
Problem Statement
In this scenario, we are tasked with designing a data structure named TwoSum
that primarily functions to handle a continuous stream of integers. The primary role of this data structure involves two major activities:
- Adding an integer to the collection.
- Checking if any two distinct integers within the collection sum up to a specified value.
The TwoSum
class provides two key methods:
- add(int number): This method is responsible for integrating a new integer into the
TwoSum
data structure. - find(int value): This method determines whether there are any two integers present in the data structure that, when added together, match the given summation target (value). The output is a boolean indicating the presence (true) or absence (false) of such a pair.
The lifecycle of the TwoSum
data structure begins when it's initialized, starting with an empty internal structure. As integers are added, this structure grows, thereby gradually increasing the complexity of the task carried out by the find
method.
Examples
Example 1
Input:
["TwoSum", "add", "add", "add", "find", "find"]
Output:
[null, null, null, null, true, false]
Explanation:
TwoSum twoSum = new TwoSum(); twoSum.add(1); // [] --> [1] twoSum.add(3); // [1] --> [1, 3] twoSum.add(5); // [1, 3] --> [1, 3, 5] twoSum.find(4); // 1 + 3 = 4, return true twoSum.find(7); // No two integers sum up to 7, return false
Constraints
-10⁵ <= number <= 10⁵
-2³¹ <= value <= 2³¹ - 1
- At most
10⁴
calls will be made toadd
andfind
.
Approach and Intuition
The key tasks for the TwoSum
class are managing the stream of integers efficiently and allowing for a rapid check to determine if a pair with the required sum exists. The premise here is based on managing trade-offs between:
- Time complexity of adding an integer.
- Time complexity of finding if a particular sum exists.
From the operations described, here's a breakdown of each:
Implementing the
add
function: When an integer is added, it needs to be stored in a manner that later facilitates quick look-ups for thefind
operation. A logical approach might involve using a hash table (dictionary in Python) where each key represents the integer and the value represents its frequency in the data structure. This enables efficient constant time O(1) addition and retrieval operations.Implementing the
find
function: To find two numbers that sum to a specified value efficiently, a single pass through keys of our hash table is utilized. For every key in the hash table:- Compute what the second number needs to be (i.e.,
target - current number
). - Check if this "needed number" exists in our hash structure.
- If it does, and it’s different from the current number, return true.
- If it’s the same number, ensure its count is at least two before returning true.
- Compute what the second number needs to be (i.e.,
This process typically involves a time complexity of O(N), where N is the number of distinct entries in our data structure.
Special attention must be given to cases where the same number is needed twice to form the target sum. For example, for a target sum of 2
and current number 1
, we must ensure that 1
appears at least twice.
By using hash maps, we leverage the average-case constant time complexity for addition and checking operations, ideally suited given the constraints of high-frequency operations. The design balances quick insertion with efficient querying.
Solutions
- C++
class PairSum {
private:
unordered_map<long long, int> counts;
public:
PairSum() {}
void addNumber(int num) {
if (counts.find(num) != counts.end())
counts[num]++;
else
counts[num] = 1;
}
bool check(int target) {
for (const auto &item : counts) {
long long needed = static_cast<long long>(target) - item.first;
if (needed != item.first) {
if (counts.find(needed) != counts.end())
return true;
} else {
if (item.second > 1) return true;
}
}
return false;
}
};
The problem involves designing a data structure that supports adding numbers and can verify if a pair with a given sum exists within these added numbers, using C++.
- Data Structure Used:
unordered_map<long long, int> counts
. This hashmap stores numbers as keys and their frequencies as values.
Methods in the Data Structure:
addNumber(int num):
- Updates the hashmap by incrementing the count of the number added. If the number is not in the map, it initializes its count to 1.
check(int target):
- Iterates through each item in the hashmap to determine if a pair summing up to the target exists.
- For each number, it calculates the required complement by subtracting the current number from the target.
- Checks two conditions:
- If the complement is different from the number, checks if the complement exists in the map.
- If the complement is the same as the number, checks if this number appears at least twice.
- Returns
true
if any of the above conditions are satisfied, otherwisefalse
after all checks.
This structure efficiently adds numbers and checks for the existence of two sum pairs by optimizing search operations through the use of a hashmap, leveraging the average O(1) time complexity for insertions and lookups.
- Java
import java.util.HashMap;
class PairSum {
private HashMap<Integer, Integer> countMap;
public PairSum() {
this.countMap = new HashMap<>();
}
public void addElement(int num) {
if (this.countMap.containsKey(num)) this.countMap.put(
num,
this.countMap.get(num) + 1
);
else this.countMap.put(num, 1);
}
public boolean checkPair(int target) {
for (HashMap.Entry<Integer, Integer> element : this.countMap.entrySet()) {
int needed = target - element.getKey();
if (needed != element.getKey()) {
if (this.countMap.containsKey(needed)) return true;
} else {
if (element.getValue() > 1) return true;
}
}
return false;
}
}
The provided Java code implements a data structure called PairSum
designed to facilitate efficient addition of numbers and then allow checking if there exist any two unique or identical numbers which sum to a given target value. This task solution involves utilizing a HashMap
to store each number added and their respective count of occurrences.
Initialization:
- The
PairSum
class features a constructor that initializes aHashMap
calledcountMap
, storing numbers as keys and their occurrences as values.
- The
Add Element:
- The
addElement
method takes an integernum
as an argument. This method checks if the number already exists incountMap
. If it does, it increments its count; otherwise, it adds the number to the map with a count of one.
- The
Check Pair:
- The
checkPair
method determines whether there exists a pair of numbers incountMap
that adds up to a specifiedtarget
value. - For each number in the
countMap
, it calculates the needed complementary number to meet the target sum by subtracting the current number from the target. - It checks two conditions:
- If the complementary number is different from the current number and exists in
countMap
, it returnstrue
indicating a successful pair match. - If the complementary number is the same as the current number, it checks if the count is greater than one; if so, it also returns
true
.
- If the complementary number is different from the current number and exists in
- If no such conditions meet for any number, the method returns
false
, indicating no pair exists that sums to the target.
- The
This structure makes it particularly efficient for operations where frequent additions and pair-checking are required, such as real-time data analysis or computational finance scenarios preferring quick response over bulky operations.
- C
typedef struct {
int id;
int occurrences;
UT_hash_handle hashHandler;
} ElementFrequency;
typedef struct {
ElementFrequency* elementsFrequency;
} ArrayElementCounter;
ArrayElementCounter* createCounter() {
ArrayElementCounter* counter = (ArrayElementCounter*)malloc(sizeof(ArrayElementCounter));
counter->elementsFrequency = NULL;
return counter;
}
void addElement(ArrayElementCounter* counter, int element) {
ElementFrequency* elementFreq;
HASH_FIND_INT(counter->elementsFrequency, &element, elementFreq);
if (elementFreq == NULL) {
elementFreq = (ElementFrequency*)malloc(sizeof(ElementFrequency));
elementFreq->id = element;
elementFreq->occurrences = 0;
HASH_ADD_INT(counter->elementsFrequency, id, elementFreq);
}
elementFreq->occurrences++;
}
bool findIfPossibleSum(ArrayElementCounter* counter, int sum) {
for (ElementFrequency* current = counter->elementsFrequency; current != NULL; current = current->hashHandler.next) {
long required = (long)sum - current->id;
if (required != (int)required) continue;
ElementFrequency* foundElement;
HASH_FIND_INT(counter->elementsFrequency, &required, foundElement);
if (foundElement != NULL) {
if (foundElement->id == current->id && current->occurrences > 1) {
return true;
} else if (foundElement->id != current->id) {
return true;
}
}
}
return false;
}
void freeCounter(ArrayElementCounter* counter) {
ElementFrequency *current, *temp;
HASH_ITER(hashHandler, counter->elementsFrequency, current, temp) {
HASH_DEL(counter->elementsFrequency, current);
free(current);
}
free(counter);
}
The code provided focuses on designing a data structure to efficiently manage and check sums of pairs in a dynamic set of integers using the C programming language. The main features of this implementation involve creating, adding elements to, finding possible sums within, and freeing a custom data structure.
The data structure is made up of two main structs:
ElementFrequency
: This stores each unique integer and its occurrence count in the dataset.ArrayElementCounter
: This struct acts as a container that holds a hash table ofElementFrequency
elements.
The four main functions described in the code operate as follows:
createCounter
: Initializes anArrayElementCounter
. This function dynamically allocates memory for anArrayElementCounter
and sets its internal hash table to NULL, readying it for elements to be added.addElement
: Adds an integer to theArrayElementCounter
. If the integer is new, it is added to the hash table with an occurrence count initialized to 0, which is then incremented. For existing integers, the occurrence count is simply incremented.findIfPossibleSum
: Determines if there exists any two distinct elements within the hash table that sum up to a specified value. This function uses the HASH_FIND_INT macro to search for the required complementary element that would form the desired sum with the current element. If a suitable pair is found (considering edge cases where the two elements might be the same), it returns true; otherwise, it continues searching until the hash table is exhausted.freeCounter
: Properly releases all memory allocated for theArrayElementCounter
and its elements. It iterates through the hash table, removing and freeing each element before finally freeing theArrayElementCounter
itself.
This solution is efficient in terms of both time and space complexities, making good use of the uthash library to handle the underlying hash table operations, such as adding and searching for elements.
- JavaScript
const NumberPair = function () {
this.counter = {};
};
NumberPair.prototype.addNumber = function (num) {
if (num in this.counter) this.counter[num]++;
else this.counter[num] = 1;
};
NumberPair.prototype.checkPairSum = function (targetSum) {
for (let x in this.counter) {
let y = targetSum - x;
if (y != x) {
if (y in this.counter) return true;
} else {
if (this.counter[x] > 1) return true;
}
}
return false;
};
This article provides a solution summary for a problem that involves designing a data structure to efficiently add numbers and check if there exists any pair of numbers that sum up to a specified target. The implementation is in JavaScript.
The data structure, NumberPair
, includes two major functionalities:
- Add numbers to the data structure.
- Check if there exists a pair of numbers within the structure whose sum equals a specified target.
The implementation involves the following steps:
- Initialize
NumberPair
with acounter
object to keep track of numbers and their respective counts. - Implement the
addNumber
method, which adds a number to thecounter
. If the number is already present, increase its count by one; otherwise, set its count to one. - Implement the
checkPairSum
method:- Iterate through each number,
x
, in thecounter
. - Calculate the complementary number
y
astargetSum - x
. - For pairs
(x, y)
wherex
is not equal toy
, check ify
exists in thecounter
. If so, returntrue
. - For pairs where
x
equalsy
, ensure there are at least two occurrences ofx
in thecounter
to form a valid pair. If so, returntrue
.
- Iterate through each number,
- If no valid pair is found by the end of the iteration, return
false
.
Features of this implementation include:
- Efficient addition of numbers.
- Quick pair sum checking using a hash table for constant-time look-ups.
- Handling duplicates by tracking the frequency of each number.
- Python
class PairSum(object):
def __init__(self) -> None:
self.value_counter = {}
def add_value(self, number: int) -> None:
if number in self.value_counter:
self.value_counter[number] += 1
else:
self.value_counter[number] = 1
def check_pair(self, target: int) -> bool:
for key in self.value_counter:
complement = target - key
if key != complement:
if complement in self.value_counter:
return True
elif self.value_counter[key] > 1:
return True
return False
The problem titled "Two Sum III - Data structure design" involves creating a data structure that supports adding values and checking if any pair of numbers sum to a given target. The provided Python implementation addresses this challenge with the following functionality:
Initialization: An instance of
PairSum
class initializes with a dictionaryvalue_counter
to keep track of occurrences of each value added.Adding a Value:
- Call
add_value
with an integer. - Check if this number is already in
value_counter
. - If present, increment its count.
- If not, insert it into the dictionary with a count of one.
- Call
Checking for a Pair:
- Call
check_pair
with a target sum integer. - Iterate through the keys (unique numbers) in
value_counter
. - Calculate the complement of each key as the difference between the target and the current key.
- If the complement differs from the key, check if the complement exists in
value_counter
. - If the same number is needed twice (e.g., 5 + 5 = 10), then check if its count in
value_counter
is greater than one. - If any such pair exists that satisfies the condition, return
True
. - If no suitable pair is found after iterating through all the numbers, return
False
.
- Call
Output:
- The method
add_value
enhances the data structure dynamically with each added value ensuring quick lookup times. - The
check_pair
function efficiently checks the possibility of achieving the target sum, making optimal use of thevalue_counter
for quick access and verification.
- The method
This implementation efficiently supports the operations necessary for solving the "Two Sum III" problem using a hash map to store and verify the existence of number pairs that sum to a specified target.
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