Distribute Repeating Integers - Leetcode Solution

Problem Description: Given an array of size 2N, where N is a positive integer, you need to find out if any sum of any N integers in the array is equal to the sum of any other N integers in the array.

Solution: We can solve this problem by using a hash map. We start by iterating over all possible sums of N integers in the array. For each sum, we check if it exists in our hash map. If it does, it means we have found two sets of N integers with equal sums and we can return true. If it doesn't exist, we add it to the hash map.

To generate all possible sums of N integers, we can use a recursive function that takes as input the current sum, the index of the current element in the array, the current set of selected elements, and the hash map. If the size of the current set is equal to N, we add the current sum to the hash map. Otherwise, we iterate over all remaining elements in the array and recursively call the function with the updated sum and set.

Here is the Python code implementation:

class Solution:
    def canPartition(self, nums: List[int]) -> bool:
        n = len(nums)
        if n % 2 != 0:
            return False
        
        def generate_sums(sum_so_far, index, selected, sums_dict):
            if len(selected) == n // 2:
                if sum_so_far in sums_dict:
                    return True
                sums_dict[sum_so_far] = True
                return False
            
            for i in range(index, n):
                selected.append(i)
                if generate_sums(sum_so_far + nums[i], i+1, selected, sums_dict):
                    return True
                selected.pop()
            
            return False
        
        sums_dict = {}
        return generate_sums(0, 0, [], sums_dict)

Time Complexity: The function to generate all possible sums takes O(2^N) time in the worst case, as we need to generate all possible subsets of size N. Each subset requires adding N integers, which takes O(N) time. Therefore, the overall time complexity of the algorithm is O(N*2^N).

Space Complexity: We need a hash map to store all possible sums, which requires O(2^N) space. Additionally, we use recursion to generate all possible sums, which requires O(N) space on the call stack. Therefore, the overall space complexity of the algorithm is O(N*2^N).

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