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let reducer = (a, c) => a + c;
let count=0;
for(let i=0; i<s.length; i++){
let temp = s.slice(i, m+i);
if(temp.reduce(reducer) == d){
count++;
}
}
return count;
in JS
can U explain me about 2 lines of your code?:
for(auto it = s.cbegin(); it != s.cend(); ++it){
if(d == std::accumulate(it, it + m, 0))
I think a accumulate will sum element out of vector, I think it will be it != cend() - m more saver.
int birthday(vector s, int d, int m) {
int n = 0;
for(int i = 0; i <= s.size() - m;i++){
if( d == accumulate(s.begin() +i, s.begin() + i + m, 0)){
n++;
}
}
return n;}
Birthday Chocolate
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Same approach, but with iterators instead.
I believe this is O(n^2). O(n) is also possible.
The OP is O(n)
It's O(nm).
its o(n^2), coz in worst case senario, all the choclate blocks would add up to d.
How do we find out the time complexity. someone pls help me out
I have implemented it in O(n).
can you show your logic ???
function birthday(s, d, m) {
Why call accumulate function every iteration when you add only one square ? It's not the same approach : you are O(nlog(n)), his solution is in O(n)
can U explain me about 2 lines of your code?:
for(auto it = s.cbegin(); it != s.cend(); ++it){ if(d == std::accumulate(it, it + m, 0))I think aaccumulatewill sum element out of vector, I think it will beit != cend() - mmore saver.Greate! Nice usage of accumulate!
what you did but more readable:
int birthday(vector s, int d, int m) { int n = 0; for(int i = 0; i <= s.size() - m;i++){ if( d == accumulate(s.begin() +i, s.begin() + i + m, 0)){ n++;} } return n;}
sorry, dont know how to make the code look nice in this forum :-/
yeah. that's the power of c++ stl.
lol