Added two pointer solution for two sum problem (#3468)

other/two_pointer.py

+"""
+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.
+
+You may assume that each input would have exactly one solution, and you
+may not use the same element twice.
+
+This is an alternative solution of the two-sum problem, which uses a
+map to solve the problem. Hence can not solve the issue if there is a
+constraint not use the same index twice. [1]
+
+Example:
+Given nums = [2, 7, 11, 15], target = 9,
+
+Because nums[0] + nums[1] = 2 + 7 = 9,
+return [0, 1].
+
+[1]: 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) = }")