apghr A compact C++ solution:
#include <iostream> int main(){ int c, n, max = 0; std::cin.ignore(); while(std::cin >> n) max < n ? c = !!(max = n) : c += max == n; std::cout << c; return 0; }
v1nea Like this a lot. Thanks.
mohammedavez85 `c
include
include
include
include
include
include
include
int birthdayCakeCandles(int n, int ar_size, int* ar) { // Complete this function int i,j,temp,count=0,b; for(i=0;i
b=ar[0]; for(i=0;i<n;i++){ if(b == ar[i]){ count++; } }return count; }
int main() { int n,ar_i; scanf("%i", &n); int *ar = malloc(sizeof(int) * n); for(ar_i = 0; ar_i < n; ar_i++){ scanf("%i",&ar[ar_i]); } int result = birthdayCakeCandles(n, n, ar); printf("%d\n", result); return 0; } BRO THIS CODE SAYS TIME LIMIT EXCEEDED PLEASE HELP
Yakubk_98 use c++ and you'll be fine
int birthdayCakeCandles(int n, vector <int> ar) { // Complete this function int max = ar[0]; int count = 0; for(int i=0; i<n; i++) if(ar[i] > max) max = ar[i]; for(int i = 0; i < n; i++) if (ar[i] == max) count++; return count; }
sandeepkataria this code was working on some test case but not all please help me
vaishaligarg2012 /*You should do like this... */
int max=ar[0]; int count=0; for(int i=0;i<n;i++){ if(ar[i]>max){ max=ar[i]; } } for(int i=0;i<n;i++){ if(ar[i]==max){ count++; } } return count;
jdanielys07 In java, maybe this way can help you! :)
if(ar!=null && ar.length>0 && n>0) { int maxHeight=0, numMaxHeight=1; for(int i=0; i<ar.length; i++) { if(maxHeight<ar[i]) { maxHeight=ar[i]; numMaxHeight=1; }else if(maxHeight==ar[i]) { numMaxHeight++; } } return numMaxHeight; } return 0;
Stelios10 just an optimization, even though in this case your code is fine: if you try an array with negatives, then you ll never get inside your if-statements. so the numMaxHeight will never change, so you dont really compute it in such cases. so this solution for an array with negatives, where the max negative appears more than once, will fail. eg [-1, -1]
proposed solution: set numMaxHeight = 0 on initialization
shrikanth_n23 You can simplify it like this:
int max=ar[0]; int count=0; for(int i=0;imax){ max=ar[i]; count = 1; //Reinitialize count if you found a new max }
else if(ar[i]==max){ count++; }} return count;
I this case you are traversing the vector only once.
pkarif4249816 [deleted]pkarif4249816 [deleted]soumabratabhatt1 i did the same thing.But it is showing me 4 test cases failed.
jord_fumar Be aware of time complexity. In my testcase 4, n was 10000. If you use a sorting algorithm with n^2, you will get a runtime error.
The way I structured my code is:
- Sort the array from min to max (quickSort)
- Parse from the very end of the array and iterate count++.
- When your values are no longer identical, stop.
jagmeetsingh I tried this
static int birthdayCakeCandles(int[] a) { Arrays.sort(a); //double pivot QuicKSort int count = 0; int i = a.length - 1; int tallest = a[i]; while (a[i--] == tallest) { count++; } return count; }
It is failing the test case
100000 followed by 100000 times 9999999 (as height of candles(array values)).
x_ireesh_x Pass all Testcases..
static int birthdayCakeCandles(int[] ar) { Arrays.sort(ar); int result = 1; int decriment = 2; while((decriment <= ar.length) && (ar[ar.length-1] == ar[ar.length-decriment])){ result++; decriment++; } return result; }
bertgayus if you use quicksort u already have O(nlogn) which is worse than just iteratring through the array and saving the highest value and then just iterate a second time incrementing a counter everytime the value is similar to the saaved one. then complexity is O(2n) = O(n). With n = 100 you would have ~ 600 steps(n*logn = 100 * 6 = 600) with your solution and just 200 steps with mine
luckysam09 ar[i]-max==0 in second if worked for me
generalom1234 sandeep did u use unsigned long or any other data type which allows you to take in large amounts of data?
r_umans Since height of the candle is always at least 1, you can safely assume max = 0; Then loop all at once, you dont need a double loop. (Java syntax)
If(max < ar[i] ) { max = ar[i]; count =1; } else if (max==ar[i]) count++;
x_ireesh_x [deleted]x_ireesh_x [deleted]x_ireesh_x [deleted]
Anoop_SS You don't need the second loop. Checking of equal can be done in the else of
if(ar[i] > max)
josephawilson86 Ask yourself a few questions:
- What do I need to return?
-The number of highest candles
- How would I get that number?
-I need to know what the highest candle is.
-I need to know how many of them there are.
- How would I know which is the highest candle?
-Simple comparison should work for knowing what the highest is. Just store it in a var and reset when I find one that's bigger.
- How would I track how many of they there are?
-Store the quantity in a var, increment it whenever I find one of the Highest candles and reset it each time I find a new highest candle.
qu4ku i'm not into c++, but this looks savage :)
trungskigoldberg Can you explain more about the complicated, savage version of that ternery statement? I really want to know this nifty code!
apghr while(std::cin >> n) max < n ? c = !!(max = n) : c += max == n;
can be translated as:
while(std::cin >> n){ // for each candle n check: if(max < n) { // does n set a new record in height? max = n; // if that's so, then n is the new max height c = 1; // and the counter c must be set to 1 again. "!!" is a cheap trick to convert any value different from 0 into 1 } else { // otherwise check if the new candle is as tall as max if (max == n) c++; // in that case, add 1 to counter (otherwise add 0) } }
arpita414 Too good
oswald200801 Nice one !!
aishwarya217 Good!
adityabcs93 nice one
lehieuminh231 nice. thank you
tarsum Well I came up with similar algorithm, so thumbs up. Also seems like solution in editorial goes though array twice, and this goes only once.
Vishwas92 hats off..
ichidan Nice one. I arrived at same algo in C, minus the one liner ternary. Looks neat, I like it. Although I'd be curious if there is not a small performance loss of doing:
c = !!(expr);
vs
(expr); c = 1;
the former may require additional conditional logic depending on architecture (for instance x86 - http://riffwiki.com/MOV_(x86_instruction) - the MOV instruction doesn't affect ZeroFlag, so I'd guess CPU would have to evaluate whether your expr results in a non-zero value. It's obvious to us as we can see the wider context of the algorithm, but not the CPU)
kurian_benoy thanks
viigihabe It is not CPU that analyses your c++, the compiler does. CPU doesn't "thin" but compiler does. It is really pointless to "optimize" your code this way because compiler is required to emit assembly that does what is ment but in which order and through which instructions, it is not specified. You better belive it does good job. You express c++ although it is slower than ++c: there is no way the compiler would use slower version if it doesn't afect the program flow. With no optimization flags may be.
lakshay91dutt [deleted]
rahulsahu_blog This is brilliant!
gharibakhamis nice why didn't i think of this sooner
parvpareek TYSM!! Helped me a lot..
sabahat_usman_su I did the same thing but my code fails for some tests :O
jagmeetsingh I tried same in java. It worked! Thanks (:
int temp, max = 0, result = 0, n; for (int i = 0; i < arCount; i++) { n = scanner.nextInt(); temp = (max < n) ? (result = ((max = n) >= 0) ? 1 : 1) : (result += (n == max) ? 1 : 0); }
Ray1984 Very nice -- thanks for nudging me to think about it as a streaming algorithm :-)
colonel_bishop It's cool! Thanks!
ScienceD I dont understand how cin.ignore(); works... And how you can feed input into loop condition...
thermosphere453 cin.ignore() ignores the input of number of candles as its not needed in his algorithm and the input in loop works as long as the input isnt 0 or null, while loop works for any number in the test condition.
kachi_kinjal can u give the solution for c programming.
oswald200801 int main(){ int n; scanf("%d",&n); int *height = malloc(sizeof(int) * n); int max=height[0]; int count=0; for(int height_i = 0; height_i < n; height_i++){ scanf("%d",&height[height_i]); } for(int i=1; imax){ max=height[i]; } } for(int k=0; k
unicpawan can u please tell me why this code is not able to pass all the test case
// the code goes here..
int main(){ int n,i,max=0,frequency=0;
int ar =(int) malloc(sizeof(int) * n);
scanf("%d", &n); for( i = 0; i < n; i++){ scanf("%d",&ar[i]); } for(i=0;i<n;i++){ if(ar[i]>max){ max=ar[i]; frequency=1; } else if(ar[i]==max){ frequency++; } } printf("%d",frequency); return 0;}
bharat8apr include
include
include
include
include
include
include
int birthdayCakeCandles(int n,int a[]) { long long int i,max=0,d=0; for(i=0;imax) { d=1; max=a[i]; } else if(a[i]==max) { d++; } } return d;
}
int main() { int n; scanf("%i", &n); int *ar = malloc(sizeof(int) * n); for(int ar_i = 0; ar_i < n; ar_i++){ scanf("%i",&ar[ar_i]); } int result = birthdayCakeCandles(n,ar); printf("%d\n", result); return 0; }
kuppesh_ds Return frequency to calling function instead of printig here.
vishalgwalani6 if you have case like 20,40,90,90,90 so you see according to your code in max variable value will 20 and frequency will be 1 and after the loop ends your result will be 4 instead it should be 3
shac_bandagi your else if part should be in for loop and in if statement remove that frequency set otherwise max number counted twice.
anurag0430 your are returning count of candles of same height. We have to return the number of candle he/she can blow out.
gaurav_877 you havent calculated the final max value and processed it before hand for comparision
maheshm121 what is max before entering the for loop??
gjaiswal108 because you have allocated memory to pointer ar using malloc function before taking the value of n from user. you should first take input n from user and then allocate memory to the pointer, i.e. first use scanf("%d",&n); then write int ar=(int)malloc(sizeof(int)*n);
hope, you understand.
kmschaal902 compact indeed, great job!
andrew_hedy77 Yeah, yours is way cleaner than mine
int birthdayCakeCandles(int n, vector <int> ar) { int count = 1; sort(ar.begin(), ar.end()); reverse(ar.begin(), ar.end()); for (int i = 0; i < n; i++){ if (ar[i] == ar[i+1]){ count++; } else break; } return count; }
amitsoni90 Why we need vector here?
cse_115 good solution
wem18 Solid solution. Wouldn't it be more efficient to do a for loop such as
//call sort //set int max = ar[n] for(int i = n; i>0; i++) { if(ar[i]==max) { count++; } else{ break; } return count; }
So you can avoid calling both sort and reverse? Not trying to correct you, genuine question
abhi69sindh it will not pass all the testcases
3001pratap [deleted]
shantanu_knight Great one
chandu007 awesome
karthikeyan90422 What is the use of cin.ignore() please tell me
lyj1186669164 That's so nice.Thanks
VIDYARANIGIDDE what is the role of cin.ignore() ?? what if we do not write that line?
cu_16bcs2416 wao! stupendous code!
benaffleks Can someone explain to me :
max < n ? c = !!(max = n) : c += max == n;
breakbadsp bro use python3 :
def birthdayCakeCandles(ar): return ar.count(max(ar))
No looping, no veriables, no mess
nsuniln1 [deleted]
ilya_zarembsky Cute
ryanfehr18 A small Java solution in O(n) time with O(1) space:
//Java 8 /* Initial Thoughts: We can keep a running max and update it if we find something larger, if we find something smaller we just keep looking and if we find something equal then we increment a counter variable Time Complexity: O(n) //We must check the height of every candle Space Complexity: O(1) //We only store a max and a frequency */ import java.io.*; import java.util.*; import java.text.*; import java.math.*; import java.util.regex.*; public class Solution { public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); int tallest = 0; int frequency = 0; for(int i=0; i < n; i++){ int height = in.nextInt(); if(height > tallest){ tallest = height; frequency = 1; } else if(height == tallest) frequency++; } System.out.println(frequency); } }
You can find more HackerRank solutions like this here
Dsmeenakshi if height is less than tallest means...wat wil happen
ryanfehr18 As mentioned above:
/* if we find something smaller we just keep looking */
Basically, we don't care about candles < tallest
Dsmeenakshi thank you dude
wyrlvillazorda it's great..thank you!
randy_evered int max = 0; for (int v : a) if (v > max) max = v; int count = 0; for (int v : a) if (v >= max) count++; return count;sntz1 you can avoid doing two for loops by doing something along the lines of:
int max = 0; int count = 0; for (int v : a) { if (v > max) {max = v; count = 1;} else if (v == max) count++; } return count;
You can avoid iterating over the collection twice
SupreethMC Nice approch !
borisaltynnik Why not to use first array element like initial value of max ? Does it make any sense ?
static int birthdayCakeCandles(int n, int[] ar) { int max=ar[0]; int num=1; for (int i=1;i<n;i++){ if (ar[i]>max){ max=ar[i]; num=1; } else if (ar[i]==max) num++; } return num; }
iamkarthik49 Yeah ! I did the same thing but complicated it... Your's looks clean and clear thanks for the solution... :) .
santunu23 Nice one thanks for sharing,actually I have been working on this solution for a hour,finaly it work thanks dude.
Coder_Zuki No Probs Dude
kvnfrederiksen0 HAH! I did the exact same thing. Nice.
int n = in.nextInt(); int j = 0; int max = 0; for(int i = 0; i < n; i++){ int x = in.nextInt(); if(x > max){ j = 1; max = x; } else if(x == max) j++; } System.out.print(j);
shantanu_bugadi simply wow.. what i got by this code is , try to do most of it simultaneously....
3001pratap static int birthdayCakeCandles(int n, int[] ar) { int count=0; Arrays.sort(ar); int tallest = ar[n-1]; for(int i=0; i<n; i++) { if(ar[i] == tallest) count++; } return count; }
fhrazib I think this can be improved a little bit using a break. We don't need to traverse all the array to get the total count when the array is sorted (here in ascending order) -
Arrays.sort(a); int tallestOne = a[n-1]; int count = 1; for(int i=n-2; i>=0; i--){ if(a[i]==tallestOne) { count++; } else break; }
hollybush Amazing solution mate!
zakiralig184 Cool!
pep1439 use
tallest = Integer.MIN_VALUE
instead, for general use.
stephen_kalletta very clean. I was sorting it and grabbing the last values, but I like this much better. thanks for sharing
Coder_Zuki Did This...! Seems Simple right?
static int birthdayCakeCandles(int n, int[] ar) { // Complete this function int size=ar.length; int tall=ar[0]; int count=0;
for(int i=0 ; i<size ; i++) { if(ar[i]>tall) { tall=ar[i]; } } for(int j=0;j<size;j++) { if(ar[j]==tall) count++; } return count; }
j_zambre66 [deleted]
j_zambre66 Can be done it like this.
int max = 0, count = 1; for (int v : ar) { if (v == max) count++; if (v > max) max = v; } return count;
ATVex Counting the tallest candles.
n = int(input().strip()) height = [int(height_temp) for height_temp in input().strip().split(' ')] print(height.count(max(height)))
abhishek106 python is the best!
rudra_utpal Any problem with Java @abhishek106
import java.util.*; public class Solution { public static void main(String[] args) { Scanner scan=new Scanner(System.in); int n=scan.nextInt(); ArrayList<Integer> al=new ArrayList<Integer>(); for(int i=0;i<n;i++) al.add(scan.nextInt()); System.out.println(Collections.frequency(al,Collections.max(al,null))); } }
aaron_gomez1 While this works, you loop through the complete collection of values 3 times. First while reading it in the scanner. Second while scanning for the max value in the collection after reading it in and third to count the number of times it was found.
fernando_becerr1 anyway the time complexity is O(n+n+n) = O(n)
riaan_rvr I'm not too familiar with the BigO notation, so an honest question how does it become n+n+n shouldn't it be 3n rather as you are doing full loops over all elements?
hua_duy In BigOh notation, I believe they drop the constants in front of n because exponents matter so much more that the constants are negligible.
egrodo1 [deleted]egrodo1 Why do that instead of just?
n = int(raw_input().strip()) candles = map(int,raw_input().strip().split(' ')) print candles.count(max(candles))
randy_evered Wow, egrodo1, Python for the win!
Modelmat I basically did the same thing!
pheepia7788 nice code! as I am a beginner, I think very long and messy code.... lol
Dulguun_Otgon [deleted]helderIO [deleted]
dandaraviteja n = int(input().strip()) ar = (list(map(int, input().strip().split(' ')))) print(ar.count(max(ar)))
paccel22 Thank you! I hadn't realized Python had a count method for lists that basically reduces the problem to 1 step.
Anubhav_singh what is the code of "count" function?? bcos when I am using loop instead of count fuction, its taking too much time to run. pls help??
ARXINO I am using loop but it is the same thing with ar.count(max(ar))
def birthdayCakeCandles(n, ar): max_height=0 double=0 for x in range(n): if ar[x]>max_height: max_height=ar[x] for y in range(n): if ar[y] == max_height: double+=1 return double
sehan2 from collections import * def birthdayCakeCandles(n, ar): res = Counter(ar) return res.most_common()[0][1]
croqaz Exactly like I did! Nice!
rowlandoti This is not a correct solution but I do not know why it is passing. You should:
- Get the max element
- Return or print its count/frequency.
pawelwiszniewski In some test cases, the highest (max) element is the one most common - like in example case, where 3 is the most common element. It also passes some test cases. The correct solution is the one with Counter.
vinasatinfo just simple
return Counter(ar).most_common()[0][1]actually :-)
Spiznak My solution finishd in about half the time, I would assume because I'm looping through it only once. Depends on whether you want a quick and dirty solution (mine) or one that's more Pythonic (in your case, can be more difficult to read):
n = input() arr = input().split() theMax, count = 0, 0 for x in arr: x = int(x) if x > theMax: theMax = x count = 1 elif x == theMax: count += 1 print(count)
icehongssii data=[10,3,333,333,333,0,13] tmp=data[0] #max data tmp2=0 #max data counter for i in range(len(data)-1): for j in range(i+1,len(data)): if(tmp<=data[j]): tmp=data[j] for k in range(len(data)): if(data[k]==tmp): tmp2+=1 print(tmp2)
this was my code. to get max_height, i use the loop twice. so it didn't work. but like you said, its working with single loop! thanks and it is so amazing that you used single looop... never imagined it
gb3010 I use max function in list. 7 out of 8 test cases work. I am not sure what's wrong with my code. Can you please have a look ?
n1=input().strip() # Input age value b1=input().strip().split() # Input list of candle heights def birthdayCakeCandles(n,b): c=0 z=max(b) for i in b: if i == z: c += 1 return(c) y=birthdayCakeCandles(n1,b1) print(y)
nicolasmontoya similar:
input() arr = list(map(int, input().split())) print(arr.count(max(arr)))
pawelwiszniewski EDIT: My mistake, i've missread the description :(. So the rest of the comment is invalid.
This doesn't work. It only counts how many times the highest element from the list (max(arr)) is in the list. In some test cases it works. In some it doesn't -> it's not a correct solution
brevnovak i don't understand. it's what we are looking for, is it not?? ;)
pawelwiszniewski You're right, I've missread the description.
wanwanzhang Hi I also use the same way to count in Python print(arr.count(max(arr))), but the result is always 2, none, which show me the worng answer. I do not know why it show me (2, none) instead of just 2.
EeXoR_Daniel_Ng python FTW:>
jeffmagill C# solution...
int tallest = height.Max(); int count = height.Count(c => c == tallest); Console.WriteLine(count);
indyHarcourt Or
int result = ar.Where(i => i == ar.Max()).Count();
unSatisfied That's O(n^2) since you're finding the max value on each iteration of the where statement. The best solution would be to use a single for/foreach loop, but jeffmagill's solution (looping through the list twice) is still much more efficient than this.
shatrudhankr [deleted]laysa_uchoa best comment
daeden I don't think you deserver a downvote, its solution i came up with the returned failed tests (tests timed out) for me.
I wish it was explained where other than comments why this wasn't a valid result.
disabajn Python:
return ar.count(max(ar))
Dalenguyen Nice to have modern JavaScript:
function birthdayCakeCandles(n, ar) { // Complete this function var max = Math.max(...ar); var result = ar.filter(c => c === max); return result.length; }
andrewseaton nice. I looped through the array to count occurances of "max". Using filter makes it so much simpler.
aJaxs 这个filter很机智
sameer_jain1 Exactly what I am thinking, but not sure why this question appearnign wrong at the moment, it doesnt have all the parameters so its throwing error to me
netris [deleted]netris You could even do this in a single line of code, like so:
function birthdayCakeCandles(ar) {
return ar.filter(i => i === Math.max(...ar)).length;
}But I found that this causes recursion issues. This works perfectly fine though:
function birthdayCakeCandles(ar) {
var max = Math.max(...ar);
return ar.filter(i => i === max).length;
}brecht_pynoo The first one is indeed overly complex, as it would search the max at each iteration
I used the reduce function instead, which allows for a single iteration, provided an extra local variable, like so
function birthdayCakeCandles(ar) { let max = 0; return ar.reduce( (count, i) => i === max ? count + 1 : i > max ? (max = i) && 1 : count, 0 ); }
n_haque Oh filter and result.length is so much simpler! Mine was a little longer
function birthdayCakeCandles(ar) { let tallest = Math.max(...ar) let candleCount = 0 ar.forEach((candle) => candle === tallest ? candleCount++ : null) return candleCount }
Elvis2597 def birthdayCakeCandles(n, ar): return max(Counter(ar).items())[1]
saurabh_kumar3 I used this:===>
TreeMap mapCandleCount=new TreeMap(); for(int i=0; i } int first = mapCandleCount.lastEntry().getValue(); return first;
bendjoudifirst [deleted]
arsenij_smth A bit different one:
def birthdayCakeCandles(n, ar): return ar.count(max(ar))
pawelwiszniewski [deleted]
owenklaiss [deleted]
marinskiy Here is Python 3 solution from my HackerrankPractice repository:
n = int(input()) ar = list(map(int, input().split())) print(ar.count(max(ar)))
Feel free to ask if you have any questions :)
voodoorrj1990 [deleted]
akomawar Solution in C
#include <math.h> #include <stdio.h> #include <string.h> #include <stdlib.h> #include <assert.h> #include <limits.h> #include <stdbool.h> int birthdayCakeCandles(int n, int ar_size, int* ar) { int i,b=0,max=0; for(i=0;i<n;i++){ if(ar[i]>max){ max=ar[i]; // printf("%d\n",max); } } for(i=0;i<n;i++){ if(max==ar[i]){ b+=1; } } return b; } int main() { int n; scanf("%i", &n); int *ar = malloc(sizeof(int) * n); for(int ar_i = 0; ar_i < n; ar_i++){ scanf("%i",&ar[ar_i]); } int result = birthdayCakeCandles(n, n, ar); printf("%d\n", result); return 0; }
sportmew Can you tell why this code of mine is not passing all test cases?
include
include
include
include
include
include
include
int birthdayCakeCandles(int n, int ar_size, int* ar) { // Complete this function int t = ar[0]; int s=0; for(int i=0;i=t) {s++; } } return s; }
int main() { int n; scanf("%i", &n); int *ar = malloc(sizeof(int) * n); for(int ar_i = 0; ar_i < n; ar_i++){ scanf("%i",&ar[ar_i]); } int result = birthdayCakeCandles(n, n, ar); printf("%d\n", result); return 0; }
akomawar - I feel there is no need of pointing first element of array to any variable.
- We have size of the array as 'n'. Use that instead of t. and increamet the value of i from 0 to n.
- Then find the maximum height amongst the candle.
- Finally, find the occurance of the maximum height and return that value.
for reference please review my code above.
sportmew [deleted]sportmew Can you tell me why this code of mine is unable to pass all test cases?
akomawar - if you see the array has 4 elements - '3 2 1 3' according to your code t=3
- Then you are checking wheather 3>=3 if yes s=0+1=1 or if no, do nothing. then, check 3>=2, 3>=1, 3>=3.
Now if you see this code works only if your first element(number) in your array is highest always.. If not it will not work.
Take this example where you have 4 elements - '1 2 4 4' now what see what your code does,
- t=1
- if 1>=1 which is yes so, s=0+1=1
- then, if 1>=2 which is No so, s=1 remains same.
- again, if 1>=4 which is No so, s=1 remains same.
- again, if 1>=4 which is No so, s=1 remains same.
your answer s=1 where as the answer should have been 2 as maximum height is 4 and it occurs twice. Hope that this is useful.
sportmew Thank you for answering in such detail. You gave an example above considering 4 elements - '1 2 4 4'. I think you have explaned it wrongly. What my code does is this:
t = 1
- if 1 >= 1 (which is true) then s = 0+1 = 1
- if 2 >= 1 (which is true) then s = 1+1 = 2
- if 4 >= 1 (which is true) then s = 2+1 = 3
- if 4 >= 1 (which is true) then s = 3+1 = 4
So the answer by my code is 4 which is definitely wrong but now with your example I totally understood where I was wrong. Thank you
akomawar ohoo yaa!!! But I think you got what I meant to say...
anuragajoshi I am little confused. Thinking practically, if the candles are placed in order(liner, in a cloumn), upon blow, only the candles, with taller ancestors won't blow. First candle will be blown no matter what. Shouldn't the solution be something like
n = int(input().strip()) height = [int(height_temp) for height_temp in input().strip().split(' ')] min = height[0] count = 0 i = 0 while i < n: if height[i] >= min: count = count + 1 min = height[i] i = i+ 1 print(count)
ilya_zarembsky Don't think too much about the specifics of the situation, just think about it as an awkward/imprecise way of wrapping the abstract question 'find the number of times that the largest element occurs in this array'
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