Create beautiful-arrangement

dynamic_programming/arithmetic_slices.py

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+"""
+An integer array is called arithmetic if it
+consists of at least three elements and if
+the difference between any two consecutive
+elements is the same.
+
+Given an integer array nums,
+return the number of
+arithmetic subarrays of nums.
+
+A subarray is a contiguous
+subsequence of the array.
+
+"""
+
+
+class ArithmeticSlices:
+    def numberofarithmeticslices(self, nums):
+        """
+        This defines a class and a function.
+        `nums` is input list of integers.
+        """
+        n = len(nums)
+
+        """
+        We store the length of the
+        array nums in variable n
+        """
+        if n < 3:
+            return 0
+
+        total = 0
+        curr = 0
+
+        """
+        An *arithmetic slice* must have **at least 3 numbers**.
+
+        So, if the array has fewer than 3 elements,
+        no arithmetic slices are possible — immediately return `0`.
+        """
+
+        for i in range(2, n):
+            if nums[i] - nums[i - 1] == nums[i - 1] - nums[i - 2]:
+                curr += 1
+                total += curr
+            else:
+                curr = 0
+
+        """
+        We start iterating from index `2`
+        because we need **three elements**
+        (`nums[i-2], nums[i-1], nums[i]`)
+        to check if they form an arithmetic pattern.
+
+<<<<<<< HEAD
+        So at each step, 
+        we are looking at a triplet ending at index `i`.
+=======
+        So at each step,
+        we’re looking at a triplet ending at index `i`.
+>>>>>>> ca34f0a649ef94c8b778a4babe040fcaa143a56e
+        """
+
+        return total
+
+
+"""
+test_cases = [
+        #  Basic cases
+        ([1, 2, 3, 4], 3),      # [1,2,3], [2,3,4], [1,2,3,4]
+        ([1, 3, 5, 7, 9], 6),   # all diffs = 2;
+                                total slices = 6
+
+        #  Edge cases
+        ([1, 2], 0),            # less than 3 elements → 0
+        ([1, 1, 1], 1),         # [1,1,1] itself is arithmetic
+        ([1], 0),               # single element
+        ([], 0),                # empty array
+        ]
+"""

dynamic_programming/beautiful_arrangement.py

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+"""
+Suppose you have n integers labeled 1 through n.
+A permutation of those n integers
+perm (1-indexed) is considered a
+"beautiful arrangement" if for every i (1 <= i <= n),
+either of the following is true:
+
+-> perm[i] is divisible by i.
+-> i is divisible by perm[i].
+Given an integer n, return the number of the
+"beautiful arrangements" that you can construct.
+
+"""
+# Solution using Backtracking
+
+
+class BeautifulArrange:
+    # function call; n is the size of the permutation (numbers 1..n)
+    def countarrangement(self, n: int) -> int:
+        self.count = 0
+        """
+        We initialize a counter to record how
+            many valid arrangements we find.
+        Using self.count lets the nested
+            function modify it without nonlocal.
+        """
+
+        used = [False] * (n + 1)
+        """
+        A boolean list to mark which numbers have
+        already been placed in the permutation.
+        """
+
+        def backtrack(pos):
+            """
+            Define the recursive backtracking function.
+            pos is the current position in the
+                permutation we are filling (1-indexed).
+            We try to assign a number to position pos.
+            """
+            if pos > n:
+                self.count += 1
+                # We found a complete valid arrangement, so increment the total count.
+                return
+            for num in range(
+                1, n + 1
+            ):  # Try every candidate number num for the current position pos.
+                """
+                Two checks in one:
+                1. not used[num] — the number num has
+                    not been placed yet (we can use it).
+                2. (num % pos == 0 or pos % num == 0) —
+                    the beautiful-arrangement condition:
+                    either num divides pos or pos divides num.
+                If both are true, num is a valid choice for pos.
+
+                """
+                if not used[num] and (num % pos == 0 or pos % num == 0):
+                    used[num] = True
+                    backtrack(pos + 1)
+                    used[num] = False
+
+        backtrack(1)
+        return self.count