Welcome to Subscribe On Youtube
2103. Rings and Rods
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 colorposition pair that is used to describe each ring where:
 The first character of the
i^{th}
pair denotes thei^{th}
ring's color ('R'
,'G'
,'B'
).  The second character of the
i^{th}
pair denotes the rod that thei^{th}
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]
wherei
is even is either'R'
,'G'
, or'B'
(0indexed).rings[i]
wherei
is odd is a digit from'0'
to'9'
(0indexed).
Solutions
Solution 1: Bit Manipulation
We can use an array $mask$ of length $10$ to represent the color situation of the rings on each rod, where $mask[i]$ represents the color situation of the ring on the $i$th rod. If there are red, green, and blue rings on the $i$th rod, then the binary representation of $mask[i]$ is $111$, that is, $mask[i] = 7$.
We traverse the string $rings$. For each color position pair $(c, j)$, where $c$ represents the color of the ring and $j$ represents the number of the rod where the ring is located, we set the corresponding binary bit of $mask[j]$, that is, $mask[j]  = d[c]$, where $d[c]$ represents the binary bit corresponding to color $c$. 
Finally, we count the number of elements in $mask$ that are $7$, which is the number of rods that have collected all three colors of rings.
The time complexity is $O(n)$, and the space complexity is $O(\Sigma)$, where $n$ represents the length of the string $rings$, and $\Sigma$ represents the size of the character set.

class Solution { public int countPoints(String rings) { int[] d = new int['Z']; d['R'] = 1; d['G'] = 2; d['B'] = 4; int[] mask = new int[10]; for (int i = 0, n = rings.length(); i < n; i += 2) { int c = rings.charAt(i); int j = rings.charAt(i + 1)  '0'; mask[j] = d[c]; } int ans = 0; for (int x : mask) { if (x == 7) { ++ans; } } return ans; } }

class Solution { public: int countPoints(string rings) { int d['Z']{['R'] = 1, ['G'] = 2, ['B'] = 4}; int mask[10]{}; for (int i = 0, n = rings.size(); i < n; i += 2) { int c = rings[i]; int j = rings[i + 1]  '0'; mask[j] = d[c]; } return count(mask, mask + 10, 7); } };

class Solution: def countPoints(self, rings: str) > int: mask = [0] * 10 d = {"R": 1, "G": 2, "B": 4} for i in range(0, len(rings), 2): c = rings[i] j = int(rings[i + 1]) mask[j] = d[c] return mask.count(7)

func countPoints(rings string) (ans int) { d := ['Z']int{'R': 1, 'G': 2, 'B': 4} mask := [10]int{} for i, n := 0, len(rings); i < n; i += 2 { c := rings[i] j := int(rings[i+1]  '0') mask[j] = d[c] } for _, x := range mask { if x == 7 { ans++ } } return }

function countPoints(rings: string): number { const idx = (c: string) => c.charCodeAt(0)  'A'.charCodeAt(0); const d: number[] = Array(26).fill(0); d[idx('R')] = 1; d[idx('G')] = 2; d[idx('B')] = 4; const mask: number[] = Array(10).fill(0); for (let i = 0; i < rings.length; i += 2) { const c = rings[i]; const j = rings[i + 1].charCodeAt(0)  '0'.charCodeAt(0); mask[j] = d[idx(c)]; } return mask.filter(x => x === 7).length; }

impl Solution { pub fn count_points(rings: String) > i32 { let mut d: [i32; 90] = [0; 90]; d['R' as usize] = 1; d['G' as usize] = 2; d['B' as usize] = 4; let mut mask: [i32; 10] = [0; 10]; let cs: Vec<char> = rings.chars().collect(); for i in (0..cs.len()).step_by(2) { let c = cs[i] as usize; let j = (cs[i + 1] as usize)  ('0' as usize); mask[j] = d[c]; } mask .iter() .filter(&&x x == 7) .count() as i32 } }