LeetCode - Candy

Problem description

[description](https://leetcode.com/problems/candy/

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?

Example 1:

Input: [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: [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.

Analysis

A simple idea is to use two array, one from left to right, and one from right to left. And we find the candies should be allocate for two directions. And the max in the same position for these two arrays should be the final allocated candies.

public class Solution {
    public int candy(int[] ratings) {
        int sum = 0;
        int[] left2right = new int[ratings.length];
        int[] right2left = new int[ratings.length];
        Arrays.fill(left2right, 1);
        Arrays.fill(right2left, 1);
        for (int i = 1; i < ratings.length; i++) {
            if (ratings[i] > ratings[i - 1]) {
                left2right[i] = left2right[i - 1] + 1;
            }
        }
        for (int i = ratings.length - 2; i >= 0; i--) {
            if (ratings[i] > ratings[i + 1]) {
                right2left[i] = right2left[i + 1] + 1;
            }
        }
        for (int i = 0; i < ratings.length; i++) {
            sum += Math.max(left2right[i], right2left[i]);
        }
        return sum;
    }
}

To improve, we can use one array to store these information.

public class Solution {
    public int candy(int[] ratings) {
        int[] candies = new int[ratings.length];
        Arrays.fill(candies, 1);
        for (int i = 1; i < ratings.length; i++) {
            if (ratings[i] > ratings[i - 1]) {
                candies[i] = candies[i - 1] + 1;
            }
        }
        int sum = candies[ratings.length - 1];
        for (int i = ratings.length - 2; i >= 0; i--) {
            if (ratings[i] > ratings[i + 1]) {
                candies[i] = Math.max(candies[i], candies[i + 1] + 1);
            }
            sum += candies[i];
        }
        return sum;
    }
}

Another one pass solution is find the valley and peak of the array.

public class Solution {
    public int count(int n) {
        return (n * (n + 1)) / 2;
    }
    public int candy(int[] ratings) {
        if (ratings.length <= 1) {
            return ratings.length;
        }
        int candies = 0;
        int up = 0;
        int down = 0;
        int old_slope = 0;
        for (int i = 1; i < ratings.length; i++) {
            int new_slope = (ratings[i] > ratings[i - 1]) ? 1 : (ratings[i] < ratings[i - 1] ? -1 : 0);
            if ((old_slope > 0 && new_slope == 0) || (old_slope < 0 && new_slope >= 0)) {
                candies += count(up) + count(down) + Math.max(up, down);
                up = 0;
                down = 0;
            }
            if (new_slope > 0)
                up++;
            if (new_slope < 0)
                down++;
            if (new_slope == 0)
                candies++;

            old_slope = new_slope;
        }
        candies += count(up) + count(down) + Math.max(up, down) + 1;
        return candies;
    }
}

What to improve

• if we can’t get the best solution, do the most trivial one.