677. Map Sum Pairs
Description
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 theMapSum
object.void insert(String key, int val)
Inserts thekey-val
pair into the map. If thekey
already existed, the originalkey-value
pair will be overridden to the new one.int sum(string prefix)
Returns the sum of all the pairs' value whosekey
starts with theprefix
.
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
andprefix
consist of only lowercase English letters.1 <= val <= 1000
- At most
50
calls will be made toinsert
andsum
.
Solution
map-sum-pairs.py
class TrieNode:
def __init__(self):
self.children = collections.defaultdict(TrieNode)
self.value = 0
class Trie:
def __init__(self):
self.root = TrieNode()
def insert(self, key, val):
curr = self.root
for c in key:
curr = curr.children[c]
curr.value = val
def sumVal(self, prefix):
curr = self.root
for c in prefix:
curr = curr.children[c]
return self.dfs(curr)
def dfs(self, curr):
total = curr.value
for char in curr.children:
if curr.children[char]:
total += self.dfs(curr.children[char])
return total
class MapSum:
def __init__(self):
self.trie = Trie()
def insert(self, key: str, val: int) -> None:
self.trie.insert(key, val)
def sum(self, prefix: str) -> int:
return self.trie.sumVal(prefix)
# Your MapSum object will be instantiated and called as such:
# obj = MapSum()
# obj.insert(key,val)
# param_2 = obj.sum(prefix)