How do you solve Two Sum II - Input Array Is Sorted on LeetCode?

Two Sum II - Input Array Is Sorted (LeetCode #167) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The problem can be solved efficiently using the two-pointer technique due to the sorted nature of the input array.

What is the brute force solution for Two Sum II - Input Array Is Sorted?

The brute force approach for Two Sum II - Input Array Is Sorted is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks every possible pair of numbers to see if they sum up to the target. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Two Sum II - Input Array Is Sorted?

Two Sum II - Input Array Is Sorted uses the following concepts: Array, Two Pointers, Binary Search. The recommended patterns to study are: Two Pointers, Array.

What are the interview tips for Two Sum II - Input Array Is Sorted?

Always consider the properties of the input data, such as whether it is sorted. Explain your thought process clearly while coding; it helps the interviewer understand your approach. Practice optimizing your initial brute-force solutions to demonstrate your problem-solving skills.

What are common mistakes when solving Two Sum II - Input Array Is Sorted?

Not recognizing that the input array is sorted, leading to inefficient solutions. Failing to return the correct indices (1-indexed) as specified in the problem.

#167

Two Sum II - Input Array Is Sorted

Medium

ArrayTwo PointersBinary SearchTwo PointersArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

Using the two-pointer technique takes advantage of the sorted nature of the array. We can efficiently find the two numbers by adjusting pointers based on their sum compared to the target.

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Algorithm

3 steps

1Step 1: Initialize two pointers, left at the start (index 0) and right at the end (last index) of the array.
2Step 2: While left is less than right, calculate the sum of the elements at these pointers.
3Step 3: If the sum equals the target, return [left + 1, right + 1]. If the sum is less than the target, increment the left pointer. If the sum is greater, decrement the right pointer.

solution.py13 lines

1# Full working Python code
2
3def two_sum_optimal(numbers, target):
4    left, right = 0, len(numbers) - 1
5    while left < right:
6        current_sum = numbers[left] + numbers[right]
7        if current_sum == target:
8            return [left + 1, right + 1]
9        elif current_sum < target:
10            left += 1
11        else:
12            right -= 1
13

ℹ

Complexity note: The time complexity is O(n) because we traverse the array at most once with two pointers. The space complexity is O(1) as we only use a constant amount of space.

1The problem can be solved efficiently using the two-pointer technique due to the sorted nature of the input array.
2Understanding the relationship between the sum of two numbers and their indices helps in deciding how to move the pointers.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.