`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!

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.