two_pointer.py

"""
Two Pointer Technique for Two Sum Problem (Sorted Array)

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

Assumptions:
- The input array is sorted in non-decreasing order.
- Each input has exactly one solution.
- The same element cannot be used twice.

Approach:
- Initialize two pointers:
    left pointer at the beginning (index 0)
    right pointer at the end (index n-1)
- Compare sum of elements at both pointers:
    - If sum == target → return indices
    - If sum < target → move left pointer forward
    - If sum > target → move right pointer backward

Time Complexity: O(n)
Space Complexity: O(1)

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

Reference:
https://github.com/TheAlgorithms/Python/blob/master/other/two_sum.py
"""

from __future__ import annotations


def two_pointer(nums: list[int], target: int) -> list[int]:
    """
    >>> two_pointer([2, 7, 11, 15], 9)
    [0, 1]
    >>> two_pointer([2, 7, 11, 15], 17)
    [0, 3]
    >>> two_pointer([2, 7, 11, 15], 18)
    [1, 2]
    >>> two_pointer([2, 7, 11, 15], 26)
    [2, 3]
    >>> two_pointer([1, 3, 3], 6)
    [1, 2]
    >>> two_pointer([2, 7, 11, 15], 8)
    []
    >>> two_pointer([3 * i for i in range(10)], 19)
    []
    >>> two_pointer([1, 2, 3], 6)
    []
    """
    i = 0
    j = len(nums) - 1

    while i < j:
        if nums[i] + nums[j] == target:
            return [i, j]
        elif nums[i] + nums[j] < target:
            i = i + 1
        else:
            j = j - 1

    return []


if __name__ == "__main__":
    import doctest

    doctest.testmod()
    print(f"{two_pointer([2, 7, 11, 15], 9) = }")