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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
intbirthdayCakeCandles(intn,vector<int>ar){// Complete this functionintmax=ar[0];intcount=0;for(inti=0;i<n;i++)if(ar[i]>max)max=ar[i];for(inti=0;i<n;i++)if(ar[i]==max)count++;returncount;}
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
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
Birthday Cake Candles
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Like this a lot. Thanks.
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int birthdayCakeCandles(int n, int ar_size, int* ar) { // Complete this function int i,j,temp,count=0,b; for(i=0;i
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
use c++ and you'll be fine
this code was working on some test case but not all please help me
/*You should do like this... */
In java, maybe this way can help you! :)
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
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 }
} return count;
I this case you are traversing the vector only once.
i did the same thing.But it is showing me 4 test cases failed.
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:
I tried this
It is failing the test case
100000 followed by 100000 times 9999999 (as height of candles(array values)).
Pass all Testcases..
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
ar[i]-max==0 in second if worked for me
sandeep did u use unsigned long or any other data type which allows you to take in large amounts of data?
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)
Ask yourself a few questions:
-The number of highest candles
-I need to know what the highest candle is.
-I need to know how many of them there are.
-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.
-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.