677. Map Sum Pairs

Problem

Design a map that allows you to do the following:

• Maps a string key to a given value.

• Returns the sum of the values that have a key with a prefix equal to a given string.

Implement the MapSum class:

• MapSum() Initializes the MapSum object.

• void insert(String key, int val) Inserts the key-val pair into the map. If the key already existed, the original key-value pair will be overridden to the new one.

• int sum(string prefix) Returns the sum of all the pairs' value whose key starts with the prefix.

  Example 1:

Input
["MapSum", "insert", "sum", "insert", "sum"]
[[], ["apple", 3], ["ap"], ["app", 2], ["ap"]]
Output
[null, null, 3, null, 5]

Explanation
MapSum mapSum = new MapSum();
mapSum.insert("apple", 3);  
mapSum.sum("ap");           // return 3 (apple = 3)
mapSum.insert("app", 2);    
mapSum.sum("ap");           // return 5 (apple + app = 3 + 2 = 5)

  Constraints:

• 1 <= key.length, prefix.length <= 50

• key and prefix consist of only lowercase English letters.

• 1 <= val <= 1000

• At most 50 calls will be made to insert and sum.

Solution (Java)

class MapSum {
    static class Node {
        int val;
        Node[] child;

        Node() {
            child = new Node[26];
            val = 0;
        }
    }

    private final Node root;

    public MapSum() {
        root = new Node();
    }

    public void insert(String key, int val) {
        Node curr = root;
        for (char c : key.toCharArray()) {
            if (curr.child[c - 'a'] == null) {
                curr.child[c - 'a'] = new Node();
            }
            curr = curr.child[c - 'a'];
        }
        curr.val = val;
    }

    private int sumHelper(Node root) {
        int o = 0;
        for (int i = 0; i < 26; i++) {
            if (root.child[i] != null) {
                o += root.child[i].val + sumHelper(root.child[i]);
            }
        }
        return o;
    }

    public int sum(String prefix) {
        Node curr = root;
        for (char c : prefix.toCharArray()) {
            if (curr.child[c - 'a'] == null) {
                return 0;
            }
            curr = curr.child[c - 'a'];
        }
        int sum = curr.val;
        return sum + sumHelper(curr);
    }
}

/**
 * Your MapSum object will be instantiated and called as such:
 * MapSum obj = new MapSum();
 * obj.insert(key,val);
 * int param_2 = obj.sum(prefix);
 */

Explain:

nope.

Complexity:

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