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  1. Prepare
  2. Algorithms
  3. Greedy
  4. Marc's Cakewalk
  5. Discussions

Marc's Cakewalk

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434 Discussions

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  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can watch explanation here : https://youtu.be/RRJoL2vDmZM

    long marcsCakewalk(vector<int> calorie) {
    sort(calorie.begin(), calorie.end(), [](int l, int r){return l > r;});
    long ans = 0;
    for(int i = 0; i < calorie.size(); i++){
        ans += pow(2, i) * calorie[i];
    }
    return ans;
}

  • rahulbhadu51
     + 0 comments

    include

    using namespace std; int main(){ long int n; cin >> n; long long int p = 0; int arr[n]; for(int i = 0; i < n; i++){ cin >> arr[i]; }

    // Sort in descending order
sort(arr, arr + n, greater<int>());

for(int i = 0; i < n; i++){
    p += pow(2, i) * arr[i];
}

cout << p;

return 0;

    }

  • emhhacker
     + 0 comments

    My Java solution with o(n log n) time complexity and o(1) space complexity:

    public static long marcsCakewalk(List<Integer> calorie) {
        // Sort the calorie array in descending order
        Collections.sort(calorie, Collections.reverseOrder());
        
        // Calculate the total miles to walk using powers of 2 (bit-shifting)
        long totalMiles = 0;
        for (int i = 0; i < calorie.size(); i++) {
            long currentMilesToWalk = (1L << i) * calorie.get(i); // More efficient power of 2 calculation
            totalMiles += currentMilesToWalk;
        }
        
        return totalMiles;
    }

  • psaumet_dvp
     + 0 comments
    def marcsCakewalk(calorie):
    return sum(2**i x for i, x in enumerate(sorted(calorie, reverse=True)))

  • bawa101001
     + 0 comments

    Here is my single line Python solution : 

    def marcsCakewalk(calorie):
    # Write your code here    
    return sum((2**i) * calorie[i] for i in range(len(calorie))) if calorie.sort(reverse=True) or True else 0