feat(algorithms): find sum of three

Author: BrianLusina

algorithms/two_pointers/find_sum_of_three/README.md

@@ -0,0 +1,13 @@
+# Find Sum of Three
+
+Given an array of integers, nums, and an integer value, target, determine if there are any three integers in nums whose 
+sum is equal to the target, that is, nums[i] + nums[j] + nums[k] == target. Return TRUE if three such integers exist in 
+the array. Otherwise, return FALSE.
+
+> Note: A valid triplet consists of elements with distinct indexes. This means, for the triplet nums[i], nums[j], and 
+> nums[k], i ≠ j, i ≠ k and j ≠ k.
+
+Constraints:
+- 3 ≤ nums.length ≤ 500
+- -10^3 ≤ nums[i] ≤ 10^3
+- −10^3 ≤ target ≤ 10^3

algorithms/two_pointers/find_sum_of_three/__init__.py

@@ -0,0 +1,35 @@
+from typing import List
+from copy import copy
+
+
+def find_sum_of_three(nums: List[int], target: int) -> bool:
+    """
+    Checks if there are any three integers in nums whose sum is equal to the target
+    Args:
+        nums (list): list of integers
+        target (int): target to equate to
+    Returns:
+        bool: True if there are three integers that sum up to target, False otherwise
+    """
+    # note that using a copy of the original array leads to unnecessary memory consumption
+    numbers = copy(nums)
+    # Sorts in place which incurs a Time complexity of O(nlog(n))
+    numbers.sort()
+
+    for idx, num in enumerate(numbers):
+        if idx > 0 and num == numbers[idx-1]:
+            continue
+
+        left, right = idx+1, len(numbers) - 1
+
+        while left < right:
+            current_sum = num + numbers[left] + numbers[right]
+
+            if current_sum > target:
+                right -= 1
+            elif current_sum < target:
+                left += 1
+            else:
+                return True
+
+    return False

algorithms/two_pointers/find_sum_of_three/test_find_sum_of_three.py

@@ -0,0 +1,43 @@
+import unittest
+from . import find_sum_of_three
+
+
+class FindSumOfThreeTestCases(unittest.TestCase):
+    def test_1(self):
+        """should return False for nums=[1,-1,0], target=-1"""
+        nums = [1, -1, 0]
+        target = -1
+        actual = find_sum_of_three(nums, target)
+        self.assertFalse(actual)
+
+    def test2(self):
+        """should return True for nums=[3,7,1,2,8,4,5], target=10"""
+        nums = [3,7,1,2,8,4,5]
+        target = 10
+        actual = find_sum_of_three(nums, target)
+        self.assertTrue(actual)
+
+    def test_3(self):
+        """should return False for nums=[3,7,1,2,8,4,5], target=21"""
+        nums = [3,7,1,2,8,4,5]
+        target = 21
+        actual = find_sum_of_three(nums, target)
+        self.assertFalse(actual)
+
+    def test_4(self):
+        """should return True for nums=[-1,2,1,-4,5,-3], target=-8"""
+        nums = [-1,2,1,-4,5,-3]
+        target = -8
+        actual = find_sum_of_three(nums, target)
+        self.assertTrue(actual)
+
+    def test_5(self):
+        """should return True for nums=[-1,2,1,-4,5,-3], target=0"""
+        nums = [-1,2,1,-4,5,-3]
+        target = 0
+        actual = find_sum_of_three(nums, target)
+        self.assertTrue(actual)
+
+
+if __name__ == '__main__':
+    unittest.main()

algorithms/two_pointers/three_sum/README.md

@@ -5,23 +5,30 @@ and nums[i] + nums[j] + nums[k] == 0.
 
 Notice that the solution set must not contain duplicate triplets.
 
+```plain
 Example 1:
-
 Input: nums = [-1,0,1,2,-1,-4]
 Output: [[-1,-1,2],[-1,0,1]]
 Explanation:
 nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
 nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
 nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
 The distinct triplets are [-1,0,1] and [-1,-1,2].
-Notice that the order of the output and the order of the triplets does not matter.
+```
+> Notice that the order of the output and the order of the triplets does not matter.
+
+```plain
 Example 2:
 
 Input: nums = [0,1,1]
 Output: []
 Explanation: The only possible triplet does not sum up to 0.
+```
+
+```plain
 Example 3:
 
 Input: nums = [0,0,0]
 Output: [[0,0,0]]
-Explanation: The only possible triplet sums up to 0.
+Explanation: The only possible triplet sums up to 0.
+```