`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

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.

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;
    }

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!

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 :-)

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.

No idea, I probably did that on a hunch. I liked the code of python guy below my post. There is no need for second for loop. Anyway, the time complexities are linear for both.