Leetcode 2103: Rings and Rods

Pierre-Marie Poitevin
2 min readJan 21, 2022

In this problem, we are given a list of rings going into rods, and we should return how many rods have rings with all the possible colors.

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.

To solve this problem, we mostly need to reformat the input to something that will solve our problem easily. I introduced arr so that the information is better organized. arr[i][j] is true if there is a ring of color j in the rod i.

To do that we need to convert colors to integers [0, 2]. So I decided arbitrarily that B is 0, R is 1, and G is 2.

Our algorithm is the following. First we fill arr by reading the string pair of characters by pair of characters. Then we check for each rod if the 3 colors are there. Finally we return the result.

Jamboard of solution

This is the code I obtained in C++:

class Solution {

int charToInt(char c) {
switch(c) {
case 'B':
return 0;
case 'R':
return 1;
case 'G':
return 2;
return -1;

int countPoints(string rings) {
bool arr[10][3]{};
int n = rings.size() / 2;
for (int i = 0; i < n; i++) {
int color = charToInt(rings.at(2 * i));
int ring = rings.at(2 * i + 1) - '0';
arr[ring][color] = true;
int res = 0;
for (int i = 0; i < 10; i++) {
int count = 0;
for (int j = 0; j < 3; j++) {
if (arr[i][j]) {
if (count == 3) {
return res;

Happy coding :)