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LESSON 257 OF 409 · HASH TABLE

Two Sum IV - O(n) Time Complexity in C++

Practice O(n) Read the full explanation

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input will have at most one solution, and you may not use the same index twice.

In case no solution exists, return [-1, -1]

Example:

Input: nums = [2, 7, 11, 15], target = 9
Output: [0, 1]
Explanation: nums[0] + nums[1] = 2 + 7 = 9

Note:

Your algorithm should run in O(n) time and use O(n) space.


Hints:

This the type of problem where we optimize the brute force approach to get the optimal one. The brute force approach took O(n^2) time because we had two nested for loops, one for i and another one j. In order to get O(n), we have to reduce this to only one for loop, so we should fix only one index, let it be j, the index of the second number. We will try out each possible j, and for each of them we should check if there is an index i < j such that nums[i] + nums[j] == target: 
for each j : 0 -> n - 1:
	// We know nums[j]. Find some index i < j such that nums[i] + nums[j] == target.
How can we find that index in O(1)?

nums[i] + nums[j] == target ⇔ nums[i] == target - nums[j]. So for each j, we should look for an element equal to target - nums[j] from the left of j. Is there a data structure that allows us to look for an element in a collection in constant time? How can we use it here?

That data structure is the Hash Map. We can have a Hash Map where we insert all the elements nums[j] we iterate on during the for loop. In this way, for each j, we can look in the Hash Map to see if there is some element of value target - nums[j] and if so, we found a solution: 
for each j : 0 -> n - 1:
	- Look for target - nums[j] in the Hash Map. If there exists, we found the solution.
	- Otherwise, insert nums[j] into the Hash Map and continue with the for loop.
I know the code above is not perfect. Hash Maps work with pairs (key => value), we cannot just insert a nums[j] into a Hash Map. What should we actually insert in this hash map? What are the keys and what are the values?

The problem with the code above is that we just check if there is an element equal to target - nums[j]. If so, we have a solution, but that's not enough. The problem asks us to return the indices of those elements and we only know of them: j. Can we adapt the solution such that if for some j there is an element equal to target - nums[j], we can also get its index?

The pairs in the Hash Map should be nums[j] => j so that for each element we insert, we can also get its index: 
for each j : 0 -> n - 1:
	if target - nums[j] in indexOfValue:
		return [indexOfValue[target - nums[j]], j]
	indexOfValue[nums[j]] = j
return [-1, -1]

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    C++ · HASH TABLE

    Two Sum in C++, explained

    Understanding the Problem

    The core challenge of this problem is to find two indices in an array such that the sum of the elements at these indices equals a given target. This problem is significant in various applications, such as financial transactions, where you need to find pairs of transactions that sum up to a specific amount.

    Potential pitfalls include assuming multiple solutions or using the same index twice, which the problem explicitly forbids.

    Approach

    To solve this problem, we can start with a naive approach and then optimize it:

    Naive Approach

    The naive approach involves using two nested loops to check all possible pairs of indices. This approach has a time complexity of O(n^2), which is not optimal for large arrays.

    Optimized Approach

    To achieve O(n) time complexity, we can use a hash map (unordered_map in C++). The idea is to iterate through the array and for each element, check if the complement (target - current element) exists in the hash map. If it does, we have found our solution. If not, we add the current element and its index to the hash map.

    Algorithm

    Here is a step-by-step breakdown of the optimized algorithm:

    1. Initialize an empty hash map.
    2. Iterate through the array.
    3. For each element, calculate its complement (target - current element).
    4. Check if the complement exists in the hash map.
    5. If it exists, return the indices of the complement and the current element.
    6. If it does not exist, add the current element and its index to the hash map.
    7. If no solution is found by the end of the iteration, return [-1, -1].

    Code Implementation

    
#include <iostream>
#include <unordered_map>
#include <vector>

std::vector<int> twoSum(const std::vector<int>& nums, int target) {
    // Create a hash map to store the value and its index
    std::unordered_map<int, int> indexOfValue;
    
    // Iterate through the array
    for (int j = 0; j < nums.size(); ++j) {
        int complement = target - nums[j];
        
        // Check if the complement exists in the hash map
        if (indexOfValue.find(complement) != indexOfValue.end()) {
            // If found, return the indices
            return {indexOfValue[complement], j};
        }
        
        // Otherwise, add the current element and its index to the hash map
        indexOfValue[nums[j]] = j;
    }
    
    // If no solution is found, return [-1, -1]
    return {-1, -1};
}

int main() {
    std::vector<int> nums = {2, 7, 11, 15};
    int target = 9;
    std::vector<int> result = twoSum(nums, target);
    
    std::cout << "[" << result[0] << ", " << result[1] << "]" << std::endl;
    return 0;
}

    Complexity Analysis

    The time complexity of this approach is O(n) because we only iterate through the array once. The space complexity is also O(n) due to the hash map storing up to n elements.

    Edge Cases

    Potential edge cases include:

    • An empty array: The function should return [-1, -1].
    • An array with no valid pairs: The function should return [-1, -1].
    • Negative numbers and zero: The function should handle these correctly.

    Testing

    To test the solution comprehensively, consider the following test cases:

    1. nums = [2, 7, 11, 15], target = 9 -> Output: [0, 1]
2. nums = [3, 2, 4], target = 6 -> Output: [1, 2]
3. nums = [3, 3], target = 6 -> Output: [0, 1]
4. nums = [], target = 0 -> Output: [-1, -1]
5. nums = [1, 2, 3], target = 7 -> Output: [-1, -1]

    Thinking and Problem-Solving Tips

    When approaching such problems, consider the following tips:

    • Understand the problem constraints and requirements thoroughly.
    • Start with a brute force solution to understand the problem better.
    • Think about data structures that can optimize your solution.
    • Practice similar problems to improve your problem-solving skills.

    Conclusion

    In this blog post, we discussed the Two Sum problem, its significance, and various approaches to solve it. We provided a detailed explanation of the optimized solution using a hash map and analyzed its complexity. Understanding and solving such problems is crucial for improving algorithmic thinking and coding skills.

    Additional Resources

    For further reading and practice, consider the following resources:

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