Formatted question description: https://leetcode.ca/all/2103.html

# 2103. Rings and Rods (Easy)

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`i`

pair denotes the^{th}`i`

ring's^{th}**color**(`'R'`

,`'G'`

,`'B'`

). - The
**second**character of the`i`

pair denotes the^{th}**rod**that the`i`

ring is placed on (^{th}`'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:1Explanation:- 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:1Explanation:- 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:0Explanation: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**).

**Similar Questions**:

## Solution 1. Counting

```
// OJ: https://leetcode.com/problems/rings-and-rods/
// Time: O(N)
// Space: O(1)
class Solution {
public:
int countPoints(string s) {
bool cnt[10][3] = {};
for (int i = 0; i < s.size(); i += 2) {
int color = s[i] == 'R' ? 0 : (s[i] == 'G' ? 1 : 2), index = s[i + 1] - '0';
cnt[index][color] = true;
}
int ans = 0;
for (int i = 0; i < 10; ++i) {
ans += cnt[i][0] && cnt[i][1] && cnt[i][2];
}
return ans;
}
};
```

## Solution 2. Counting with Bitmask

```
// OJ: https://leetcode.com/problems/rings-and-rods/
// Time: O(N)
// Space: O(1)
class Solution {
public:
int countPoints(string s) {
char cnt[10] = {};
for (int i = 0; i < s.size(); i += 2) {
int color = s[i] == 'R' ? 0 : (s[i] == 'G' ? 1 : 2), index = s[i + 1] - '0';
cnt[index] |= 1 << color;
}
int ans = 0;
for (int i = 0; i < 10; ++i) ans += cnt[i] == 0b111;
return ans;
}
};
```