135. Candy

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.length
• 1 <= n <= 2 * 104
• 0 <= 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.

• class Solution {
public int candy(int[] ratings) {
int n = ratings.length;
int up = 0;
int down = 0;
int peak = 0;
int candies = 1;
for (int i = 1; i < n; i++) {
if (ratings[i - 1] < ratings[i]) {
up++;
peak = up + 1;
down = 0;
candies += peak;
} else if (ratings[i] == ratings[i - 1]) {
peak = 0;
up = 0;
down = 0;
candies++;
} else {
down++;
up = 0;
candies += down + (peak > down ? 0 : 1);
}
}
return candies;
}
}

• 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;
}
};

• 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))


• 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
}

• 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;
}


• 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;
}
}