We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Practice
  2. Algorithms
  3. Implementation
  4. Breaking the Records
  5. Discussions

Breaking the Records

Sort 744 Discussions, By:

votes

  • lahouari + 9 comments

    My java submission:

    static int[] getRecord(int[] s)
{
    int highest, lowest;
    highest = lowest = s[0];
    
    int[] result = new int[2];
    
    for (int s_i = 1; s_i < s.length; s_i++)
    {
        if (s[s_i] > highest)
        {
            highest = s[s_i];
            ++result[0];
        }
        else if (s[s_i] < lowest)
        {
            lowest = s[s_i];
            ++result[1];
        }
    }
    
    return result;
}
    • JulioLeme + 2 comments

      Show, my solution is very like with yours, nice to see the similarities, see it:

          static int[] getRecord(int[] s){
        int[] result = new int[2];
        int countMax=0, countMin=0 ,max = s[0], min = s[0];
        for(int i=1; i< s.length; i++){
           if(s[i] > max){ max = s[i];  countMax++; }
           if(s[i] < min){ min = s[i];  countMin++; }
        }
        result[0] = countMax; result[1] = countMin;
        return result;
    }
      • work30 + 2 comments

        "else if" is slightly better than if, if because both cannot be true Also pre-increment slightly better than post-increment I think?

        HTH

        • enricogiurin + 0 comments

          or you can add a continue after the first if

        • Anime_Love + 0 comments

          please explain where the hell pre increment is better?

      • umangsomtiya8083 + 1 comment

        include

        include

        using namespace std;

        int main() { int n; cin>>n; int i,j,score[n],min,max,u=0,v=0; for(i=0;i>score[n]; } max=score[0]; min=score[0]; for(j=0;j 

            }
     for(j=0;j<n;j++)
    {
        if( min>score[j])
        {
            v++;
            min=score[j];
        }

    }


cout<<u<<" "<<v;

        } this code is not giving correct answer and logic is simmilar to your code. can you plz explain why?

        • PedroStu + 1 comment

          Why are you doing it with interger array? Try it using vector, as given in the question. If you change the format of the answer/method, you are skipping the learning part. Vectors are really very powerful as compared to any other array. I know this will give you many other troubles like the old 'Segmentation Fault'. But this is what is expected once you are advancing in the programming world. this can be a way:

          vector<int> breakingRecords(vector<int> scores) 
{
vector<int> Minmax(2);  //VERY CRUCIAL FOR AVOIDING SEGMENTATION FAULT

int i=0;
long long int Min=0, Max=0;

Max = scores.at(i);
Min = scores.at(i);i++;

while (i< scores.size())
{
if(scores.at(i) > Max)
{
    Max = scores.at(i);Minmax.at(0)++;
}
else if(scores.at(i) < Min)
{
    Min = scores.at(i);Minmax.at(1)++;
}
i++;
}

return(Minmax);
}
          • tjoberoi + 0 comments

            Yes, you are right. Thanks for your precious advice :)

    • alphazygma + 0 comments

      Heh heh heh, my solution is very similar to yours and Julio's

      static int[] getRecord(int[] s){
    if (s.length == 0) {
        return new int[2];
    }

    int lowest       = s[0];
    int highest      = s[0];
    int lowestCount  = 0;
    int highestCount = 0;
    for (int score : s) {
        if (score < lowest) {
            lowest = score;
            lowestCount++;

        } else if (score > highest) {
            highest = score;
            highestCount++;
        }
    }

    return new int[]{highestCount, lowestCount};
}
    • hiljusti + 3 comments

      Mine (in JS) was very similar: 

      function getRecord(games) {
    let most = games[0];
    let least = games[0];
    
    let dMost = 0;
    let dLeast = 0;
    
    for (const score of games) {
        if (most < score) {
            most = score;
            dMost++;
        }
        else if (score < least) {
            least = score;
            dLeast++;
        }
    }
    
    return [dMost, dLeast];
}
      • armadillo_ko167 + 0 comments

        it's min (in js)

        function breakingRecords(score) {
		        const scoreLength = score.length;

	        	let mostScore = score[0];
	        	let worstScore = score[0];
	        	let mostScoreIncreased = 0;
	        	let worstScoreDecreased = 0;

	            for(let i = 0; i < scoreLength; i++){
	            	const curScore = score[i];

	            	if(mostScore < curScore){
	            		mostScore = curScore;
			            mostScoreIncreased++;
		            }
		            if(worstScore > curScore){
	            		worstScore = curScore;
			            worstScoreDecreased++;
		            }
	            }
	            return [mostScoreIncreased, worstScoreDecreased];
	        }
      • anishd19 + 0 comments
        [deleted]
      • afreen08neela + 1 comment
        [deleted]
    • cu_16bcs2416 + 1 comment

      you could have used break in first if stmt as is would reduce time complexity!

      • enricogiurin + 0 comments

        You mean a continue, not a break.

    • rabiyaaruba + 0 comments

      evn...i hv done d same code...bt i'm getting error.....at the result initialization.....can u plz help me..!

    • Thinkster + 0 comments

      Ruby Implementation:

      def breaking_records(scores)
  min_el = max_el = scores[0]
  min_count, max_count = 0, 0
	
  (1...scores.length).each do |score|
    min, max = scores[0..score].min, scores[0..score].max
    if min < min_el
      min_el, min_count = min, (min_count += 1)
    elsif max > max_el
      max_el, max_count = max, (max_count += 1)
    end
  end
	
  [max_count, min_count]
end
    • coder_aky + 2 comments

      100 % working ,Simple and easy :-)

      In Python 3 :

      n,a= int(input()),list(map(int, input().split()))
max=min=a[0];c1=c2=0
for i in range(1,n):
    if a[i]>max:
        c1+=1;max=a[i]
    if a[i]<min:
        c2+=1;min=a[i]
print(c1,c2,sep=' ')
      • evterpa + 0 comments

        Why using the index? It's python - for score in scores :)

      • atiratkar + 0 comments

        That's good, but you should not be using min and max as variable names because they are predefined functions in Python.

    • CyberZii + 0 comments

      Beautiful!

    • shreeshreya17 + 0 comments

      what's the need of taking array for result..is there not a way to print min and max directly.

  • marinskiy + 5 comments

    Here is Python 3 solution from my HackerrankPractice repository:

    def breaking_records(score):
    min = max = score[0]
    min_count = max_count = 0
    for i in score[1:]:
        if i > max:
            max_count += 1
            max = i
        if i < min:
            min_count += 1
            min = i
    return max_count, min_count


n = int(input())
score = list(map(int, input().split()))
print(*breaking_records(score))

    Feel free to ask if you have any questions :)

    • mahamadbilal696 + 2 comments

      why you added * in the print statement.... please explain me.

      print(*breaking_records(score))

      • TheSoleCodr + 0 comments
        [deleted]
      • TheSoleCodr + 0 comments

        Addding * in print statement will print your answer as

        1 2 3 4 5

        and will print your answer like this when you do not use it

        [ 1, 2, 3, 4, 5 ]

    • Thinkster + 0 comments

      How about Ruby Implementation?

      def breaking_records(scores)
  min_el = max_el = scores[0]
  min_count = max_count = 0
	
  (1...scores.length).each do |score|
    min, max = scores[0..score].min, scores[0..score].max
    if min < min_el
      min_el, min_count = min, (min_count += 1)
    elsif max > max_el
      max_el, max_count = max, (max_count += 1)
    end
  end
	
  [max_count, min_count]
end
    • 4573r14 + 0 comments

      I liked this solution

    • krishnansr_siva + 0 comments

      I feel that the below solution is more 'pythonic' but obivously not well optimized

      def breaking_records(score):

      hs = [ i for i,item in enumerate(scores[1:]) if item>max(scores[0:i+1]) ]

      ls = [ i for i,item in enumerate(scores[1:]) if item

      return [len(hs), len(ls)]

    • robmcortez + 0 comments

      Just FYI, min and max are built in functions in python. You probably don't want to use them for variable names.

  • abhiarora1137 + 1 comment

    My code is working properly in editor, but i'm getting error here , "result.join() is not a function". Code language Javascript.

    • santosh532 + 1 comment

      Looks like you have to return in array format eg: [max, min]

      • seantbmorgan + 0 comments

        Good lookin' out.

  • itsbaldeep + 2 comments

    JavaScript

    // Preparing variables
let [ hi, lo ] = [ scores[0], scores[0] ];
let [ max, min ] = [ 0, 0 ];

// Calculating
for (let i = 1; i < scores.length; i++) {
	if (scores[i] > hi) { hi = scores[i]; max++; }
	if (scores[i] < lo) { lo = scores[i]; min++; }
}

//Returning
return [ max, min ];
    • mdjonesonly310 + 0 comments

      Beautiful

    • gary_l_hewitt + 1 comment

      The only advantage is that it also gives the record of each time the record is broken.

      function breakingRecords(scores) {
var h = [scores[0]]; 
var l = [scores[0]];
scores.forEach( e=> {
if (e>h.reduce((a,c)=>Math.max(a,c))) h.push(e);
if (e<l.reduce((a,c)=>Math.min(a,c))) l.push(e);
});

return [h.length-1,l.length-1];
}
      • NaveenKashyap + 1 comment

        if you are in an interview you don't wanna use built in functions

        • gary_l_hewitt + 1 comment

          Why not?

          • NaveenKashyap + 1 comment

            If you use a library function then it defeats the primary objective of testing your competency.

            • gary_l_hewitt + 0 comments

              Perhaps competency is using certain language features? Not trying to be a troll but one could say that syntax like:

              let [ hi, lo ] = [ scores[0], scores[0] ];

              is as much built-in as Math.max() or Array.reduce() which access javascript type-primitive methods.

              The loop method is much more efficient, however, since it maintains the min & max rather than recalculating it every time w the reduce(). I think I managed to create a very nice example of Schlemiel the painter's algorithm

  • helggg + 1 comment

    python:

    def breakingRecords(scores):
    m = l = scores[0]
    cm = cl = 0
    for i in scores:
        if i > m:
            cm += 1
            m = i
        if i < l:
            cl += 1
            l = i
    return cm, cl
    • vaishnavi_msn + 3 comments

      ive written a similar one but this code is solving only 11/15 cases. for n>1000 its not able to solve.idk why

      def breakingRecords(scores):
    p=q=scores[0]
    g=l=0
    for i in scores:
        if(i>p):
            p=i
            g=g+1
        if(i<q):
            q=i
            l=l+1
    g=str(g)
    l=str(l)
    z=[g,l]
    return(''.join(z))
      • catch22kn + 0 comments

        I had a similar issue. For some reason, changing my returning of a formatted string to:

        return highCount, lowCount

        worked for the ones that were giving me an issue.

      • yypark_mgiy + 0 comments
        [deleted]
      • ammadrocks + 0 comments

        same here i am also using python 3

Load more conversations