What is the Integer Partition Problem?

The Integer Partition Problem asks how many ways we can express a given integer ( n ) as a sum of positive integers. For instance, the number 4 can be partitioned into the following sums:

• 4
• 3 + 1
• 2 + 2
• 2 + 1 + 1
• 1 + 1 + 1 + 1

Thus, there are 5 different ways to partition 4. The task of finding the total number of different partitions of a number is what makes this problem both interesting and challenging.

Mathematical Representation

Formally, we represent the number of partitions of a number ( n ) as ( p(n) ). Thus, for ( n = 4 ), we have ( p(4) = 5 ).

Breaking Down the Problem

To solve the Integer Partition Problem, we can utilize dynamic programming (DP), which allows us to store intermediate results and build upon them. We will develop a top-down and a bottom-up approach.

Bottom-Up Dynamic Programming Approach

1. Initialize an Array: Create an array dp of size ( n+1 ) where dp[i] will hold the number of partitions of integer ( i ). Initialize dp[0] to 1 since there's one way to partition zero - by not using any number.

2. Iterate Over Each Integer: For each i from 1 to ( n ), iterate over all integers ( j ) from 1 to ( i ). Update our dp array using:

   [ dp[i] += dp[i - j] ]

   This means that for each integer j added to the partition, we ensure to count all the partitions of the remaining sum ( i - j ).

Example: Calculating Partitions of 5

Let’s take ( n = 5 ):

• Start with an array initialized as:

  [ dp = [1, 0, 0, 0, 0, 0] ]

• First pass (i = 1):

  • Update for ( j = 1 ):

    [ dp[1] = dp[1] + dp[0] = 1 \implies [1, 1, 0, 0, 0, 0] ]

• Second pass (i = 2):

  • Update for ( j = 1 ):

    [ dp[2] = dp[2] + dp[1] = 1 \implies [1, 1, 1, 0, 0, 0] ]

  • Update for ( j = 2 ):

    [ dp[2] = dp[2] + dp[0] = 2 \implies [1, 1, 2, 0, 0, 0] ]

Continuing this way, we build the final dp array for all values up to 5, resulting in ( dp[5] = 7 ). The partitions of 5 are:

• 5
• 4 + 1
• 3 + 2
• 3 + 1 + 1
• 2 + 2 + 1
• 2 + 1 + 1 + 1
• 1 + 1 + 1 + 1 + 1

Top-Down Recursive Approach with Memoization

The top-down approach involves defining a function that recursively calculates the number of partitions while storing results in a hash map to avoid redundant calculations.

def count_partitions(n, max_num):
    if n == 0:
        return 1
    if n < 0 or max_num == 0:
        return 0

    return (count_partitions(n - max_num, max_num) + 
            count_partitions(n, max_num - 1))

# Using memoization
memo = {}
def memoized_count_partitions(n, max_num):
    if (n, max_num) in memo:
        return memo[(n, max_num)]
    if n == 0:
        return 1
    if n < 0 or max_num == 0:
        return 0

    memo[(n, max_num)] = (memoized_count_partitions(n - max_num, max_num) + 
                          memoized_count_partitions(n, max_num - 1))
    return memo[(n, max_num)]

By calling memoized_count_partitions(5, 5), you can efficiently compute the number of partitions for 5.

Complexity Analysis

The bottom-up technique runs in ( O(n^2) ) time complexity due to the nested iterations, while the space complexity is ( O(n) ). The top-down approach, with memoization, has a time complexity of ( O(n \cdot n) ), as it explores each number with every distinct max_num.

As you dive deeper into problems like the Integer Partition Problem, the strategies and approaches outlined here will be invaluable for your dynamic programming toolkit, especially during interviews.