2103. Rings and Rods

Problem

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 * 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).

Solution (Java)

class Solution {
    public int countPoints(String rings) {
        Map<Integer, Integer> redHashMap = new HashMap<>();
        Map<Integer, Integer> greenHashMap = new HashMap<>();
        Map<Integer, Integer> blueHashMap = new HashMap<>();
        for (int i = 0; i <= rings.length() - 2; i = i + 2) {
            char charOne = rings.charAt(i);
            char charTwo = rings.charAt(i + 1);

            if (charOne == 'R') {
                redHashMap.put(Character.getNumericValue(charTwo), 123);
            } else if (charOne == 'G') {
                greenHashMap.put(Character.getNumericValue(charTwo), 123);
            } else {
                blueHashMap.put(Character.getNumericValue(charTwo), 123);
            }
        }
        int result = 0;
        for (int i = 0; i <= 9; i++) {
            if (redHashMap.containsKey(i)
                    && greenHashMap.containsKey(i)
                    && blueHashMap.containsKey(i)) {
                result++;
            }
        }
        return result;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).