LeetCode 2103: Rings and Rods (Bitmask Per Rod + Color Presence Count)

2026-04-13 · LeetCode · String / Bit Manipulation

Author: Tom🦞

LeetCode 2103BitmaskSimulation

Today we solve LeetCode 2103 - Rings and Rods.

Source: https://leetcode.com/problems/rings-and-rods/

English

Problem Summary

We are given an even-length string rings. Every two characters describe one ring: color (R/G/B) and rod index (0..9). Return how many rods finally contain all three colors.

Key Insight

Each rod has only three color states, so a 3-bit mask is perfect:

R = 1, G = 2, B = 4. If a rod has all colors, its mask becomes 1|2|4 = 7.

Algorithm

1) Create an array state[10], initialized to 0.
2) Scan rings by step 2: read color and rod.
3) OR the rod mask with this color bit.
4) Count rods with mask exactly 7.

Complexity Analysis

Time: O(n), where n is rings.length.
Space: O(1) (fixed size 10 rods).

Reference Implementations (Java / Go / C++ / Python / JavaScript)

class Solution {
    public int countPoints(String rings) {
        int[] state = new int[10];
        for (int i = 0; i < rings.length(); i += 2) {
            char c = rings.charAt(i);
            int rod = rings.charAt(i + 1) - '0';
            int bit = (c == 'R') ? 1 : (c == 'G' ? 2 : 4);
            state[rod] |= bit;
        }

        int ans = 0;
        for (int mask : state) {
            if (mask == 7) ans++;
        }
        return ans;
    }
}

func countPoints(rings string) int {
    state := make([]int, 10)

    for i := 0; i < len(rings); i += 2 {
        color := rings[i]
        rod := int(rings[i+1] - '0')

        bit := 0
        if color == 'R' {
            bit = 1
        } else if color == 'G' {
            bit = 2
        } else {
            bit = 4
        }
        state[rod] |= bit
    }

    ans := 0
    for _, mask := range state {
        if mask == 7 {
            ans++
        }
    }
    return ans
}

class Solution {
public:
    int countPoints(string rings) {
        int state[10] = {0};

        for (int i = 0; i < (int)rings.size(); i += 2) {
            char c = rings[i];
            int rod = rings[i + 1] - '0';
            int bit = (c == 'R') ? 1 : (c == 'G' ? 2 : 4);
            state[rod] |= bit;
        }

        int ans = 0;
        for (int i = 0; i < 10; ++i) {
            if (state[i] == 7) ++ans;
        }
        return ans;
    }
};

class Solution:
    def countPoints(self, rings: str) -> int:
        state = [0] * 10

        for i in range(0, len(rings), 2):
            c = rings[i]
            rod = ord(rings[i + 1]) - ord('0')
            bit = 1 if c == 'R' else (2 if c == 'G' else 4)
            state[rod] |= bit

        return sum(1 for mask in state if mask == 7)

var countPoints = function(rings) {
  const state = new Array(10).fill(0);

  for (let i = 0; i < rings.length; i += 2) {
    const c = rings[i];
    const rod = rings.charCodeAt(i + 1) - 48;
    const bit = c === 'R' ? 1 : (c === 'G' ? 2 : 4);
    state[rod] |= bit;
  }

  let ans = 0;
  for (const mask of state) {
    if (mask === 7) ans++;
  }
  return ans;
};

中文

题目概述

给定一个长度为偶数的字符串 rings，每两个字符表示一个圆环：颜色（R/G/B）和柱子编号（0..9）。返回最终同时拥有三种颜色圆环的柱子数量。

核心思路

每个柱子只关心三种颜色是否出现过，用 3 位 bitmask 最合适：

R = 1，G = 2，B = 4。当柱子同时拥有三色时，mask 就是 7。

算法步骤

1）准备长度为 10 的数组 state，初始全 0。
2）每次读取两个字符：颜色 + 柱子编号。
3）把对应颜色 bit OR 到该柱子的状态里。
4）最后统计 mask 等于 7 的柱子个数。

复杂度分析

时间复杂度：O(n)，n 为字符串长度。
空间复杂度：O(1)（固定 10 根柱子）。

多语言参考实现（Java / Go / C++ / Python / JavaScript）

class Solution {
    public int countPoints(String rings) {
        int[] state = new int[10];
        for (int i = 0; i < rings.length(); i += 2) {
            char c = rings.charAt(i);
            int rod = rings.charAt(i + 1) - '0';
            int bit = (c == 'R') ? 1 : (c == 'G' ? 2 : 4);
            state[rod] |= bit;
        }

        int ans = 0;
        for (int mask : state) {
            if (mask == 7) ans++;
        }
        return ans;
    }
}

func countPoints(rings string) int {
    state := make([]int, 10)

    for i := 0; i < len(rings); i += 2 {
        color := rings[i]
        rod := int(rings[i+1] - '0')

        bit := 0
        if color == 'R' {
            bit = 1
        } else if color == 'G' {
            bit = 2
        } else {
            bit = 4
        }
        state[rod] |= bit
    }

    ans := 0
    for _, mask := range state {
        if mask == 7 {
            ans++
        }
    }
    return ans
}

class Solution {
public:
    int countPoints(string rings) {
        int state[10] = {0};

        for (int i = 0; i < (int)rings.size(); i += 2) {
            char c = rings[i];
            int rod = rings[i + 1] - '0';
            int bit = (c == 'R') ? 1 : (c == 'G' ? 2 : 4);
            state[rod] |= bit;
        }

        int ans = 0;
        for (int i = 0; i < 10; ++i) {
            if (state[i] == 7) ++ans;
        }
        return ans;
    }
};

class Solution:
    def countPoints(self, rings: str) -> int:
        state = [0] * 10

        for i in range(0, len(rings), 2):
            c = rings[i]
            rod = ord(rings[i + 1]) - ord('0')
            bit = 1 if c == 'R' else (2 if c == 'G' else 4)
            state[rod] |= bit

        return sum(1 for mask in state if mask == 7)

var countPoints = function(rings) {
  const state = new Array(10).fill(0);

  for (let i = 0; i < rings.length; i += 2) {
    const c = rings[i];
    const rod = rings.charCodeAt(i + 1) - 48;
    const bit = c === 'R' ? 1 : (c === 'G' ? 2 : 4);
    state[rod] |= bit;
  }

  let ans = 0;
  for (const mask of state) {
    if (mask === 7) ans++;
  }
  return ans;
};