Description
There are n children standing in a line. Each child is assigned a rating value given in the integer array ratings.
You are giving candies to these children subjected to the following requirements:
- Each child must have at least one candy.
- Children with a higher rating get more candies than their neighbors.
Return the minimum number of candies you need to have to distribute the candies to the children.
Example 1:
Input: ratings = [1,0,2]
Output: 5
Explanation: You can allocate to the first, second and third child with 2, 1, 2 candies respectively.
Example 2:
Input: ratings = [1,2,2]
Output: 4
Explanation: You can allocate to the first, second and third child with 1, 2, 1 candies respectively.
The third child gets 1 candy because it satisfies the above two conditions.
Constraints:
n == ratings.length1 <= n <= 2 * 1040 <= ratings[i] <= 2 * 104
Solutions
Solution 1: Two traversals
We initialize two arrays left and right, where left[i] represents the minimum number of candies the current child should get when the current child's score is higher than the left child's score, and right[i] represents the minimum number of candies the current child should get when the current child's score is higher than the right child's score. Initially, left[i]=1, right[i]=1.
We traverse the array from left to right once, and if the current child's score is higher than the left child's score, then left[i]=left[i-1]+1; similarly, we traverse the array from right to left once, and if the current child's score is higher than the right child's score, then right[i]=right[i+1]+1.
Finally, we traverse the array of scores once, and the minimum number of candies each child should get is the maximum of left[i] and right[i], and we add them up to get the answer.
Time complexity O(n), space complexity O(n). Where n is the length of the array of scores.
PythonJavaC++GoTypeScriptRustC#
class Solution:
def candy(self, ratings: List[int]) -> int:
n = len(ratings)
left = [1] * n
right = [1] * n
for i in range(1, n):
if ratings[i] > ratings[i - 1]:
left[i] = left[i - 1] + 1
for i in range(n - 2, -1, -1):
if ratings[i] > ratings[i + 1]:
right[i] = right[i + 1] + 1
return sum(max(a, b) for a, b in zip(left, right))(code-box)
class Solution {
public int candy(int[] ratings) {
int n = ratings.length;
int[] left = new int[n];
int[] right = new int[n];
Arrays.fill(left, 1);
Arrays.fill(right, 1);
for (int i = 1; i < n; ++i) {
if (ratings[i] > ratings[i - 1]) {
left[i] = left[i - 1] + 1;
}
}
for (int i = n - 2; i >= 0; --i) {
if (ratings[i] > ratings[i + 1]) {
right[i] = right[i + 1] + 1;
}
}
int ans = 0;
for (int i = 0; i < n; ++i) {
ans += Math.max(left[i], right[i]);
}
return ans;
}
}(code-box)
class Solution {
public:
int candy(vector<int>& ratings) {
int n = ratings.size();
vector<int> left(n, 1);
vector<int> right(n, 1);
for (int i = 1; i < n; ++i) {
if (ratings[i] > ratings[i - 1]) {
left[i] = left[i - 1] + 1;
}
}
for (int i = n - 2; ~i; --i) {
if (ratings[i] > ratings[i + 1]) {
right[i] = right[i + 1] + 1;
}
}
int ans = 0;
for (int i = 0; i < n; ++i) {
ans += max(left[i], right[i]);
}
return ans;
}
};(code-box)
func candy(ratings []int) int {
n := len(ratings)
left := make([]int, n)
right := make([]int, n)
for i := range left {
left[i] = 1
right[i] = 1
}
for i := 1; i < n; i++ {
if ratings[i] > ratings[i-1] {
left[i] = left[i-1] + 1
}
}
for i := n - 2; i >= 0; i-- {
if ratings[i] > ratings[i+1] {
right[i] = right[i+1] + 1
}
}
ans := 0
for i, a := range left {
b := right[i]
ans += max(a, b)
}
return ans
}(code-box)
function candy(ratings: number[]): number {
const n = ratings.length;
const left = new Array(n).fill(1);
const right = new Array(n).fill(1);
for (let i = 1; i < n; ++i) {
if (ratings[i] > ratings[i - 1]) {
left[i] = left[i - 1] + 1;
}
}
for (let i = n - 2; i >= 0; --i) {
if (ratings[i] > ratings[i + 1]) {
right[i] = right[i + 1] + 1;
}
}
let ans = 0;
for (let i = 0; i < n; ++i) {
ans += Math.max(left[i], right[i]);
}
return ans;
}(code-box)
impl Solution {
pub fn candy(ratings: Vec<i32>) -> i32 {
let n = ratings.len();
let mut left = vec![1; n];
let mut right = vec![1; n];
for i in 1..n {
if ratings[i] > ratings[i - 1] {
left[i] = left[i - 1] + 1;
}
}
for i in (0..n - 1).rev() {
if ratings[i] > ratings[i + 1] {
right[i] = right[i + 1] + 1;
}
}
ratings.iter()
.enumerate()
.map(|(i, _)| left[i].max(right[i]) as i32)
.sum()
}
}(code-box)
public class Solution {
public int Candy(int[] ratings) {
int n = ratings.Length;
int[] left = new int[n];
int[] right = new int[n];
Array.Fill(left, 1);
Array.Fill(right, 1);
for (int i = 1; i < n; ++i) {
if (ratings[i] > ratings[i - 1]) {
left[i] = left[i - 1] + 1;
}
}
for (int i = n - 2; i >= 0; --i) {
if (ratings[i] > ratings[i + 1]) {
right[i] = right[i + 1] + 1;
}
}
int ans = 0;
for (int i = 0; i < n; ++i) {
ans += Math.Max(left[i], right[i]);
}
return ans;
}
}(code-box)
