677. Map Sum Pairs

MediumDesignTrieHash TableString
Leetcode Link

Problem Description

The problem requires designing a data structure that can map strings to integer values and efficiently calculate the sum of values for strings that start with a certain prefix. This data structure must support two operations:

  1. Insert a key-value pair. If the key already exists, update its value.
  2. Compute the sum of values for all keys that begin with a specific prefix.

This is useful in scenarios where you need to store key-value pairs and perform prefix sum queries, like in an autocomplete feature where you might want to know the aggregate score of suggestions based on a certain prefix.


The intuition behind this solution involves using two main concepts: a hash map and a trie (prefix tree). The hash map (defaultdict) is used to store key-value pairs, ensuring that we can update the value associated with a key quickly, in constant time (O(1)).

However, a hash map alone is not efficient for prefix-related queries, which is where the trie comes in. A trie is a tree-like data structure particularly suited for handling prefix related operations. Each node in a trie represents a letter, and each path from the root to a node represents a prefix formed by the corresponding letters.

Here's how the two structures work together in the solution:

  • Whenever we insert a new key-value pair, we first calculate the difference (x) between the new value and the existing value associated with the key in the hash map (which is 0 if the key is new). This difference is then used to update values in the trie. We update every node in the path of the key in the trie by this difference so that every prefix that includes this key has the correct sum.

  • The insert operation updates the hash map with the new value for the key and traverses the trie to reflect these changes across all relevant prefixes.

  • The sum operation simply traverses the trie along the path formed by the prefix to find the sum of all values associated with the keys that start with that prefix. The val stored at each trie node is used to calculate this sum, which is the cumulative value of all strings passing through that node.

This approach leads to an efficient solution since the insert operation is only as slow as the length of the word being inserted, and the sum operation is only as slow as the length of the prefix we're querying, both being O(m) where m is the length of the respective strings.

Solution Approach

The solution is implemented using a combination of a Trie and a hash map. Here's how the main components of the solution work:

  1. Trie Node: The Trie class represents nodes of the trie with two properties:

    • children: An array of 26 elements representing the 26 lowercase English letters.
    • val: An integer value storing the cumulative sum at the current node.
  2. Trie Insertion: insert method of the Trie class is used to add a string w along with an integer x indicating the difference to be added at each node. For each character in w, we calculate its index and proceed down the trie, creating new nodes if necessary. The val at each node is incremented by x.

  3. Trie Search: search method is for computing the sum for a given prefix. It traverses the trie using each character of the prefix. If at any point the next character of the prefix does not exist in the trie, the method returns 0. If the prefix is found in the trie, it returns the val at the last node of the prefix.

  4. MapSum Class: This class utilizes a hash map (self.d) to keep track of the most recent values for each key.

  5. Initialization: In the constructor of MapSum, we initialize the hash map (self.d) and the trie (self.tree).

  6. Inserting Key-Value Pair: The insert method first calculates the difference x between the new value and the current value of the key in the hash map. It then updates the hash map with the new value for the key. Lastly, it calls the insert method on the trie to adjust the cumulative sums for the nodes in the path of the inserted key.

  7. Sum of Prefix: The sum method uses the search method on the trie to return the sum of all values in the map that have a key with the given prefix.

This solution ensures that values are only updated in relevant paths of the trie and that the trie reflects the correct cumulative sums for each prefix at any given time. The insert and sum operations take O(m) time where m is the length of the input string, and this makes the operations efficient, especially considering multiple queries for different prefixes.

Level Up Your
Algo Skills

Example Walkthrough

Let's demonstrate the solution approach using a small example. Suppose we are developing an autocomplete system and we want to use our data structure to store the strings along with their associated scores.

First, we initialize our data structure.

1# Initialization of data structure
2mapsum = MapSum()
  1. We insert the key-value pair ('apple', 3) into our MapSum.
1# Insert key-value pair ('apple', 3)
2mapsum.insert('apple', 3)

In the insert method, since 'apple' is a new key, its difference x is 3 (since the old value is implicitly 0). The hash map self.d is updated to store {'apple': 3}, and the trie nodes for 'a', 'p', 'p', 'l', and 'e' are created or updated with the cumulative sum so far. Each of the nodes along the path will have a value of 3 added to their val.

  1. Now, let's insert another key-value pair ('app', 2).
1# Insert key-value pair ('app', 2)
2mapsum.insert('app', 2)

The difference x this time is 2. The hash map now stores {'apple': 3, 'app': 2}. The trie is updated such that the nodes for the path 'app' now have an additional 2 added to their val.

  1. If we now query the sum of keys that start with the prefix 'ap',
1# Calculate the sum of values for keys starting with 'ap'
2total_sum = mapsum.sum('ap')  # Should return 5, since 'apple' and 'app' start with 'ap'

The trie is traversed through the path of 'a' to 'p' summing up the values. Since 'apple' and 'app' both start with 'ap', their values are added, which is 3 + 2 = 5. Thus, total_sum would be 5.

  1. Let's update the value for 'app' by inserting a new value for it.
1# Update the existing key 'app' with a new value
2mapsum.insert('app', 4)

Initially, the value in the hash map for 'app' was 2, so the difference x is now 4 - 2 = 2. The hash map is updated to {'apple': 3, 'app': 4}, and the trie nodes on the path for 'app' have an additional 2 added to their val. The node for 'apple' does not change since we're only updating the prefix 'app'.

  1. Querying the sum of keys with the prefix 'ap' again,
1# Re-calculate the sum of values for keys starting with 'ap' after the update
2total_sum = mapsum.sum('ap')  # Should return 7, because we now have 'apple': 3 and 'app': 4

Upon calling the sum method of MapSum, the traversal in the trie for 'ap' now sums up to 7, as both 'apple' and 'app' are updated.

Through this example, we can see how the combination of a hash map and a trie makes it efficient to not only insert and update key-value pairs, but also compute sums based on prefixes with time complexity tied to the length of the input strings.

Python Solution

1from collections import defaultdict
2from typing import List
5class TrieNode:
6    def __init__(self):
7        # Initialize children for each letter of the alphabet
8        self.children: List[TrieNode | None] = [None] * 26
9        # Value to be accumulated at each node
10        self.value: int = 0
12    def insert(self, word: str, val: int):
13        # Start at the root node
14        node = self
15        # Iterate over each character in the word
16        for char in word:
17            index = ord(char) - ord('a')  # Calculate index in the alphabet
18            # If the child doesn't exist, create a new TrieNode
19            if node.children[index] is None:
20                node.children[index] = TrieNode()
21            node = node.children[index]
22            # Add the value to the node's value
23            node.value += val
25    def search(self, word: str) -> int:
26        # Start at the root node
27        node = self
28        # Iterate over each character in the word
29        for char in word:
30            index = ord(char) - ord('a')  # Calculate index in the alphabet
31            # If a child doesn't exist, return 0 as we have no match
32            if node.children[index] is None:
33                return 0
34            node = node.children[index]
35        # Return the accumulated value
36        return node.value
39class MapSum:
40    def __init__(self):
41        # Dictionary to keep track of inserted keys and their values
42        self.dictionary = defaultdict(int)
43        # Trie data structure to keep track of the prefixes and cumulative sum
44        self.trie_root = TrieNode()
46    def insert(self, key: str, val: int) -> None:
47        # Calculate the difference from the current value to update
48        delta = val - self.dictionary[key]
49        # Update or set the key's value
50        self.dictionary[key] = val
51        # Insert the key with the updated difference into the trie
52        self.trie_root.insert(key, delta)
54    def sum(self, prefix: str) -> int:
55        # Find the sum for the prefix in the trie
56        return self.trie_root.search(prefix)
59# The MapSum object usage
60# map_sum = MapSum()
61# map_sum.insert(key, value)
62# total = map_sum.sum(prefix)

Java Solution

1class Trie {
2    // Each Trie node keeps an array of its children for each letter of the alphabet
3    private Trie[] children = new Trie[26];
4    // Value to store the sum of the inserted values for words that pass through the node
5    private int value;
7    // Inserts a word with its corresponding value into the Trie
8    public void insert(String word, int valueToAdd) {
9        Trie node = this;
10        for (int i = 0; i < word.length(); ++i) {
11            int index = word.charAt(i) - 'a'; // Convert char to index (0-25)
12            if (node.children[index] == null) {
13                node.children[index] = new Trie();
14            }
15            node = node.children[index];
16            node.value += valueToAdd; // Update the value at each node in the path
17        }
18    }
20    // Searches the Trie for a word and returns its value
21    public int search(String word) {
22        Trie node = this;
23        for (int i = 0; i < word.length(); ++i) {
24            int index = word.charAt(i) - 'a';
25            if (node.children[index] == null) {
26                return 0;
27            }
28            node = node.children[index];
29        }
30        return node.value;
31    }
34class MapSum {
35    // Dictionary to keep track of the original key-value pairs
36    private Map<String, Integer> dictionary = new HashMap<>();
37    // Trie structure to store prefixes and their corresponding summed values
38    private Trie trie = new Trie();
40    public MapSum() {
41        // Constructor is empty because no initialization is needed
42    }
44    // Inserts a key-value pair into the MapSum
45    public void insert(String key, int value) {
46        // Compute the difference of the new value and the previous value of the key
47        int delta = value - dictionary.getOrDefault(key, 0);
48        dictionary.put(key, value); // Update the dictionary
49        trie.insert(key, delta); // Insert the delta to update values along the path in Trie
50    }
52    // Returns the sum of all values that have a key with the prefix specified
53    public int sum(String prefix) {
54        return trie.search(prefix);
55    }
59 * The MapSum class can be instantiated and used as shown in the example below:
60 * 
61 * MapSum mapSum = new MapSum();
62 * mapSum.insert("key", 3);
63 * int sum = mapSum.sum("ke");
64 * 
65 * This will return the sum of all the values corresponding to keys that start with "ke".
66 */

C++ Solution

1#include <vector>
2#include <string>
3#include <unordered_map>
5// Trie node implementation for storing the string as key and the sum as value.
6class TrieNode {
8    TrieNode() : children(26, nullptr), value(0) {
9    }
11    // Inserts a string into the Trie and updates the nodes with the given value.
12    void insert(const std::string& word, int valueToAdd) {
13        TrieNode* node = this;
14        for(char ch : word) {
15            ch -= 'a';
16            if (!node->children[ch]) {
17                node->children[ch] = new TrieNode();
18            }
19            node = node->children[ch];
20            node->value += valueToAdd;
21        }
22    }
24    // Searches for a string in the Trie and returns the sum of all values of nodes corresponding to the prefix.
25    int search(const std::string& word) {
26        TrieNode* node = this;
27        for (char ch : word) {
28            ch -= 'a';
29            if (!node->children[ch]) {
30                return 0;
31            }
32            node = node->children[ch];
33        }
34        return node->value;
35    }
38    std::vector<TrieNode*> children; // Children of this node in the trie.
39    int value; // Value associated with the node.
42// MapSum class using TrieNode to store keys and values, and support increment of values.
43class MapSum {
45    // Initializes the MapSum object.
46    MapSum(): root(new TrieNode()) {
47    }
49    // Inserts the key-value pair into the map. If the key already exists, value is replaced.
50    void insert(const std::string& key, int val) {
51        int delta = val;
52        if (keysCount.find(key) != keysCount.end()) {
53            delta -= keysCount[key];
54        }
55        keysCount[key] = val;
56        root->insert(key, delta);
57    }
59    // Returns the sum of all values that have the given prefix.
60    int sum(const std::string& prefix) {
61        return root->search(prefix);
62    }
65    std::unordered_map<std::string, int> keysCount; // Tracks the latest value of keys to calculate deltas.
66    TrieNode* root; // The trie's root node.
70 * Your MapSum object will be instantiated and called as such:
71 * MapSum* obj = new MapSum();
72 * obj->insert(key, val);
73 * int param_2 = obj->sum(prefix);
74 */

Typescript Solution

1interface TrieNode {
2    children: (TrieNode | null)[];
3    value: number;
6// A Trie data structure for storing strings and their values
7const trieRoot: TrieNode = {
8    children: new Array(26).fill(null),
9    value: 0
12// Adds a word to the Trie and updates the values along the path
13function insertIntoTrie(word: string, value: number, node: TrieNode): void {
14    for (const char of word) {
15        const index = char.charCodeAt(0) - 'a'.charCodeAt(0);
16        if (node.children[index] === null) {
17            node.children[index] = { children: new Array(26).fill(null), value: 0 };
18        }
19        node = node.children[index]!;
20        node.value += value;
21    }
24// Searches the Trie and returns the total value for the given prefix
25function searchTrie(prefix: string, node: TrieNode): number {
26    for (const char of prefix) {
27        const index = char.charCodeAt(0) - 'a'.charCodeAt(0);
28        if (!node.children[index]) {
29            return 0;
30        }
31        node = node.children[index]!;
32    }
33    return node.value;
36// A map to store the key-value pairs for MapSum operations
37const mapSumDictionary: Map<string, number> = new Map<string, number>();
39// Inserts a key with its value into MapSum and updates the Trie
40function insertMapSum(key: string, value: number): void {
41    const diff = value - (mapSumDictionary.get(key) || 0);
42    mapSumDictionary.set(key, value);
43    insertIntoTrie(key, diff, trieRoot);
46// Computes the sum of all values in MapSum that have the given prefix
47function sumMapSum(prefix: string): number {
48    return searchTrie(prefix, trieRoot);
51// Usage example:
52// insertMapSum("apple", 3);
53// console.log(sumMapSum("ap")); // Output will be 3
54// insertMapSum("app", 2);
55// console.log(sumMapSum("ap")); // Output will be 5

Time and Space Complexity

Trie Class

  • The Trie class insert method has a time complexity of O(L), where L is the length of the word w. This is because it touches each character in the word once.
  • The space complexity of the insert method is also O(L), considering the worst-case scenario where each letter of the word corresponds to a new Trie node.
  • The search method in the Trie class has a time complexity of O(L) as well, as it only traverses the word once.

MapSum Class

  • The insert method in the MapSum class has a time complexity of O(L) for the same reasons as the Trie insert method, where L is the length of the key. The updating of the dictionary self.d has a constant time complexity O(1), but since we also call the Trie insert method, the overall time complexity remains O(L).

  • The space complexity of insert depends on whether new Trie nodes are created. In the worst case, it is O(L), where L is the length of the key. The dictionary self.d has a space complexity of O(N), where N is the number of unique keys inserted since each key-value pair is stored in the dictionary.

  • The sum method has a time complexity of O(L), which is required to traverse the Trie. After reaching the end of the prefix in the Trie, the value is obtained in O(1) time. There is no significant space complexity for the sum operation, so it is O(1).

Overall, the space complexity of the MapSum class as a whole is driven by the space used to store all the unique keys in the dictionary and Trie nodes, which sums up to O(N * L) where N is the number of unique keys and L is the average length of the keys.

Become an
Algo Monster

Got a question? Ask the Teaching Assistant anything you don't understand.

Still not clear? Ask in the Forum,  Discord or Submit the part you don't understand to our editors.

TA 👨‍🏫