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2103. Rings and Rods

中文文档

Description

There are n rings and each ring is either red, green, or blue. The rings are distributed across ten rods labeled from 0 to 9.

You are given a string rings of length 2n that describes the n rings that are placed onto the rods. Every two characters in rings forms a color-position pair that is used to describe each ring where:

  • The first character of the ith pair denotes the ith ring's color ('R', 'G', 'B').
  • The second character of the ith pair denotes the rod that the ith ring is placed on ('0' to '9').

For example, "R3G2B1" describes n == 3 rings: a red ring placed onto the rod labeled 3, a green ring placed onto the rod labeled 2, and a blue ring placed onto the rod labeled 1.

Return the number of rods that have all three colors of rings on them.

 

Example 1:

Input: rings = "B0B6G0R6R0R6G9"
Output: 1
Explanation: 
- The rod labeled 0 holds 3 rings with all colors: red, green, and blue.
- The rod labeled 6 holds 3 rings, but it only has red and blue.
- The rod labeled 9 holds only a green ring.
Thus, the number of rods with all three colors is 1.

Example 2:

Input: rings = "B0R0G0R9R0B0G0"
Output: 1
Explanation: 
- The rod labeled 0 holds 6 rings with all colors: red, green, and blue.
- The rod labeled 9 holds only a red ring.
Thus, the number of rods with all three colors is 1.

Example 3:

Input: rings = "G4"
Output: 0
Explanation: 
Only one ring is given. Thus, no rods have all three colors.

 

Constraints:

  • rings.length == 2 * n
  • 1 <= n <= 100
  • rings[i] where i is even is either 'R', 'G', or 'B' (0-indexed).
  • rings[i] where i is odd is a digit from '0' to '9' (0-indexed).

Solutions

Using hash table.

Python3

class Solution:
    def countPoints(self, rings: str) -> int:
        mp = defaultdict(set)
        for i in range(1, len(rings), 2):
            c = int(rings[i])
            mp[c].add(rings[i - 1])
        return sum(1 for v in mp.values() if len(v) == 3)

Java

class Solution {
    public int countPoints(String rings) {
        Map<Integer, Set<Character>> mp = new HashMap<>();
        for (int i = 1; i < rings.length(); i += 2) {
            int c = rings.charAt(i) - '0';
            mp.computeIfAbsent(c, k -> new HashSet<>()).add(rings.charAt(i - 1));
        }
        int ans = 0;
        for (Set<Character> e : mp.values()) {
            if (e.size() == 3) {
                ++ans;
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int countPoints(string rings) {
        unordered_map<int, unordered_set<char>> mp;
        for (int i = 1; i < rings.size(); i += 2)
        {
            int c = rings[i] - '0';
            mp[c].insert(rings[i - 1]);
        }
        int ans = 0;
        for (int i = 0; i < 10; ++i)
            if (mp[i].size() == 3)
                ++ans;
        return ans;
    }
};

TypeScript



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