`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

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)
    }
}

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.

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

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; }

Return frequency to calling function instead of printig here.

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

your else if part should be in for loop and in if statement remove that frequency set otherwise max number counted twice.

your are returning count of candles of same height. We have to return the number of candle he/she can blow out.

you havent calculated the final max value and processed it before hand for comparision

what is max before entering the for loop??

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.

Why we need vector here?

good solution

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

C++ allows for integers to be treated as logical values. If the integer is non-zero, it's "true" and if the integer is zero it's "false". The ! is the logical-not operator and the language standard says !0 will be equal to 1.

A similar rule holds for max == n. Whenever, that expression evaluates to "true" it will also represent the numerical value 1.

Assignments, such as max = n, also have the side-effect of "returning" the assigned value (n in this case).

Therefore, this is a really terse way of writing:

if (max < n)  {
	max = n;
	c = 1;
} else if (max == n) {
    c += 1;
}

wow this is crazy, thank you

i did the same. python is so useful when it comes to counting the number of occurences

No looping?

So, how were "max" and "count" calculated?

This is wrong if you do an interview. They most likely want to see a solution which is O(n) and yours is O(n^2).

This is mine, which is O(n):

def birthdayCakeCandles(ar):
    max_h = ar[0]
    max_n = 1
    for c in ar[1:]:
        if c > max_h:
            max_h = c
            max_n = 1
        elif c == max_h:
            max_n += 1
    print(max_n)

U are great bro.

so simple !!!!

Thank you!

I did by sorting the list

def birthdayCakeCandles(ar):

ar.sort(reverse=True)
for i in range(len(ar)-1):
    if (ar[i]==ar[i+1]):
        k=i+1
return k+1

You can go through this tutorial-

Here is the video explanation of my solution -

https://youtu.be/1gxFE9EfanE

most welcome. :)

If you dont mind can you please provide a feedback on my video by commenting or liking & disliking. It will help me and others too to find the solution over internet.

Do not forget to subscribe my youtube channel for hackerrank solutions .:)

Hi,

your solution may be correct but may not be efficient .you can optimized your solution further.

Here is the video explanation of my solution in O(n) time -

https://www.youtube.com/watch?v=1gxFE9EfanE&list=PLSIpQf0NbcCltzNFrOJkQ4J4AAjW3TSmA

I would really appreciate your comments and feedback on my video.

As mentioned above:

/*
if we find something smaller
we just keep looking 
*/

Basically, we don't care about candles < tallest

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

Yeah ! I did the same thing but complicated it... Your's looks clean and clear thanks for the solution... :) .

Nice one thanks for sharing,actually I have been working on this solution for a hour,finaly it work thanks dude.

This is not works, when a number in the array (last number of array) is bigger than then previous ones the result is not ok.

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;
    }

Give count=1;

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)));
    }
}

Wow, egrodo1, Python for the win!

I basically did the same thing!

nice code! as I am a beginner, I think very long and messy code.... lol

Map object doesn't have the count attribute. You will get an error

Thank you! I hadn't realized Python had a count method for lists that basically reduces the problem to 1 step.

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??

Exactly like I did! Nice!

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.

just simple

return Counter(ar).most_common()[0][1]

actually :-)

Why is there a [0][1]?

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

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)

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

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.

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.

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.

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 );
}

I did exactly what you did, but half the test failed and couldn't figure out why :P

That oneliner kind of works, but it's extremely slow because it recalculates Math.max(ar) for every single item.

You can avoid that by storing the value of Math max like this:

const birthdayCakeCandles = a => 
(f => a.filter(v => v === f).length)(Math.max(...a));

That way it's still short, but also blazing fast. (Math max is implicitly stored as f here).

most welcome. if you dont mind.Would really appreciate your feedback like, dislike , comment etc. on my video.

It's not complex, you just need to store it. Instead of doing the same calcultion every time.

Issue was with Reversing the Array. Looks like it gets confused which array its revering. Wrapping in brackets was not enough. It seems like JS can't sort it correctly for some reason. I checked and every entry is type number so I'm not sure why .sort().reverse() doesn't work and why it needs to be in a bracket.

function birthdayCakeCandles(ar) {
    const size = ar.length;
    const candles = ar;
    const tallest = (candles.sort((a, b) => {
            return b - a;
    }))[0];

    const output = candles.reduce((acc, n) => {
            acc = n === tallest ? acc + 1 : acc;
            return acc;
    }, 0);

    return output > size ? size : output;
}

The question is how to proceed. The test cases were too large and causing a time out. Please help.