README_EN.md

2103. Rings and Rods

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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(len(v) == 3 for v in mp.values())

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;
    }
};

Go

func countPoints(rings string) int {
	mp := make(map[byte]map[byte]bool)
	for i := 1; i < len(rings); i += 2 {
		c := rings[i]
		if len(mp[c]) == 0 {
			mp[c] = make(map[byte]bool)
		}
		mp[c][rings[i-1]] = true
	}
	ans := 0
	for _, v := range mp {
		if len(v) == 3 {
			ans++
		}
	}
	return ans
}

TypeScript

function countPoints(rings: string): number {
    const helper = (c: string) => c.charCodeAt(0) - 'A'.charCodeAt(0);
    const n = rings.length;
    const target = (1 << helper('R')) + (1 << helper('G')) + (1 << helper('B'));
    const count = new Array(10).fill(0);
    for (let i = 0; i < n; i += 2) {
        count[rings[i + 1]] |= 1 << helper(rings[i]);
    }
    return count.reduce((r, v) => (r += v === target ? 1 : 0), 0);
}

Rust

impl Solution {
    pub fn count_points(rings: String) -> i32 {
        let rings = rings.as_bytes();
        let target = (1 << b'R' - b'A') + (1 << b'G' - b'A') + (1 << b'B' - b'A');
        let n = rings.len();
        let mut count = [0; 10];
        let mut i = 0;
        while i < n {
            count[(rings[i + 1] - b'0') as usize] |= 1 << rings[i] - b'A';
            i += 2;
        }
        count.iter().filter(|&v| *v == target).count() as i32
    }
}

C

int countPoints(char *rings) {
    int target = (1 << ('R' - 'A')) + (1 << ('G' - 'A')) + (1 << ('B' - 'A'));
    int count[10] = {0};
    for (int i = 0; rings[i]; i += 2) {
        count[rings[i + 1] - '0'] |= 1 << (rings[i] - 'A');
    }
    int ans = 0;
    for (int i = 0; i < 10; i++) {
        if (count[i] == target) {
            ans++;
        }
    }
    return ans;
}

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