[ Leetcode ] 1431. Kids With the Greatest Number of Candies - Java

 

Easy

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Given the array candies and the integer extraCandies, where candies[i] represents the number of candies that the ith kid has.

For each kid check if there is a way to distribute extraCandies among the kids such that he or she can have the greatest number of candies among them. Notice 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: Kid 1 has 2 candies and if he or she receives all extra candies (3) 
 will have 5 candies --- the greatest number of candies among the kids. 
 Kid 2 has 3 candies and if he or she receives at least 2 extra candies will have 
 the greatest number of candies among the kids. 
 Kid 3 has 5 candies and this is already the greatest number of candies among the kids. Kid 4 has 1 candy and even if he or she receives all extra candies will only have 4 candies. Kid 5 has 3 candies and if he or she receives at least 2 extra candies will have the greatest number of candies among the kids.
 

 

Example 2:

 
 Input: candies = [4,2,1,1,2], extraCandies = 1 
 Output: [true,false,false,false,false] 
 Explanation: There is only 1 extra candy, therefore only kid 1 will have 
 the greatest number of candies among the kids regardless of who takes the extra candy.
 

Example 3:

 
 Input: candies = [12,1,12], extraCandies = 10 
 Output: [true,false,true]
 

 

Constraints:

• 2 <= candies.length <= 100
• 1 <= candies[i] <= 100
• 1 <= extraCandies <= 50

 

 

Solution

 

 
 import java.util.ArrayList;
 import java.util.List;
 class Solution {
    public List<Boolean> kidsWithCandies(int[] candies, int extraCandies) {
        List<Boolean> list = new ArrayList<>();
        for(int i=0; i< candies.length; i++){
            for(int j=0; j<candies.length; j++){
                if(candies[i]+extraCandies < candies[j]){
                    list.add(false);
                    break;
                }
                if(j == candies.length -1){
                    list.add(true);
                }   
            }
        }
        return list;
    }
 }
 

 

 

++ 100%

이중for문보다 Max값을 찾은 뒤 비교하는게 더 속도가 빠르다

 
 import java.util.ArrayList;
 import java.util.List;
 class Solution {
    public List<Boolean> kidsWithCandies(int[] candies, int extraCandies) {

        int maxCount = 0;
        for(int i = 0; i < candies.length; i++){
            maxCount = Math.max(maxCount, candies[i]);
        }
        
        List<Boolean> list = new ArrayList<>();
        for(int i = 0; i < candies.length; i++){
            if(candies[i]+extraCandies >= maxCount){
                list.add(true);
            } else {
                list.add(false);
            }
        } 
        return list;
    }
    
 }
 

근데 여전히 메모리를 좀 많이 쓴다

뭐가 문제인가

속도와 메모리를 같이 잡을 순 없나