There are n kids with candies. You are given an integer array candies, where each candies[i] represents the number of candies the ith kid has, and an integer extraCandies, denoting the number of extra candies that you have.
Return a boolean array result of length n, where result[i] is true if, after giving the ith kid all the extraCandies, they will have the greatest number of candies among all the kids, or false otherwise.
Note that multiple kids can have the greatest number of candies.
Example 1:
Input: candies = [2,3,5,1,3], extraCandies = 3
Output: [true,true,true,false,true]
Explanation: If you give all extraCandies to:
Example 2:
Input: candies = [4,2,1,1,2], extraCandies = 1
Output: [true,false,false,false,false]
Explanation: There is only 1 extra candy.
Kid 1 will always have the greatest number of candies, even if a different kid is given the extra candy.
Example 3:
Input: candies = [12,1,12], extraCandies = 10
Output: [true,false,true]
Constraints:
n == candies.length
2 <= n <= 100
1 <= candies[i] <= 100
1 <= extraCandies <= 50
class Solution:
def kidsWithCandies(self, candies: List[int], extraCandies: int) -> List[bool]:
max_num = max(candies)
result = list(candy + extraCandies >= max_num for candy in candies)
return result
Among logics with O(N), tried to do as less operations as possible.
So considered a method that
1. don't need another list
2. don't + extraCandies and revoke it
class Solution:
def kidsWithCandies(self, candies: List[int], extraCandies: int) -> List[bool]:
maxCandy = max(candies)
result = [True if candies[i]+extraCandies>=maxCandy else False for i in range(len(candies))]
return result
Is range(len(List)) better than List iteration??
Quite satisfied with my answer this time lol
but I still don't know why lighter solutions are lighter.
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