Two Sum III - Data structure design

Two Sum III - Data structure design - Problem

Array Hash Table Two Pointers Design Data Stream Easy

Design a data structure that accepts a stream of integers and checks if it has a pair of integers that sum up to a particular value.

Implement the TwoSum class:

TwoSum() - Initializes the TwoSum object, with an empty array initially.
void add(int number) - Adds number to the data structure.
boolean find(int value) - Returns true if there exists any pair of numbers whose sum is equal to value, otherwise returns false.

Input & Output

Example 1 — Basic Operations

$ Input: operations = ["TwoSum", "add", "add", "add", "find", "find"], values = [null, 1, 3, 2, 4, 7]

› Output: [null, null, null, null, true, false]

💡 Note: Initialize TwoSum, add numbers 1, 3, 2. find(4) returns true because 1+3=4. find(7) returns false because no pair sums to 7.

Example 2 — Duplicate Numbers

$ Input: operations = ["TwoSum", "add", "add", "find"], values = [null, 2, 2, 4]

› Output: [null, null, null, true]

💡 Note: Add 2 twice, then find(4) returns true because 2+2=4. Need to handle duplicate numbers correctly.

Example 3 — No Valid Pair

$ Input: operations = ["TwoSum", "add", "add", "find"], values = [null, 1, 5, 3]

› Output: [null, null, null, false]

💡 Note: Add 1 and 5, then find(3) returns false because no combination of 1 and 5 equals 3.

Constraints

-10⁵ ≤ number ≤ 10⁵
-2×10⁵ ≤ value ≤ 2×10⁵
At most 10⁴ calls to add and find

Visualization

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Asked in

Li LinkedIn 15 A Amazon 12 G Google 8 M Microsoft 6

The key insight is to use a hash map to store numbers with their frequencies, enabling O(1) complement lookups during find operations. The optimal hash map approach achieves O(1) add and O(n) find operations. Time: O(n) for find, Space: O(n) for storage.

Common Approaches

✓ Brute Force with Array

⏱️ Time: O(n²) Space: O(n)

Maintain a simple array of numbers. For each find operation, use nested loops to check all possible pairs. This is straightforward but becomes slow as more numbers are added.

Hash Map Optimization

⏱️ Time: O(n) Space: O(n)

Store numbers in a hash map with their frequencies. For each find operation, iterate through unique numbers and check if their complement exists. Handle the case where a number needs to pair with itself.

Brute Force with Array — Algorithm Steps

Store each added number in an array
For find, check every pair (i,j) where i < j
Return true if any pair sums to target

Visualization

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Step-by-Step Walkthrough

Store Numbers

Add numbers to array as they come

Check Pairs

For find, check all i,j combinations

Return Result

True if any pair sums to target

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>

typedef struct {
    int* numbers;
    int size;
    int capacity;
} TwoSum;

TwoSum* twoSumCreate() {
    TwoSum* obj = (TwoSum*)malloc(sizeof(TwoSum));
    obj->capacity = 10;
    obj->numbers = (int*)malloc(obj->capacity * sizeof(int));
    obj->size = 0;
    return obj;
}

void twoSumAdd(TwoSum* obj, int number) {
    if (obj->size == obj->capacity) {
        obj->capacity *= 2;
        obj->numbers = (int*)realloc(obj->numbers, obj->capacity * sizeof(int));
    }
    obj->numbers[obj->size++] = number;
}

bool twoSumFind(TwoSum* obj, int value) {
    int n = obj->size;
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            if (obj->numbers[i] + obj->numbers[j] == value) {
                return true;
            }
        }
    }
    return false;
}

void parseOperations(const char* str, char operations[][20], int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',' || *p == '"') p++;
        if (*p == ']' || *p == '\0') break;
        
        int i = 0;
        while (*p && *p != '"' && *p != ',' && *p != ']') {
            operations[*size][i++] = *p++;
        }
        operations[*size][i] = '\0';
        (*size)++;
        
        while (*p && *p != ',' && *p != ']') p++;
    }
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        
        if (strncmp(p, "null", 4) == 0) {
            arr[(*size)++] = 0; // null represented as 0
            p += 4;
        } else {
            arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
        }
    }
}

int main() {
    char line1[1000], line2[1000];
    fgets(line1, sizeof(line1), stdin);
    fgets(line2, sizeof(line2), stdin);
    
    char operations[100][20];
    int values[100];
    int opSize, valSize;
    
    parseOperations(line1, operations, &opSize);
    parseArray(line2, values, &valSize);
    
    TwoSum* obj = NULL;
    printf("[");
    
    for (int i = 0; i < opSize; i++) {
        if (i > 0) printf(",");
        
        if (strcmp(operations[i], "TwoSum") == 0) {
            obj = twoSumCreate();
            printf("null");
        } else if (strcmp(operations[i], "add") == 0) {
            twoSumAdd(obj, values[i]);
            printf("null");
        } else if (strcmp(operations[i], "find") == 0) {
            bool result = twoSumFind(obj, values[i]);
            printf("%s", result ? "true" : "false");
        }
    }
    
    printf("]\n");
    
    if (obj) {
        free(obj->numbers);
        free(obj);
    }
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Each find operation requires checking all n×(n-1)/2 pairs with nested loops

⚠ Quadratic Growth

Space Complexity

O(n)

Array stores up to n numbers

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force with Array — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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