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 * n1 <= n <= 100rings[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 $i$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.
1
2
3
4
5
6
7
8
9
| 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)
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
| 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;
}
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
| 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);
}
};
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
| 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
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
| 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;
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
| 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
}
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
| 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;
}
|
Solution 2#
1
2
3
4
5
6
7
| 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;
}
|
