Leetcode #677: Map Sum Pairs
In this guide, we solve Leetcode #677 Map Sum Pairs in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Design, Trie, Hash Table, String
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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)
Python Solution
class Trie:
def __init__(self):
self.children: List[Trie | None] = [None] * 26
self.val: int = 0
def insert(self, w: str, x: int):
node = self
for c in w:
idx = ord(c) - ord('a')
if node.children[idx] is None:
node.children[idx] = Trie()
node = node.children[idx]
node.val += x
def search(self, w: str) -> int:
node = self
for c in w:
idx = ord(c) - ord('a')
if node.children[idx] is None:
return 0
node = node.children[idx]
return node.val
class MapSum:
def __init__(self):
self.d = defaultdict(int)
self.tree = Trie()
def insert(self, key: str, val: int) -> None:
x = val - self.d[key]
self.d[key] = val
self.tree.insert(key, x)
def sum(self, prefix: str) -> int:
return self.tree.search(prefix)
# Your MapSum object will be instantiated and called as such:
# obj = MapSum()
# obj.insert(key,val)
# param_2 = obj.sum(prefix)
Complexity
The time complexity is O(n). The space complexity is , where and are the number of keys and the maximum length of the keys, respectively; and is the size of the character set, which is in this problem.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.