Largest Divisible Subset – Problem Overview

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Problem Statement

Given a set of distinct positive integers, you need to find the largest subset such that every pair (i, j) in the subset satisfies:

• i % j == 0 (i is divisible by j), or equivalently,
• j % i == 0.

Your task is to return the largest subset from the given set of integers.

Example 1:

codeInput: [1, 2, 3, 4, 6, 8, 9, 12, 18]
Output: [1, 2, 4, 8]  # This is the largest subset

Example 2:

codeInput: [1, 2, 4, 8]
Output: [1, 2, 4, 8]  # This is the largest subset

Approach to Solve the Problem

This problem can be solved using a dynamic programming approach, which essentially involves:

1. Sorting the input array to make sure that we can check divisibility between the numbers in order (small to large).
2. Using dynamic programming to store the longest divisible subset for each number in the array.
3. Tracking the sequence of elements that can form the largest subset.

Let’s break down the approach into manageable steps:

Steps to Solve

1. Sort the Input Array: Sorting helps because if a number a[i] can be divisible by a number a[j], then i > j. This property allows us to check divisibility in a linear fashion after sorting.
2. Define Dynamic Programming State: Let dp[i] represent the size of the largest divisible subset that ends with the element a[i].
3. Track the Predecessors: We also need to keep track of the indices of previous elements in the subset. For each element a[i], we will check all previous elements a[j] (j < i) and see if a[i] % a[j] == 0. If true, we can extend the subset ending at a[j] to include a[i].
4. Reconstruct the Largest Subset: Once we find the largest subset length, we will reconstruct the subset by tracing back the predecessors.

Code Implementation

codedef largestDivisibleSubset(nums):
    if not nums:
        return []

    # Step 1: Sort the numbers in ascending order
    nums.sort()
    
    # Step 2: Initialize DP array and predecessor array
    n = len(nums)
    dp = [1] * n  # dp[i] will store the length of the largest subset ending at nums[i]
    prev = [-1] * n  # prev[i] will store the previous index for the subset
    
    # Step 3: Build the DP table
    max_len = 1  # The length of the largest divisible subset
    max_idx = 0  # The index of the last element in the largest subset
    
    for i in range(1, n):
        for j in range(i):
            if nums[i] % nums[j] == 0:
                # If nums[i] can be added to the subset ending at nums[j]
                if dp[i] < dp[j] + 1:
                    dp[i] = dp[j] + 1
                    prev[i] = j
                    # Update the largest subset length if needed
                    if dp[i] > max_len:
                        max_len = dp[i]
                        max_idx = i
    
    # Step 4: Reconstruct the largest divisible subset
    subset = []
    while max_idx != -1:
        subset.append(nums[max_idx])
        max_idx = prev[max_idx]
    
    return subset[::-1]  # Reverse the subset to get it in correct order

Explanation of the Code

1. Sorting: We start by sorting the nums array to make sure we can easily check divisibility in an increasing order.
2. DP and Predecessor Array:
   • dp[i] stores the length of the largest divisible subset ending at nums[i].
   • prev[i] helps us track the previous index in the subset so we can later reconstruct it.
3. Double Loop for DP Calculation: The outer loop (i) checks each element in the array, and the inner loop (j) compares it with previous elements (nums[j] for j < i). If nums[i] % nums[j] == 0, it means we can add nums[i] to the subset that ends at nums[j], and we update dp[i] accordingly.
4. Track the Largest Subset: We update max_len and max_idx whenever we find a new largest subset. The max_idx stores the index of the last element in the largest subset.
5. Reconstruction of the Subset: After building the DP array, we reconstruct the largest subset by tracing back from max_idx using the prev array.
6. Return Result: Finally, we return the reconstructed subset after reversing it because we constructed it backwards.

Time Complexity

• Sorting: The sorting operation takes O(nlog⁡n)O(n \log n)O(nlogn), where nnn is the length of the input array.
• Dynamic Programming: The nested loops iterate over the array, so the DP step takes O(n2)O(n^2)O(n2) time.
• Total Complexity: Therefore, the overall time complexity is O(n2)O(n^2)O(n2).

Space Complexity

• We use two additional arrays (dp and prev) of size nnn, so the space complexity is O(n)O(n)O(n).

Walkthrough

Let’s walk through an example with the input:

codeInput: [1, 2, 3, 4, 6, 8, 9, 12, 18]

1. Sort the array: [1, 2, 3, 4, 6, 8, 9, 12, 18]
2. Initialize dp and prev arrays:
   • dp = [1, 1, 1, 1, 1, 1, 1, 1, 1]
   • prev = [-1, -1, -1, -1, -1, -1, -1, -1, -1]
3. Fill the dp and prev arrays: After processing all pairs, the largest divisible subset will have a size of 4 (ending at index 7, corresponding to [1, 2, 4, 8]).
4. Reconstruct the subset: By following the prev pointers, we get the subset [1, 2, 4, 8].

Conclusion

The dynamic programming solution for the Largest Divisible Subset problem ensures an optimal solution with a time complexity of O(n2)O(n^2)O(n2) and space complexity of O(n)O(n)O(n). By leveraging sorting, dynamic programming, and predecessor tracking, we can efficiently find the largest subset where every pair of numbers divides each other.