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LESSON 258 OF 410 · HASH TABLE

Two Sum IV in O(n) Time Complexity using Python

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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    PYTHON · HASH TABLE

    Two Sum in Python, explained

    Problem Definition

    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.


    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 the target value. 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 value.

    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 solution and then optimize it:

    Naive Solution

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

    Optimized Solution

    To achieve an O(n) time complexity, we can use a hash map (dictionary in Python) to store the elements and their indices as we iterate through the array. This allows us to check in constant time whether the complement of the current element (i.e., target - current element) exists in the hash map.

    Algorithm

    def two_sum(nums, target):
    # Create a dictionary to store the value and its index
    index_map = {}
    
    # Iterate through the list
    for i, num in enumerate(nums):
        # Calculate the complement
        complement = target - num
        
        # Check if the complement exists in the dictionary
        if complement in index_map:
            # If found, return the indices
            return [index_map[complement], i]
        
        # Otherwise, add the current number and its index to the dictionary
        index_map[num] = i
    
    # If no solution is found, return [-1, -1]
    return [-1, -1]

    Code Implementation

    def two_sum(nums, target):
    # Create a dictionary to store the value and its index
    index_map = {}
    
    # Iterate through the list
    for i, num in enumerate(nums):
        # Calculate the complement
        complement = target - num
        
        # Check if the complement exists in the dictionary
        if complement in index_map:
            # If found, return the indices
            return [index_map[complement], i]
        
        # Otherwise, add the current number and its index to the dictionary
        index_map[num] = i
    
    # If no solution is found, return [-1, -1]
    return [-1, -1]

# Example usage
nums = [2, 7, 11, 15]
target = 9
print(two_sum(nums, target))  # Output: [0, 1]

    Complexity Analysis

    The time complexity of this solution is O(n) because we 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 one element: The function should return [-1, -1].
    • No two elements sum up to the target: The function should return [-1, -1].

    Examples:

    print(two_sum([], 9))  # Output: [-1, -1]
print(two_sum([1], 9))  # Output: [-1, -1]
print(two_sum([1, 2, 3], 7))  # Output: [-1, -1]

    Testing

    To test the solution comprehensively, include a variety of test cases:

    • Simple cases with small arrays.
    • Edge cases with empty or single-element arrays.
    • Cases where no solution exists.
    • Cases with negative numbers and zero.

    Example test cases:

    def test_two_sum():
    assert two_sum([2, 7, 11, 15], 9) == [0, 1]
    assert two_sum([3, 2, 4], 6) == [1, 2]
    assert two_sum([3, 3], 6) == [0, 1]
    assert two_sum([], 9) == [-1, -1]
    assert two_sum([1], 9) == [-1, -1]
    assert two_sum([1, 2, 3], 7) == [-1, -1]
    assert two_sum([-1, -2, -3, -4, -5], -8) == [2, 4]
    print("All test cases pass")

test_two_sum()

    Thinking and Problem-Solving Tips

    When approaching such problems:

    • Understand the problem requirements and constraints.
    • Start with a brute force solution to understand the basic logic.
    • Look for ways to optimize the solution using data structures like hash maps.
    • Consider edge cases and test your solution thoroughly.

    Conclusion

    In this blog post, we discussed the Two Sum problem, explored a naive solution, and then optimized it using a hash map to achieve O(n) time complexity. Understanding and solving such problems is crucial for developing strong problem-solving skills in programming.

    Practice is key to mastering these concepts, so keep solving similar problems and exploring different algorithms.

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