[Leet Code] Rings and Rods

  • 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').
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.
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.
Input: rings = "G4"
Output: 0
Explanation:
Only one ring is given. Thus, no rods have all three colors.
  • 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).
func countPoints(rings string) int {
hashMap := make(map[string][]string)
counter := 0
for i := 0; i < len(rings); i+= 2{
letter := string(rings[i])
number := string(rings[i+1])
hashMap[number] = append(hashMap[number], letter)
}
fmt.Println(hashMap)
for number := range hashMap{
if len(unique(hashMap[number])) == 3{
counter += 1
}
}
return counter
}
func unique(intSlice []string) []string {
keys := make(map[string]bool)
list := []string{}
for _, entry := range intSlice {
if _, value := keys[entry]; !value {
keys[entry] = true
list = append(list, entry)
}
}
return list
}

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Matthew Boyd

Matthew Boyd

Learning, and posting my findings!