README.md

677. Map Sum Pairs

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.

Solutions (Python)

1. Solution

class MapSum:

    def __init__(self):
        self.keyval = {}
        self.trie = {}

    def insert(self, key: str, val: int) -> None:
        diff = val - self.keyval.get(key, 0)
        self.keyval[key] = val
        curr = self.trie

        for ch in key:
            if ch not in curr:
                curr[ch] = {"val": 0}
            curr = curr[ch]
            curr["val"] += diff

    def sum(self, prefix: str) -> int:
        curr = self.trie

        for i, ch in enumerate(prefix):
            if ch not in curr:
                return 0
            curr = curr[ch]
            if i == len(prefix) - 1:
                return curr["val"]


# Your MapSum object will be instantiated and called as such:
# obj = MapSum()
# obj.insert(key,val)
# param_2 = obj.sum(prefix)