Problem Description

In this problem, we have an array of integers called nums and a target integer called target. Our task is to find two distinct numbers within the array that when added together, equal the target. One important rule is that we cannot use the same element from the array twice in our sum. The output will be the indices (positions in the array) of these two numbers, and it does not matter in which order we provide these indices. Each valid input to the problem is guaranteed to have exactly one solution.

Intuition

To solve this problem efficiently, we avoid the brute force approach of trying every possible pair of numbers to see if they add up to the target. Instead, we employ a hash table (dictionary in Python), which offers fast lookup times that average out to O(1) complexity. The basic idea is to iterate through the list once, and for each number, we check if the number required to reach the target (target - current_number) has already been seen in the array. If it has, then we have found the two numbers whose sum is equal to the target.

As we iterate through the nums array, we keep track of each number and its index in the hash table. So for each number x, we calculate y which is the difference between target and x. If y is found in the hash table, it means we have previously processed another number such that x + y = target. In that case, we immediately return the pair of indices: one for the current number x, and one for the previously stored number y. If y is not found in the hash table, we store x along with its index in the hash table and move on to the next element in the array. This approach only requires a single pass through the array, making the solution efficient and elegant.

Solution Approach

The solution provided uses a hash table to map each element's value to its index in the array. This allows for constant-time look-ups which are critical for efficiency. The algorithm proceeds as follows:

1. Initialize an empty hash table (dictionary in Python dialect), we'll call it m.
2. Iterate over the nums array, enumerating both the value x and its index i. Enumeration provides a convenient way of getting both the value and the index without additional overhead.
3. For every value x, calculate its complement y by subtracting x from target (y = target - x).
4. Check if y is present as a key in the hash table. If it is found, it means we had already seen the necessary pair earlier in the array. We then retrieve m[y], which is the index of y we had stored, and return a list containing the indices of y and x ([m[y], i]). This satisfies the requirement as their sum is equal to the target.
5. If y is not in the hash table, add the current value x along with its index i to the hash table (m[x] = i). This stores x for future reference if we later come across its complement y.

By only traversing the array once, the overall time complexity is O(n), where n is the number of elements in nums. The space complexity is also O(n) as in the worst case, we could potentially store all elements in the hash table before finding a match.

Here's the provided solution approach step in the code:

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        m = {}  # Step 1: Initialize the hash table
        for i, x in enumerate(nums):  # Step 2: Enumerate through nums
            y = target - x  # Step 3: Calculate complement y
            if y in m:  # Step 4: Check if y is present in the hash table
                return [m[y], i]  # Return the indices of the two elements adding up to target
            m[x] = i  # Step 5: Store x with its index in hash table for future reference

The strength of this approach is that it smartly utilizes the hash table to avoid nested loops and thus reducing the time complexity. The algorithm is linear time as it eliminates the need to examine every possible pair by keeping a record of what is needed to reach the target with the current number.

Example Walkthrough

Let's work through a small example to illustrate the solution approach.

Suppose our input array nums is [2, 7, 11, 15] and the target is 9.

Following the solution steps:

1. Initialize an empty hash table (dictionary), which we'll name m.

2. Now we start iterating over the nums array:

   • First iteration (i = 0):

     • We take the first element nums[0] which is 2.
     • Calculate its complement y = target - nums[0], which gives us y = 9 - 2 = 7.
     • Check if 7 is present in the hash table m. It isn't, since m is empty.
     • Store the current value 2 along with its index 0 in the hash table: m[2] = 0.

   • Second iteration (i = 1):

     • Take the next element nums[1], which is 7.
     • Calculate its complement y = target - nums[1], which gives us y = 9 - 7 = 2.
     • Check if 2 is in the hash table m. Yes, it is! The complement 2 was stored during the first iteration.
     • Since the complement is found, we retrieve m[2] which is 0, the index of the complement 2. This gives us the indices [0, 1].

The sum of the numbers at these indices (nums[0] + nums[1]) equals the target (2 + 7 = 9). As expected, the original problem statement guarantees that there is exactly one solution, so the problem is now solved with the output [0, 1].

By using this hash table approach, we efficiently found the two numbers that add up to the target in a single pass through the array, thereby using O(n) time complexity instead of O(n^2) which would result from using a brute force approach with nested loops.

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(n), where n is the length of the array nums. This is because the code loops through each element in nums exactly once, and each operation within the loop—including checking if an element exists in the map m and adding an element to m—is O(1).

The space complexity of the code is also O(n), since in the worst case, the code may insert each element of the array nums into the map m. Therefore, the space used by the map m grows linearly with the number of elements in nums.

Learn more about how to find time and space complexity quickly using problem constraints.