# Solution For Find Players With Zero Or One Losses

Problem Statement:
You are given a table, Scores, containing two columns: ‘Player’ and ‘Score’. The table contains data about the players and their scores in a competition. Write a SQL query to find all the players with zero or one loss. The output should be ordered by the player’s ID.

Sample Input:

Scores Table:

| Player | Score |
|——–|——-|
| 1 | 10 |
| 2 | 20 |
| 3 | 30 |
| 4 | 40 |
| 5 | 50 |
| 6 | 60 |
| 7 | 70 |

Sample Output:

Output Table:

| Player |
|——–|
| 1 |
| 2 |

Explanation:
The players with the IDs of 1 and 2 have 0 or 1 losses and are therefore included in the output.

Solution:

The problem requires us to find the players with 0 or 1 loss. We can start with aggregating the number of losses by counting the number of distinct games lost by each player and filtering the players having 0 or 1 losses.

Now, the scores table only provides the score of each game played by a player. We can calculate the number of losses by comparing each player’s highest score with the score of the other players. If the highest score of a player is less than the maximum score of any other player, we consider it a loss for that player.

The SQL query for the solution is given below:

`SELECT Player FROM (SELECT Player, COUNT(DISTINCT b.Player) AS losses FROM Scores a LEFT JOIN Scores b ON a.Player != b.Player AND a.Score < b.Score GROUP BY Player) p WHERE p.losses <= 1 ORDER BY Player;`

Explanation:

We start by creating a subquery to count the number of losses of each player. We do this by using a self-join on the Scores table where we match each row with all the other rows having the same or higher scores.

`SELECT Player, COUNT(DISTINCT b.Player) AS losses FROM Scores a LEFT JOIN Scores b ON a.Player != b.Player AND a.Score < b.Score GROUP BY Player`

Since we are only interested in the number of other players having higher scores than a player, we use the condition `a.Score < b.Score` in the join statement. Also, we use the `DISTINCT` keyword with the count function to avoid counting the same player multiple times.

We then use this subquery as a derived table named p to filter the players having 0 or 1 losses.

`WHERE p.losses <= 1`

Finally, we order the result by the player’s ID using the `ORDER BY` clause.

`ORDER BY Player;`

The output table will have the player IDs of the players with 0 or 1 losses, sorted in ascending order.

## Step by Step Implementation For Find Players With Zero Or One Losses

```There are several ways to solve this problem. One approach would be to use a HashMap to keep track of the players with zero or one losses. Another approach would be to use an array to keep track of the players with zero or one losses.

We will use the HashMap approach.

First, we create a HashMap and iterate through the list of players. For each player, we check to see if they have zero or one losses. If they do, we add them to the HashMap.

Next, we iterate through the HashMap and print out the players with zero or one losses.

import java.util.HashMap;

public class Solution {

public static void main(String[] args) {

// create a HashMap to keep track of the players with zero or one losses

HashMap map = new HashMap<>();

// iterate through the list of players

for (String player : players) {

// check to see if the player has zero or one losses

if (hasZeroOrOneLosses(player)) {

// add the player to the HashMap

map.put(player, losses);

}

}

// iterate through the HashMap and print out the players with zero or one losses

for (String player : map.keySet()) {

System.out.println(player + " has " + map.get(player) + " losses");

}

}

}```
```def findPlayers(self, players):
ans = []
for player in players:
if player == 0 or player == 1:
ans.append(player)
return ans```
```var findPlayers = function(losses, wins) {
// create a map to store the losses and wins for each player
let map = new Map();

// iterate over the losses array and store each player's losses in the map
for (let i = 0; i < losses.length; i++) {
if (!map.has(losses[i])) {
map.set(losses[i], 1);
} else {
map.set(losses[i], map.get(losses[i]) + 1);
}
}

// iterate over the wins array and store each player's wins in the map
for (let i = 0; i < wins.length; i++) {
if (!map.has(wins[i])) {
map.set(wins[i], -1);
} else {
map.set(wins[i], map.get(wins[i]) - 1);
}
}

// create an array to store the players with zero or one losses
let players = [];

// iterate over the map and push the players with zero or one losses into the array
for (let [player, losses] of map) {
if (losses <= 1) {
players.push(player);
}
}

// return the array of players
return players;
};```
```#include
#include
#include

using namespace std;

class Solution {
public:
vector findPlayers(vector& losses, vector& wins) {
vector result;
int n = losses.size();
int m = wins.size();
if (n != m) return result;
for (int i = 0; i < n; ++i) {
if (losses[i] == 0 || wins[i] == 1) {
result.push_back(i);
}
}
return result;
}
};```
```using System;
using System.Collections.Generic;
using System.Linq;

public class Solution {
public IList FindPlayers(int[] nums) {

// create a list to store the results
List results = new List();

// create a hashset to store the unique values
HashSet set = new HashSet();

// loop through the input array and add values to the hashset
for(int i = 0; i < nums.Length; i++) {
}

// loop through the hashset and check if the value - 1 is also present in the hashset
// if it is not present, then it means that the player has no losses, so add it to the results list
foreach(int val in set) {
if(!set.Contains(val - 1)) {