lahouari 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 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 "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 or you can add a continue after the first if
umangsomtiya8083 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 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); }
alphazygma 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 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 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 [deleted]
afreen08neela what is the time limit and memory limit given by hacker rank for this problem?
davidheisnam [deleted]
cu_16bcs2416 you could have used break in first if stmt as is would reduce time complexity!
enricogiurin You mean a continue, not a break.
rabiyaaruba evn...i hv done d same code...bt i'm getting error.....at the result initialization.....can u plz help me..!
Thinkster 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 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=' ')
CyberZii Beautiful!
shreeshreya17 what's the need of taking array for result..is there not a way to print min and max directly.
marinskiy 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 why you added * in the print statement.... please explain me.
print(*breaking_records(score))
TheSoleCodr [deleted]TheSoleCodr 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 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 I liked this solution
krishnansr_siva 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 Just FYI,
minandmaxare built in functions in python. You probably don't want to use them for variable names.
abhiarora1137 My code is working properly in editor, but i'm getting error here , "result.join() is not a function". Code language Javascript.
santosh532 Looks like you have to return in array format eg: [max, min]
seantbmorgan Good lookin' out.
helggg 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 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 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 [deleted]
ammadrocks same here i am also using python 3
itsbaldeep 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 Beautiful
gary_l_hewitt 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 if you are in an interview you don't wanna use built in functions
gary_l_hewitt Why not?
NaveenKashyap If you use a library function then it defeats the primary objective of testing your competency.
gary_l_hewitt 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()orArray.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
madhudskumar def getRecord(scores): ls = hs = s[0]; lc = hc = 0; for score in scores: if(score > hs): hs = score; hc += 1; if(score < ls): ls = score; lc += 1; return [hc, lc]
This worked
esyray15 You could have just wrote the
returnstatment asreturn hc, lc
david_mathis_ii why iterate through s[0] if you already assigned it? you could shave this like:
fore score in scores[1:]:
asd_shivam98 include
using namespace std; int main() { int n,s[n]; cin >> n; cin >> s[0]; int c=0,d=0; int a=s[0]; int b=s[0]; for(int i=1;i> s[i]; if(s[i]>a) { a=s[i]; c++; } else if(s[i] cout << c << " " << d; return 0; }
[deleted] My code is almost similar to yours. But there are apparently problems with mine. Could you please point them out?
n = int(raw_input().strip()) score = list(map(int, raw_input().strip().split(' '))) hb, lb = 0, -1 a = score[0] for b in score: if (b>a): a = b hb +=1 for c in score: if (c<=a): a = c lb +=1 print hb, print lb
lrkoehler77 Your not returning correctly. Do it like: return hb,lb
mahamadbilal696 that's not a problem... he assign the initial value of lb = -1 instead of lb = 0
chinmay43 My logic is almost same, But this code is giving runtime error in test case 10. what can be the issue ?
def breakingRecords(score): min_1 = score[0] max_1 = score[0] min_brk = 0 max_brk = 0 if len(score) < 2: return 0 for i in range(len(score)): if (score[i] < min_1): min_brk += 1 min_1 = score[i] elif(score[i] > max_1): max_brk += 1 max_1 = score[i] else: max_brk = max_brk min_brk = min_brk i += 1 return(max_brk, min_brk)
tylercjulian I just tested your code. The issue is with this line:
if len(score) < 2: return 0
Remove that line and your code passes the tests.
Also, for efficiency, there is no need for the
i += 1, since the for loop automatically increments i. You could also iterate directly through the scores instead of having to rely on using a range for index numbers. The parent example uses this method.chinmay43 Thanks a lot :-)
tylercjulian You are welcome :)
sshayann7 [deleted]
mumbaikar We can do directly in main function -
public static void main(String[] args) { int min_record = 0; int max_record = 0; int worst = 0; int best = 0; int curr_record = 0; Scanner in = new Scanner(System.in); int n = in.nextInt(); for(int s_i=0; s_i < n; s_i++){ curr_record = in.nextInt(); if(s_i == 0) { min_record = curr_record; max_record = curr_record; } if(s_i > 0) { if(curr_record < min_record) { min_record = curr_record; worst++; } else if (curr_record > max_record) { max_record = curr_record; best++; } } } System.out.println(best + " " + worst); System.out.println(""); in.close(); }
sgawlik I haven't found a C solution that actually uses the code as it is given. I wanted something that doesn't change any of the code but still returns the result in the format the author intended. If you follow my comments you can see how to pass values between functions in C. I hope this helps people who were also unsatisfied with solutions that were written entirely in the main function or those who've wondered what to do about result_size in the C starter code of so many of these early exercises.
#define HOWMANYRESULTS 2 int* getRecord(int s_size, int* s, int *result_size) { // set best and worst to first game int worst = s[0]; int best = s[0]; // let main know how many results you'll print *result_size = HOWMANYRESULTS; // allocate heap space for the results and initialize to 0 int *result = malloc(sizeof(int) * *result_size); result[0] = 0; result[1] = 0; // start loop from the second game for (int i = 1; i < s_size; i++) { if (s[i] > best) { // increase counter for best and reset best result[0]++; best = s[i]; } if (s[i] < worst) { // increase counter for worst and reset worst result[1]++; worst = s[i]; } } // return the address of the results on heap to main return result; }
I'm guessing that other people have also struggled with the common format for all C starter code so I have added some comments to the part of main that I found tricky:
// how many things will be printed int result_size; // call helper function, expect implementer to return // an int pointer (to the results) and set result_size int* result = getRecord(n, s, &result_size); for(int i = 0; i < result_size; i++) { // when i = 0 this evaluates to 0 which is same as // false and we don't print a space if (i) { printf(" "); } // on every round through the loop print another result printf("%d", result[i]); }
I know that allocating space on the heap is not as efficient/fast in the real world as just doing it all in main or printing in getRecord, my focus was on using the starter code and practicing with passing values between functions in C. Solution is O(n) runtime and O(1) space.
lee_rw_cs I like your code comment, I always forget to write documentation, sigh
rajputnisarg17 thanks
prusovigor You can also declare result as a static variable instead of using malloc, this will intialize it with zeros and it will still be accessible after return from function.
AzizStark int main(){ long int arra[2000],n,i,j,haa=0,great=0; scanf("%ld",&n); for(i=0;i<n;i++){ scanf("%ld", &arra[i]); } //========================================= great=arra[0]; for(i=0;i<n;i++){ for(j=1;j<n+1;j++){ if(great<arra[j]) { great=arra[j]; haa=haa+1; } } } printf("%ld", haa); //========================================= haa=0; great=arra[0]; for(i=0;i<n;i++){ for(j=0;j<n;j++){ if(great>arra[j]) { great=arra[j]; haa=haa+1; } } } printf("% ld", haa); return 0; }
Joe_08 Hello sgawlik! Could please explain how *result_size gets the value HOWMANYRESULTS? and also what is the purpose of doing so?....Thank you for your clear solution using C.
sghaida Scala Solution
var res = score.foldLeft((0,0),(score(0),score(0))){ case (acc, c) => c match { case c if c > acc._2._1 => ((acc._1._1+1,acc._1._2),(c,acc._2._2)) case c if c < acc._2._2 => ((acc._1._1, acc._1._2+1),(acc._2._1,c)) case _=> acc } } println(s"${res._1._1} ${res._1._2}")
andtorg Very nice!
That's mine with foldLeft as well: purely functional but with some boilerplate.
class Acc (val min: Int, val max: Int, val nMin: Int, val nMax: Int) object Acc { def apply(a: Int, b: Int, c: Int, d: Int) = new Acc(a,b,c,d) } def count(acc: Acc, elem: Int): Acc = if ( elem > acc.max ) Acc(acc.min, elem, acc.nMin, acc.nMax + 1) else if ( elem < acc.min ) Acc(elem, acc.max, acc.nMin + 1, acc.nMax) else Acc(acc.min, acc.max, acc.nMin, acc.nMax) val res = l.tail.foldLeft(Acc(l.head, l.head, 0, 0))(count)
liu00861 I am wondering why this failed
def breakingRecords(scores): max_score = scores[0] min_score = scores[0] high = 0 low = 0 for idx in range(1,len(scores)): if scores[idx] > max_score: high+=1 max_score = scores[idx] elif scores[idx] < min_score: low += 1 min_score = scores[idx] return str(high) + str(low)
Thanks!
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