Solution Explanation: Distribute Repeating Integers

This problem asks whether we can distribute a given array of integers (nums) to satisfy a set of customer orders (quantity), ensuring each customer receives the exact quantity of identical integers. The solution uses dynamic programming with bit manipulation for efficient subset management.

Approach:

1. Count Occurrences: First, count the occurrences of each unique integer in nums using a hash map (or Counter in Python). This reduces the problem to working with counts instead of individual numbers.

2. State Representation: The core idea is to represent subsets of customer orders using bitmasks. If there are m customers, a bitmask of m bits can represent all possible subsets. Each bit corresponds to a customer; a '1' means the customer is included in the subset, and '0' means they're not.

3. Preprocessing Subset Sums: Calculate the sum of quantities for each subset. This precomputation avoids redundant calculations in the dynamic programming step.

4. Dynamic Programming: The dynamic programming table f[i][j] represents whether it's possible to satisfy the customer orders in subset j using the counts of the first i unique integers from nums.

   • f[i][0] is always True because an empty subset is always satisfiable.
   • To determine f[i][j], we iterate through all subsets k that are subsets of j. If we can satisfy subset j XOR k using the first i-1 integers (f[i-1][j XOR k]), and the count of the i-th integer is greater than or equal to the sum of quantities in subset k (s[k]), then f[i][j] is True.

5. Final Result: f[n-1][(1 << m) - 1] (where n is the number of unique integers) indicates whether all customer orders can be satisfied using all unique integers.

Time Complexity: O(n * 3m), where n is the number of unique integers and m is the number of customers. The dominant factor is the nested loops iterating through subsets. Each bit in the mask can be either 0 or 1, or part of a larger subset, leading to 3m complexity.

Space Complexity: O(n * 2m) to store the dynamic programming table.

Code Explanation (Python):

class Solution:
    def canDistribute(self, nums: List[int], quantity: List[int]) -> bool:
        m = len(quantity)
        s = [0] * (1 << m)  # Precompute subset sums
        for i in range(1, 1 << m):
            for j in range(m):
                if i >> j & 1:
                    s[i] = s[i ^ (1 << j)] + quantity[j]
                    break
 
        cnt = Counter(nums)  # Count occurrences of each integer
        arr = list(cnt.values())  # Array of counts
        n = len(arr)
 
        f = [[False] * (1 << m) for _ in range(n)]  # DP table
        for i in range(n):
            f[i][0] = True
 
        for i, x in enumerate(arr):
            for j in range(1, 1 << m):
                k = j
                while k:
                    ok1 = j == k if i == 0 else f[i - 1][j ^ k]
                    ok2 = s[k] <= x
                    if ok1 and ok2:
                        f[i][j] = True
                        break
                    k = (k - 1) & j
 
        return f[-1][-1]
 

The other languages (Java, C++, Go, TypeScript) follow a similar structure and logic, adapting the syntax and data structures to their respective language features. The core algorithm remains consistent across all implementations.