let bestCount = [];
    let lowestCount = [];
    let bestScore = scores[0];
    let lowestScore = scores[0]

    for (let i = 0; i < scores.length; i++) {
        if (scores[i] > bestScore) {
            bestCount.push(scores[i])
            bestScore = scores[i]
        } else if (scores[i] < lowestScore) {
            lowestScore = scores[i]
            lowestCount.push(scores[i])
        }
    }
    console.log(bestCount, lowestCount)
    return [bestCount.length, lowestCount.length]

Here is my Javascript solution

const max = scores[0]; const min = scores[1];

let maxCount = 0;
let minCount = 0;

for(let i=2; i < scores.length; i++){
    if(scores[i] > max){
        maxCount += 1;
    }else if(scores[i] < min){
        minCount += 1;
    }
}

console.log(maxCount, minCount);

Here is my c++ solution, you can watch the explanation here : https://youtu.be/vnovwmHPVXE

vector<int> breakingRecords(vector<int> scores) {
    int lowest = scores[0], highest = scores[0], lowestCount = 0, highestCount = 0;
    for(int score: scores){
        if(score < lowest){lowestCount++; lowest = score;}
        if(score > highest){highestCount++; highest = score;}
    }
    return {highestCount, lowestCount};
}

Python 3

def breakingRecords(scores):

HS=scores[0]
LS=scores[0]
HC = 0
LC = 0
for score in scores:

    if score > HS:
        HS = score
        HC+=1
        print(HS)
    elif score < LS:
        LS = score
        LC+=1

return [HC, LC]

pypy3

Write your code here

def breakingRecords(scores):

hcount=0
lcount=0
temp=scores[0]
temp2=scores[0]

for i in range(1,len(scores)):
    if scores[i]>temp:
        hcount+=1
        temp=scores[i]
    elif scores[i]<temp2:
        lcount+=1
        temp2=scores[i]


return hcount , lcount********