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/*
Author: King, wangjingui@outlook.com
Date: Dec 12, 2014
Problem: Candy
Difficulty: Easy
Source: https://oj.leetcode.com/problems/candy/
Notes:
There are N children standing in a line. Each child is assigned a rating value.
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.
What is the minimum candies you must give?
Solution: You may refer to https://github.com/AnnieKim/ITint5/blob/master/031_%E5%88%86%E9%85%8D%E7%B3%96%E6%9E%9C.cpp
1. O(n) space.
2. traverse only once with O(1) space.
*/
public class Solution {
public int candy(int[] ratings) {
return candy_1(ratings);
}
public int candy_1(int[] ratings) {
int N = ratings.length;
if (N == 0) return 0;
int[] height = new int[N];
int res = 0;
height[0] = 1;
for (int i = 1; i < N; ++i) {
height[i] = 1;
if (ratings[i] > ratings[i - 1]) {
height[i] = height[i - 1] + 1;
}
}
for (int i = N - 2; i >= 0; --i) {
if (ratings[i] > ratings[i + 1]) {
height[i] = Math.max(height[i], height[i + 1] + 1);
}
}
for (int i = 0; i < N; ++i) {
res +=height[i];
}
return res;
}
public int candy_2(int[] ratings) {
int N = ratings.length;
if (N == 0) return 0;
int candy = 1, res = 1;
int maxVal = 1, maxIdx = 0;
for (int i = 1; i < N; ++i) {
if (ratings[i] >= ratings[i - 1]) {
candy = ratings[i] == ratings[i - 1] ? 1 : candy + 1;
maxVal = candy;
maxIdx = i;
} else {
if (candy == 1) {
if (maxVal <= i - maxIdx) {
++maxVal;
++res;
}
res += i - maxIdx - 1;
}
candy = 1;
}
res += candy;
}
return res;
}
}
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