Add solutions for binary search problems: sqrt and arrange coins

Implemented two functions, `sqrt` and `arrange_coins`, using binary search to solve LeetCode problems. Included comprehensive unit tests covering small, large, and edge case inputs to ensure correctness and reliability of both algorithms.

Author: nicholasadamou

Binary Search/Math/arrange_coins.py

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+# https://leetcode.com/problems/arranging-coins
+
+import unittest
+
+
+def arrange_coins(n: int) -> int:
+    left = 0
+    right = n
+
+    while left <= right:
+        k = (left + right) // 2
+
+        current = k * (k + 1) // 2
+
+        if current == n:
+            return k
+
+        if current < n:
+            left = k + 1
+        else:
+            right = k - 1
+
+    return right
+
+
+class TestArrangeCoins(unittest.TestCase):
+    def test_small_numbers(self):
+        self.assertEqual(arrange_coins(0), 0)
+        self.assertEqual(arrange_coins(1), 1)
+        self.assertEqual(arrange_coins(2), 1)
+        self.assertEqual(arrange_coins(3), 2)
+
+    def test_medium_numbers(self):
+        self.assertEqual(arrange_coins(5), 2)
+        self.assertEqual(arrange_coins(8), 3)
+        self.assertEqual(arrange_coins(10), 4)
+
+    def test_large_numbers(self):
+        self.assertEqual(arrange_coins(10000), 140)
+        self.assertEqual(arrange_coins(500000), 999)
+
+    def test_boundary_conditions(self):
+        self.assertEqual(arrange_coins(15), 5)  # Fully filled staircase
+        self.assertEqual(arrange_coins(16), 5)  # 15 + 1 extra coin
+        self.assertEqual(arrange_coins(17), 5)

Binary Search/Math/sqrt.py

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+# https://leetcode.com/problems/sqrtx
+
+import unittest
+
+
+def sqrt(x: int) -> int:
+    if x < 2:
+        return x
+
+    left = 2
+    right = x // 2
+
+    while left <= right:
+        mid = (left + right) // 2
+
+        num = mid * mid
+
+        if num == x:
+            return mid
+
+        if num < x:
+            left = mid + 1
+        else:
+            right = mid - 1
+
+    return right
+
+
+class TestSqrtFunction(unittest.TestCase):
+    def test_small_numbers(self):
+        self.assertEqual(sqrt(0), 0)  # Square root of 0 is 0
+        self.assertEqual(sqrt(1), 1)  # Square root of 1 is 1
+        self.assertEqual(sqrt(2), 1)  # Closest integer square root of 2 is 1
+
+    def test_perfect_squares(self):
+        self.assertEqual(sqrt(4), 2)  # 2^2 = 4
+        self.assertEqual(sqrt(9), 3)  # 3^2 = 9
+        self.assertEqual(sqrt(16), 4)  # 4^2 = 16
+        self.assertEqual(sqrt(25), 5)  # 5^2 = 25
+
+    def test_non_perfect_squares(self):
+        self.assertEqual(sqrt(8), 2)  # Closest integer square root of 8 is 2
+        self.assertEqual(sqrt(10), 3)  # Closest integer square root of 10 is 3
+        self.assertEqual(sqrt(27), 5)  # Closest integer square root of 27 is 5
+
+    def test_large_numbers(self):
+        self.assertEqual(sqrt(1000000), 1000)  # Perfect square
+        self.assertEqual(sqrt(1234567), 1111)  # Closest integer square root of 1234567
+
+    def test_edge_cases(self):
+        self.assertEqual(
+            sqrt(2147395599), 46339
+        )  # Largest x for which x^2 still fits in an int (32-bit)
+        self.assertEqual(
+            sqrt(2147483647), 46340
+        )  # Input is the max 32-bit integer, square root approximates