}, {3, 3, 2}}
Output: 2
             0 1
Explanation: Below are the players who need training
Player with id = 0, having power = 1 and endurance = 1 is strictly less than player with id = 2. 
Player with id = 1, having power = 2 and endurance = 2 is strictly less than player with id = 2.
Therefore, there are 2 players who need training. 

方法：这个问题可以通过使用贪心方法来解决。让X和Y成为两个玩家。如果存在Y使得X 的力量 < Y 的力量并且X 的耐力 < Y 的耐力，则玩家X需要训练。请按照以下步骤解决给定的问题。

• 将根据两个参数对两个玩家进行比较。
• 以非递减的幂顺序对数组player[][]进行排序。
• 如果两个元素具有相同的功率值，则首先考虑耐力值较大的元素。
• 从 player [][]数组的右侧迭代并跟踪最大的先前耐力。
  • 一开始，如果任何玩家有资格接受培训，则存储其id 。
  • 如果当前玩家耐力小于之前的最大耐力值，则增加玩家计数。
  • 否则，更新最大耐力值。
• 返回存储需要训练的玩家的所有id的数组。

下面是上述方法的实现。

C++

// C++ program for above approach
#include 
using namespace std;
 
bool compare(vector, vector);
 
vector CountOfPlayers(vector >& qualities)
{
    int count = 0;
    int n = qualities.size();
 
    sort(qualities.begin(), qualities.end(),
         [](vector& entry1, vector& entry2)
    {
       
             // If power value is equal
             // for both elements
             // Sort in descending order
             // according to endurance value
             if (entry1[0] == entry2[0])
                 return entry2[1] < entry1[1];
 
             else
                 return entry1[0] < entry2[0];
         });
 
    // Keep track of maximum
    // endurance value in right side
    int ma = 0;
 
    vector res;
 
    // Traverse the array players
    for (int i = n - 1; i >= 0; i--) {
 
        // If current endurance
        // value is smaller than
        // max then we will
        // increment the count
        int id = qualities[i][2];
        if (qualities[i][1] < ma) {
 
            // Adding player
            // to the final array
            res.push_back(id);
 
            int q1 = qualities[i + 1][0] - qualities[i][0];
            int q2 = ma - qualities[i][1];
 
            // Increase the count
            count++;
        }
 
        // Else update max value
        ma = max(ma, qualities[i][1]);
    }
 
    return res;
}
 
// Driver Code
int main()
{
    vector > qualities
        = { { 1, 1, 0 }, { 2, 2, 1 }, { 3, 3, 2 } };
    vector ans = CountOfPlayers(qualities);
 
    // Print total number of players
    // who need training
    cout << ans.size() << "\n";
 
    // Printing id of each player
    for (int i = 0; i < ans.size(); i++) {
        cout << ans[i] << " ";
    }
 
   return 0;
}
 
    // This code is contributed by rakeshsahni

Java

// Java program for above approach
import java.io.*;
import java.util.*;
 
class GFG {
    private static Vector
    CountOfPlayers(int[][] qualities)
    {
        int count = 0;
        int n = qualities.length;
 
        sort(qualities);
 
        // Keep track of maximum
        // endurance value in right side
        int max = 0;
 
        Vector res
            = new Vector();
 
        // Traverse the array players
        for (int i = n - 1; i >= 0; i--) {
 
            // If current endurance
            // value is smaller than
            // max then we will
            // increment the count
            int id = qualities[i][2];
            if (qualities[i][1] < max) {
 
                // Adding player
                // to the final array
                res.add(id);
 
                int q1
                    = qualities[i + 1][0]
                      - qualities[i][0];
                int q2 = max - qualities[i][1];
 
                // Increase the count
                count++;
            }
 
            // Else update max value
            max = Math.max(max, qualities[i][1]);
        }
 
        return res;
    }
 
    // Sort on the array according
    // to the above approach
    public static void sort(int arr[][])
    {
        // Using built-in sort
        // function Arrays.sort
        Arrays.sort(arr, new Comparator() {
 
            @Override
            // Compare values according to columns
            public int compare(int[] entry1,
                               int[] entry2)
            {
 
                // If power value is equal
                // for both elements
                // Sort in descending order
                // according to endurance value
                if (entry1[0] == entry2[0])
                    return entry2[1] - entry1[1];
 
                else
                    return entry1[0] - entry2[0];
            }
        });
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int[][] qualities
            = { { 1, 1, 0 },
                { 2, 2, 1 },
                { 3, 3, 2 } };
        Vector ans
            = CountOfPlayers(qualities);
 
        // Print total number of players
        // who need training
        System.out.println(ans.size());
 
        // Printing id of each player
        for (int i = 0; i < ans.size(); i++) {
            System.out.print(ans.get(i));
            System.out.print(" ");
        }
    }
}

C#

// C# program for above approach
 
using System;
using System.Collections.Generic;
 
public class GFG {
  private static List
    CountOfPlayers(List>  qualities)
  {
    int count = 0;
    int n = qualities.Count;
 
    qualities.Sort((entry1,entry2)=>{
      if (entry1[0] == entry2[0])
        return entry2[1] - entry1[1];
 
      else
        return entry1[0] - entry2[0];
    });
 
    // Keep track of maximum
    // endurance value in right side
    int max = 0;
 
    List res
      = new List();
 
    // Traverse the array players
    for (int i = n - 1; i >= 0; i--) {
 
      // If current endurance
      // value is smaller than
      // max then we will
      // increment the count
      int id = qualities[i][2];
      if (qualities[i][1] < max) {
 
        // Adding player
        // to the readonly array
        res.Add(id);
 
        int q1
          = qualities[i + 1][0]
          - qualities[i][0];
        int q2 = max - qualities[i][1];
 
        // Increase the count
        count++;
      }
 
      // Else update max value
      max = Math.Max(max, qualities[i][1]);
    }
 
    return res;
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    int[,] qualities
      = { { 1, 1, 0 },
         { 2, 2, 1 },
         { 3, 3, 2 } };
    List> qualities = new List> ();
    for(int i = 0; i < qualities.GetLength(0); i++){
      List l = new List();
      l.Add(qualities[i, 0]);
      l.Add(qualities[i, 1]);
      l.Add(qualities[i, 2]);
      qualitie.Add(l);
    }
    List ans
      = CountOfPlayers(qualitie);
 
    // Print total number of players
    // who need training
    Console.WriteLine(ans.Count);
 
    // Printing id of each player
    for (int i = 0; i < ans.Count; i++) {
      Console.Write(ans[i]);
      Console.Write(" ");
    }
  }
}
 
// This code contributed by Rajput-Ji

Javascript

 
      // JavaScript Program to implement
      // the above approach
      function CountOfPlayers(qualities)
      {
          let count = 0;
          let n = qualities.length;
 
          qualities.sort(function (entry1, entry2)
          {
 
              // If power value is equal
              // for both elements
              // Sort in descending order
              // according to endurance value
              if (entry1[0] == entry2[0])
                  return entry2[1] - entry1[1];
 
              else
                  return entry1[0] - entry2[0];
          })
 
          // Keep track of maximum
          // endurance value in right side
          let ma = 0;
 
          let res = [];
 
          // Traverse the array players
          for (let i = n - 1; i >= 0; i--) {
 
              // If current endurance
              // value is smaller than
              // max then we will
              // increment the count
              let id = qualities[i][2];
              if (qualities[i][1] < ma) {
 
                  // Adding player
                  // to the final array
                  res.push(id);
 
                  let q1 = qualities[i + 1][0] - qualities[i][0];
                  let q2 = ma - qualities[i][1];
 
                  // Increase the count
                  count++;
              }
 
              // Else update max value
              ma = Math.max(ma, qualities[i][1]);
          }
 
          return res;
      }
 
      // Driver Code
 
      let qualities
          = [[1, 1, 0], [2, 2, 1], [3, 3, 2]];
      let ans = CountOfPlayers(qualities);
 
      // Print total number of players
      // who need training
      document.write(ans.length + '<br>');
 
      // Printing id of each player
      for (let i = 0; i < ans.length; i++) {
          document.write(ans[i] + " ");
      }
 
  // This code is contributed by Potta Lokesh
  

输出

2
1 0 

时间复杂度： O(NlogN)，其中 N 是数组的大小。

空间复杂度： O(1)。