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 theith
ring's color ('R'
,'G'
,'B'
). - The second character of the
ith
pair denotes the rod that theith
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'
(0-indexed). -
rings[i]
wherei
is odd is a digit from'0'
to'9'
(0-indexed).
SOLUTION:
class Solution:
def countPoints(self, rings: str) -> int:
n = len(rings)
rods = [set(), set(), set()]
for i in range(0, n, 2):
if rings[i] == 'R':
rods[0].add(int(rings[i + 1]))
if rings[i] == 'G':
rods[1].add(int(rings[i + 1]))
if rings[i] == 'B':
rods[2].add(int(rings[i + 1]))
return len(set.intersection(*rods))
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