vasukapil2196 why is the problem in greedy algorithm section? Basically i want to ask how to figure out if it is a greedy algorithm.
combinatorial_p1 [deleted]
ggorlen It's greedy because at every step, always choosing the local highest calorie amount will yield the globally optimal solution.
coolash123 can you explain what do you mean by local highest?
Slaunger Python 3
Two-liner
input() print(sum(c * 2 ** j for (j, c) in enumerate(sorted(map(int, input().split()), reverse=True))))
NicolasBi You can one-line this beauty like this :
input(), print(sum(c * 2 ** j for (j, c) in enumerate(sorted(map(int, input().split()), reverse=True))))
Slaunger But it is an odd construction to use a tuple just to avoid a newline and a bit harder to understand. It is not a goal in itself to make one-liners IMO.
codertoaster But it's good enough to make us non-python programmers jealous lol.
edelmoral Javascript one-line:
return calorie.sort((a,b) => a-b).reverse().reduce((t,a,i) => t + a * 2**i);
KVS123 could you pls explain what reverse = True does
Slaunger sorted(iterable, reverse=True) iterates over the elements in iterable in reverse sorting order. That is, in descending order.
For instance (python 3)
print(list(sorted([1, 3, 5, 2], reverse=True)))
gives
[5, 3, 2, 1]
whereas
print(list(sorted([1, 3, 5, 2])))
gives
[1, 2, 3, 5]
KVS123 Thank you :)
mike_holcomb Using Horner's method in the lamda to avoid the repeated calls to power function.
from functools import reduce input() print(reduce(lambda x, y: 2*x + y, sorted(map(int, input().split()))))
luumanhlapdi :D
return sum(sorted(calorie)[::-1][i] *2**i for i in range(len(calorie)))sarathy_v_krish1 C++ solution :
long marcsCakewalk(vector<int> calorie) { long sum=0; sort(calorie.begin(), calorie.end(), greater<int>()); for (int i=0;i<calorie.size();i++) sum+=pow(2, i)*calorie[i]; return sum; }
joserico1 MY Python3 solution:
def marcsCakewalk(calorie): calorie.sort(reverse = True) return sum((2**i)*c for i,c in enumerate(calorie))
alfonso_delamor1 C++ Solution
long marcsCakewalk(vector<int> calorie) { sort(calorie.begin(), calorie.end(), greater<int>()); int exponent = 0; long result = 0; for(int i : calorie) result = result + (i * pow(2, exponent++)); return result; }
raffaele_provin1 Done the same in php. Thought it was the fastest way
vldvel One-liner JavaScript
calorie.sort((p,c) => c - p).reduce((p,c,i) => p + c * Math.pow(2, i), 0)
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