Description#
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:
- Kid 1, they will have 2 + 3 = 5 candies, which is the greatest among the kids.
- Kid 2, they will have 3 + 3 = 6 candies, which is the greatest among the kids.
- Kid 3, they will have 5 + 3 = 8 candies, which is the greatest among the kids.
- Kid 4, they will have 1 + 3 = 4 candies, which is not the greatest among the kids.
- Kid 5, they will have 3 + 3 = 6 candies, which is the greatest 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.
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
Solutions#
Solution 1#
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| class Solution:
def kidsWithCandies(self, candies: List[int], extraCandies: int) -> List[bool]:
mx = max(candies)
return [candy + extraCandies >= mx for candy in candies]
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| class Solution {
public List<Boolean> kidsWithCandies(int[] candies, int extraCandies) {
int mx = 0;
for (int candy : candies) {
mx = Math.max(mx, candy);
}
List<Boolean> res = new ArrayList<>();
for (int candy : candies) {
res.add(candy + extraCandies >= mx);
}
return res;
}
}
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| class Solution {
public:
vector<bool> kidsWithCandies(vector<int>& candies, int extraCandies) {
int mx = *max_element(candies.begin(), candies.end());
vector<bool> res;
for (int candy : candies) {
res.push_back(candy + extraCandies >= mx);
}
return res;
}
};
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| func kidsWithCandies(candies []int, extraCandies int) (ans []bool) {
mx := slices.Max(candies)
for _, candy := range candies {
ans = append(ans, candy+extraCandies >= mx)
}
return
}
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| function kidsWithCandies(candies: number[], extraCandies: number): boolean[] {
const max = candies.reduce((r, v) => Math.max(r, v));
return candies.map(v => v + extraCandies >= max);
}
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| impl Solution {
pub fn kids_with_candies(candies: Vec<i32>, extra_candies: i32) -> Vec<bool> {
let max = *candies.iter().max().unwrap();
candies
.iter()
.map(|v| v + extra_candies >= max)
.collect()
}
}
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| class Solution {
/**
* @param Integer[] $candies
* @param Integer $extraCandies
* @return Boolean[]
*/
function kidsWithCandies($candies, $extraCandies) {
$max = max($candies);
$rs = [];
for ($i = 0; $i < count($candies); $i++) {
array_push($rs, $candies[$i] + $extraCandies >= $max);
}
return $rs;
}
}
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| #define max(a, b) (((a) > (b)) ? (a) : (b))
/**
* Note: The returned array must be malloced, assume caller calls free().
*/
bool* kidsWithCandies(int* candies, int candiesSize, int extraCandies, int* returnSize) {
int mx = 0;
for (int i = 0; i < candiesSize; i++) {
mx = max(mx, candies[i]);
}
bool* ans = malloc(candiesSize * sizeof(bool));
for (int i = 0; i < candiesSize; i++) {
ans[i] = candies[i] + extraCandies >= mx;
}
*returnSize = candiesSize;
return ans;
}
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