LeetCode 2103. Rings and Rods Solution in Java, C++, Python & More | Explanation + Code

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

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 ith rod. If there are red, green, and blue rings on the ith 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(|Σ|), where n represents the length of the string rings, and |Σ| represents the size of the character set.

PythonJavaC++GoTypeScriptRustC

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)(code-box)

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; } }(code-box)

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); } };(code-box)

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 }(code-box)

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; }(code-box)

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 } }(code-box)

int countPoints(char* rings) { int d['Z']; memset(d, 0, sizeof(d)); d['R'] = 1; d['G'] = 2; d['B'] = 4; int mask[10]; memset(mask, 0, sizeof(mask)); for (int i = 0, n = strlen(rings); i < n; i += 2) { int c = rings[i]; int j = rings[i + 1] - '0'; mask[j] |= d[c]; } int ans = 0; for (int i = 0; i < 10; i++) { if (mask[i] == 7) { ans++; } } return ans; }(code-box)

Solution 2

TypeScript

function countPoints(rings: string): number { let c = 0; for (let i = 0; i <= 9; i++) { if (rings.includes('B' + i) && rings.includes('R' + i) && rings.includes('G' + i)) c++; } return c; }(code-box)