Java AC solution with explanation

  • 17

    The logic I have used is to construct a Map of maps, that contains all possible a/b and b/a from the given input and their values.

    For the given input
    equations = [ ["a", "b"], ["b", "c"] ]. values = [2.0, 3.0]

    The map that gets constructed is :

    [a: [b:2.0]
    b: [a:0.5], [c:3.0]
    c: [b:0.333]]

    For each key in the outer map, the value represents a map, that denotes all possible denominators for the key and the corresponding key/value.

    With this map constructed, the logic for evaluating a query is simple in a dfs style:

    To find any m/n, if the map of m contains x1, x2, x3
    m/n = m/x1 * x1/n if this gives a valid result or m/x2 * x2/n or m/x3 * x3/n

    public static double[] calcEquation(String[][] equations, double[] values, String[][] query) {
            Map<String, Map<String, Double>> numMap = new HashMap<>();
            int i = 0;
            for(String[] str : equations) {
                insertPairs(numMap, str[0], str[1], values[i]);
                insertPairs(numMap, str[1], str[0], 1.0/values[i]);
            double[] res = new double[query.length];
            i = 0;
            for(String[] q: query) {
                Double resObj = handleQuery(q[0], q[1], numMap, new HashSet<>());
                res[i++] = (resObj != null) ? resObj : -1.0;
            return res;
        public static void insertPairs(Map<String, Map<String, Double>> numMap, String num, String denom, Double value) {
            Map<String, Double> denomMap = numMap.get(num);
            if(denomMap == null) {
                denomMap = new HashMap<>();
                numMap.put(num, denomMap);
            denomMap.put(denom, value);
        public static Double handleQuery(String num, String denom, Map<String, Map<String, Double>> numMap, Set<String> visitedSet) {
            String dupeKey = num+":"+denom;
            if(visitedSet.contains(dupeKey)) return null;
            if(!numMap.containsKey(num) || !numMap.containsKey(denom)) return null;
            if(num.equals(denom)) return 1.0;
            Map<String, Double> denomMap = numMap.get(num);
            for(String key : denomMap.keySet()) {
                Double res = handleQuery(key, denom, numMap, visitedSet);
                if(res != null) {
                    return denomMap.get(key) * res;
            return null;

  • 1

    Good solution.

  • 1

    said in Java AC solution with explanation:
    Any reason to put the add and remove inside the for-loop rather than outside?



  • 0

    @bing25 Thanks, they should be outside.

  • 0

    I think it will be efficient to do this before for loop in dfs

    if (numMap.get(num).containsKey(denom)) {
    return numMap.get(num).get(denom);

  • 0

    Two questions:
    Why does visitedSet consist of dupeKeys of the form num:denom? Rather than just num.

    Might be a related question: why do you remove the dupe key after the for loop?

    Thank you!

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