How do you solve Map Sum Pairs on LeetCode?

Map Sum Pairs (LeetCode #677) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a Trie allows efficient prefix-based queries.

What is the brute force solution for Map Sum Pairs?

The brute force approach for Map Sum Pairs is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute-force approach involves storing all key-value pairs in a simple dictionary and iterating through all keys to calculate the sum of values that match the given prefix. This method is straightforward but inefficient for larger datasets. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Map Sum Pairs?

Map Sum Pairs uses the following concepts: Hash Table, String, Design, Trie. The recommended patterns to study are: Hash Map, Trie.

What are the interview tips for Map Sum Pairs?

Explain your thought process clearly. Discuss trade-offs between different data structures. Consider edge cases during your implementation.

What are common mistakes when solving Map Sum Pairs?

Not updating existing keys correctly during insertion. Forgetting to handle cases where the prefix does not exist.

#677

Map Sum Pairs

Medium

Hash TableStringDesignTrieHash MapTrie

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(n)	O(n)

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Intuition

Time O(n)Space O(n)

Using a Trie (prefix tree) allows us to efficiently store keys and calculate the sum of values for any prefix. This structure enables us to traverse the tree based on the prefix and quickly gather the sums of all values associated with keys that share that prefix.

⚙️

Algorithm

4 steps

1Step 1: Create a TrieNode class to represent each node in the Trie, storing children and a value.
2Step 2: In the MapSum class, initialize the root TrieNode and a dictionary to store key-value pairs.
3Step 3: For the insert operation, traverse the Trie according to the characters in the key, updating the value at the final node and storing the key in the dictionary.
4Step 4: For the sum operation, traverse the Trie according to the prefix and accumulate values from all descendant nodes.

solution.py29 lines

1class TrieNode:
2    def __init__(self):
3        self.children = {}
4        self.value = 0
5
6class MapSum:
7    def __init__(self):
8        self.root = TrieNode()
9        self.map = {}
10
11    def insert(self, key: str, val: int) -> None:
12        node = self.root
13        if key in self.map:
14            old_val = self.map[key]
15            val -= old_val
16        self.map[key] = val
17        for char in key:
18            if char not in node.children:
19                node.children[char] = TrieNode()
20            node = node.children[char]
21            node.value += val
22
23    def sum(self, prefix: str) -> int:
24        node = self.root
25        for char in prefix:
26            if char not in node.children:
27                return 0
28            node = node.children[char]
29        return node.value

ℹ

Complexity note: The time complexity is O(n) for insertions and O(m) for sum operations, where n is the length of the key and m is the length of the prefix. The space complexity is O(n) due to the Trie structure storing each character of the keys.

1Using a Trie allows efficient prefix-based queries.
2Updating values in a Trie can be done by adjusting node values directly.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.