Discussion on Utopian Tree Challenge

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2 weeks ago+ 0 comments

Here is my c++ solution, you can watch the explanation here : https://youtu.be/0G-kEeQC25E

Solution 1 : O(N)

int utopianTree(int n) {
    int ans = 1;
    for(int i = 1; i <= n; i++){
        if(i % 2 == 0) ans++;
        else ans *=2;
    }
    return ans;
}

Solutoin 2 : O(1)

int utopianTree(int n) {
    string s(n / 2 + 1, '1');
    if(n % 2 == 1) s += '0';
    bitset<60> ans(s); 
    return ans.to_ullong();
}

Solutoin 3 : O(1)

int utopianTree(int n) {
    int ans = (1 << ((n / 2) + 1) ) - 1;
    if(n % 2 == 1) ans*=2;
    return ans;
}

Solutoin 4 : O(1)

int utopianTree(int n) {
    return ((1 << ((n / 2) + 1) ) - 1) << n%2;
}

3 weeks ago+ 0 comments

def utopianTree(n):
    # Write your code here
    return 2 ** (math.floor((n + 1) / 2) + 1) - 1 - (n % 2)

4 weeks ago+ 0 comments

My C++ solution -

int utopianTree(int n) {
    int height = 1;
    auto is_odd = [](int num){ return (num & 1);};
    
    for (int i=1;i<=n;i++) {
        if (is_odd(i)) {
            height *= 2;
        } else {
            height += 1;
        }
    }
    
    return height;
}

3 months ago+ 0 comments

Perl:

sub utopianTree {
    my $n = shift;
    
    return ($n % 2 == 0) ? (2 ** (($n / 2) + 1)) - 1 : (2 ** ((($n + 1) / 2) + 1)) - 2;
}

3 months ago+ 1 comment

int height = 0;
    int i = 0;
    while(i<=n) {
        height = i % 2 ==0 ? height + 1 : height * 2;
        i++;
    }
    return height;

3 weeks ago+ 0 comments

def one_cycle(height_one): return (2*height_one +1)

def utopianTree(n): # Write your code here height=1

    cycles=n//2    

if cycles:
    for i in range (cycles):
        height=one_cycle(height) 
if n%2==0:
    return height
else: 

    return (2*height)

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