Create combination_sum_iv.py (#7672)

* Create combination_sum_iv.py

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* Update dynamic_programming/combination_sum_iv.py

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* Update dynamic_programming/combination_sum_iv.py

Co-authored-by: Caeden Perelli-Harris <caedenperelliharris@gmail.com>

* Update dynamic_programming/combination_sum_iv.py

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* Update combination_sum_iv.py

* Update combination_sum_iv.py

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* minor change, argument missing in function

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* Update dynamic_programming/combination_sum_iv.py

Co-authored-by: Christian Clauss <cclauss@me.com>

* minor change

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Co-authored-by: Caeden Perelli-Harris <caedenperelliharris@gmail.com>
Co-authored-by: Christian Clauss <cclauss@me.com>

dynamic_programming/combination_sum_iv.py

@@ -0,0 +1,102 @@
+"""
+Question:
+You are given an array of distinct integers and you have to tell how many
+different ways of selecting the elements from the array are there such that
+the sum of chosen elements is equal to the target number tar.
+
+Example
+
+Input:
+N = 3
+target = 5
+array = [1, 2, 5]
+
+Output:
+9
+
+Approach:
+The basic idea is to go over recursively to find the way such that the sum
+of chosen elements is “tar”. For every element, we have two choices
+    1. Include the element in our set of chosen elements.
+    2. Don’t include the element in our set of chosen elements.
+"""
+
+
+def combination_sum_iv(n: int, array: list[int], target: int) -> int:
+    """
+    Function checks the all possible combinations, and returns the count
+    of possible combination in exponential Time Complexity.
+
+    >>> combination_sum_iv(3, [1,2,5], 5)
+    9
+    """
+
+    def count_of_possible_combinations(target: int) -> int:
+        if target < 0:
+            return 0
+        if target == 0:
+            return 1
+        return sum(count_of_possible_combinations(target - item) for item in array)
+
+    return count_of_possible_combinations(target)
+
+
+def combination_sum_iv_dp_array(n: int, array: list[int], target: int) -> int:
+    """
+    Function checks the all possible combinations, and returns the count
+    of possible combination in O(N^2) Time Complexity as we are using Dynamic
+    programming array here.
+
+    >>> combination_sum_iv_dp_array(3, [1,2,5], 5)
+    9
+    """
+
+    def count_of_possible_combinations_with_dp_array(
+        target: int, dp_array: list[int]
+    ) -> int:
+        if target < 0:
+            return 0
+        if target == 0:
+            return 1
+        if dp_array[target] != -1:
+            return dp_array[target]
+        answer = sum(
+            count_of_possible_combinations_with_dp_array(target - item, dp_array)
+            for item in array
+        )
+        dp_array[target] = answer
+        return answer
+
+    dp_array = [-1] * (target + 1)
+    return count_of_possible_combinations_with_dp_array(target, dp_array)
+
+
+def combination_sum_iv_bottom_up(n: int, array: list[int], target: int) -> int:
+    """
+    Function checks the all possible combinations with using bottom up approach,
+    and returns the count of possible combination in O(N^2) Time Complexity
+    as we are using Dynamic programming array here.
+
+    >>> combination_sum_iv_bottom_up(3, [1,2,5], 5)
+    9
+    """
+
+    dp_array = [0] * (target + 1)
+    dp_array[0] = 1
+
+    for i in range(1, target + 1):
+        for j in range(n):
+            if i - array[j] >= 0:
+                dp_array[i] += dp_array[i - array[j]]
+
+    return dp_array[target]
+
+
+if __name__ == "__main__":
+    import doctest
+
+    doctest.testmod()
+    n = 3
+    target = 5
+    array = [1, 2, 5]
+    print(combination_sum_iv(n, array, target))