Map Sum Pairs

Medium

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StringsTrees

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.

Try coding it in LeetCode

Map Sum Pairs

Medium

Asked by:

4 views

Topics:

StringsTrees

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.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of possible values for the keys and values being inserted?
Are we expected to handle empty strings or null values as keys?
If a key is inserted multiple times with different values, should the sum include all previous values, or only the most recent one?
How should the `sum` function behave if the prefix is empty or not present in the map?
Are there any memory constraints I should consider given the potentially large number of key-value pairs?

Brute Force Solution

Approach

The brute force method for the map sum problem means we will recalculate the sum for a given prefix every time. It involves going through all the previously stored words to find matches for the provided prefix and add their corresponding values.

Here's how the algorithm would work step-by-step:

When we need to calculate the sum for a given prefix, we will start from scratch.
We'll check every single word that we've previously stored to see if the provided prefix exists at the beginning of the word.
If the prefix matches the beginning of a stored word, we'll add the value associated with that word to a running total.
Once we've gone through all the stored words, the final total will be our answer, which we return.

Code Implementation

class MapSum:

def __init__(self):
        self.word_value_map = {}

def insert(self, key, value):
        self.word_value_map[key] = value

def sum(self, prefix):
        total_sum = 0

        # Iterate through each word in our storage to check for matches
        for word, value in self.word_value_map.items():

            # See if the current word starts with the given prefix
            if word.startswith(prefix):

                # Add the word's value to the running total if it matches
                total_sum += value

        return total_sum

Big(O) Analysis

Time Complexity

O(n*m) – The 'insert' operation takes O(1) time, as it only involves storing the key-value pair in a hash map. The 'sum' operation iterates through all previously inserted keys (n, where n is the number of keys inserted) and checks if each key starts with the given prefix. Checking if a key of length m starts with the prefix takes O(m) time, where m is the average length of the stored strings. Therefore, the overall time complexity of the 'sum' operation is O(n*m).

Space Complexity

O(1) – The described brute force method iterates through previously stored words to calculate the sum for a given prefix. It doesn't create any additional data structures whose size depends on the number of stored words (let N be the number of stored words). The only auxiliary space used is for a few variables, like a running total, that store intermediate calculations. The space used by these variables is constant, irrespective of N. Therefore, the space complexity is O(1).

Optimal Solution

Approach

The problem requires us to store word-value pairs and calculate sums based on prefixes. Instead of recomputing sums every time, we can use a data structure that allows us to efficiently update and retrieve information about prefixes.

Here's how the algorithm would work step-by-step:

Use a dictionary to store each word and its corresponding value.
Use a second dictionary to store the prefix sums. This dictionary will hold prefixes as keys and the accumulated value sums as values.
When a new word with a value is inserted, update the word-value dictionary first.
Then, calculate prefix sums by iterating through the prefixes of the newly inserted word.
For each prefix, update the prefix sum dictionary by adding the new value to the current value of that prefix (or creating a new entry if the prefix isn't present).
To compute the sum of values for words starting with a given prefix, simply return the value stored for that prefix in the prefix sum dictionary.

Code Implementation

class MapSum:

    def __init__(self):
        self.word_values = {}
        self.prefix_sums = {}

    def insert(self, word: str, value: int) -> None:
        original_value = self.word_values.get(word, 0)
        value_difference = value - original_value

        self.word_values[word] = value

        # Iterate over all prefixes of the word.
        for i in range(1, len(word) + 1):
            prefix = word[:i]

            # Update prefix sums, adding the value difference.
            if prefix in self.prefix_sums:
                self.prefix_sums[prefix] += value_difference
            else:
                self.prefix_sums[prefix] = value_difference

    def sum(self, prefix: str) -> int:
        # Return the stored sum for the given prefix.
        return self.prefix_sums.get(prefix, 0)

# Your MapSum object will be instantiated and called as such:
# mapSum = MapSum()
# mapSum.insert("apple", 3)
# mapSum.sum("ap")
# mapSum.insert("app", 2)
# mapSum.sum("ap")

Big(O) Analysis

Time Complexity

O(k*l) – The insert operation iterates through the prefixes of the inserted word. Let 'l' be the length of the word being inserted. For each prefix, it updates the prefix sum dictionary. Therefore, inserting a single word takes O(l) time. The sum operation retrieves the value associated with the given prefix from the prefix sum dictionary, which takes O(1) time assuming the dictionary provides constant time lookups on average using a hash table. Assuming 'k' insert operations with potentially different average word length of 'l', in the worst-case the time complexity is O(k*l).

Space Complexity

O(N*L) – The solution uses two dictionaries. The first dictionary stores word-value pairs, potentially storing all N input words. The second dictionary stores prefix sums. In the worst case, each word could have a unique prefix of every length up to the word's length, L. So, in the prefix sum dictionary, we may store all possible prefixes for all words. Therefore, the prefix sum dictionary can store prefixes of length up to L for each of the N words, resulting in a space complexity of O(N*L), where L is the average length of the input strings and N is the number of words.

Edge Cases

Case	How to Handle
Empty key	Empty key should be allowed and treated as a valid, potentially meaningful prefix.
Null input for key	Throw an IllegalArgumentException or return a specific error code, as null keys are generally disallowed.
Very long key string (close to max string length)	Ensure solution does not cause StackOverflowError or excessive memory allocation during string operations or trie traversal.
Large number of inserts with the same key overwriting the value	Overwriting existing keys should work correctly, and the sum should reflect the last inserted value.
Large number of keys with common prefixes, stressing trie memory usage	The solution's memory usage should scale reasonably, ideally linearly, with the number of distinct prefixes rather than the total length of all keys.
Integer overflow when calculating the sum of values	Use a data type with a larger range (e.g., long) for storing sums to prevent integer overflow.
Non-ASCII characters in the key string	The Trie implementation needs to handle Unicode characters correctly (consider using a wider character type or appropriate encoding).
Prefix that does not exist in the map	The sum operation should return 0 when the prefix does not exist in the map.

Empty key

How to Handle:

Empty key should be allowed and treated as a valid, potentially meaningful prefix.

Null input for key

How to Handle:

Throw an IllegalArgumentException or return a specific error code, as null keys are generally disallowed.

Very long key string (close to max string length)

How to Handle:

Ensure solution does not cause StackOverflowError or excessive memory allocation during string operations or trie traversal.

Large number of inserts with the same key overwriting the value

How to Handle:

Overwriting existing keys should work correctly, and the sum should reflect the last inserted value.

Large number of keys with common prefixes, stressing trie memory usage

How to Handle:

The solution's memory usage should scale reasonably, ideally linearly, with the number of distinct prefixes rather than the total length of all keys.

Integer overflow when calculating the sum of values

How to Handle:

Use a data type with a larger range (e.g., long) for storing sums to prevent integer overflow.

Non-ASCII characters in the key string

How to Handle:

The Trie implementation needs to handle Unicode characters correctly (consider using a wider character type or appropriate encoding).

Prefix that does not exist in the map

How to Handle:

The sum operation should return 0 when the prefix does not exist in the map.