Partition Equal Subset Sum in DSA Java - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main goal of the Partition Equal Subset Sum problem?

To determine if a given set of positive integers can be split into two subsets with equal sums.

beginner

Why do we check if the total sum of the array is even in the Partition Equal Subset Sum problem?

Because if the total sum is odd, it is impossible to split the array into two equal sum subsets.

intermediate

What does the dynamic programming table represent in the Partition Equal Subset Sum solution?

It represents whether a subset with a certain sum is possible using the first few elements of the array.

beginner

In the Partition Equal Subset Sum problem, what is the target sum we try to find a subset for?

Half of the total sum of all elements in the array.

intermediate

What is the time complexity of the dynamic programming solution for Partition Equal Subset Sum?

O(n * sum/2), where n is the number of elements and sum is the total sum of the array.

What is the first step in solving the Partition Equal Subset Sum problem?

A Find the maximum element

B Sort the array

C Check if the total sum is even

D Calculate the product of elements

We first check if the total sum is even because if it is odd, partitioning into equal subsets is impossible.

If the total sum of the array is 20, what subset sum should we try to find?

A 10

B 20

C 5

D 15

We try to find a subset with sum equal to half of the total sum, which is 10.

What does a 'true' value in the DP table indicate in this problem?

A A subset with the target sum is possible

B The array is sorted

C The element is prime

D The subset sum is zero

A 'true' value means a subset with the target sum can be formed from the elements considered.

Which algorithmic technique is commonly used to solve Partition Equal Subset Sum efficiently?

A Backtracking only

B Greedy Algorithm

C Divide and Conquer

D Dynamic Programming

Dynamic Programming is used to efficiently check subset sums up to the target.

If no subset sums to half the total sum, what is the answer to the problem?

A True

B False

C Depends on array size

D Depends on element values

If no subset sums to half the total sum, the array cannot be partitioned into equal subsets.

Explain the steps to solve the Partition Equal Subset Sum problem using dynamic programming.

Think about how to represent subset sums and how to build solutions from smaller subsets.

Describe why the Partition Equal Subset Sum problem cannot be solved if the total sum is odd.

Consider the meaning of equal subsets and integer sums.

Practice

1. What does the Partition Equal Subset Sum problem ask you to determine about a given array of positive integers?

easy

A. If the array can be split into two subsets with equal sums

B. If the array contains any duplicate elements

C. If the array is sorted in ascending order

D. If the array contains only even numbers

Solution

1. Step 1: Understand the problem statement

   The problem asks whether the array can be divided into two parts where the sum of elements in both parts is the same.

2. Step 2: Identify the key condition

   This means the total sum of the array must be even and there must be a subset with sum equal to half of the total sum.

3. Final Answer:

   If the array can be split into two subsets with equal sums -> Option A

4. Quick Check:

   Partition Equal Subset Sum = Split array into two equal sum subsets [OK]

Hint: Check if total sum is even first to decide quickly [OK]

Common Mistakes:

• Thinking the problem is about sorting the array
• Confusing equal sum subsets with duplicate elements
• Ignoring the sum condition and just checking elements

2. Which of the following Java method signatures correctly represents a function to solve the Partition Equal Subset Sum problem?

easy

A. public boolean canPartition(int[] nums)

B. public int canPartition(int[] nums)

C. public void canPartition(int[] nums)

D. public String canPartition(int nums)

Solution

1. Step 1: Identify the return type

   The function should return true or false, so the return type must be boolean.

2. Step 2: Check parameter type

   The input is an array of integers, so parameter should be int[] nums.

3. Final Answer:

   public boolean canPartition(int[] nums) -> Option A

4. Quick Check:

   Boolean return + int array parameter = public boolean canPartition(int[] nums) [OK]

Hint: Return boolean for yes/no problems [OK]

Common Mistakes:

• Using int return type instead of boolean
• Using void return type which gives no answer
• Passing single int instead of int array

3. What will be the output of the following Java code snippet?

int[] nums = {1, 5, 11, 5};
System.out.println(canPartition(nums));

public static boolean canPartition(int[] nums) {
    int sum = 0;
    for (int num : nums) sum += num;
    if (sum % 2 != 0) return false;
    int target = sum / 2;
    boolean[] dp = new boolean[target + 1];
    dp[0] = true;
    for (int num : nums) {
        for (int i = target; i >= num; i--) {
            dp[i] = dp[i] || dp[i - num];
        }
    }
    return dp[target];
}

medium

A. Compilation error

B. false

C. true

D. Runtime exception

Solution

1. Step 1: Calculate total sum and check parity

   Sum = 1+5+11+5 = 22, which is even, so continue.

2. Step 2: Use dynamic programming to check subset sum

   The dp array tracks sums achievable. The subset {11, 5, 5} sums to 21, but target is 11. Actually, subset {11} and {1,5,5} both sum to 11, so dp[target] becomes true.

3. Final Answer:

   true -> Option C

4. Quick Check:

   Subset sum equals half total sum = true [OK]

Hint: Check if half sum subset exists using dp array [OK]

Common Mistakes:

• Not checking if total sum is even first
• Confusing dp array update direction
• Assuming subset must be contiguous

4. Identify the bug in this Java method for Partition Equal Subset Sum:

public boolean canPartition(int[] nums) {
    int sum = 0;
    for (int num : nums) sum += num;
    if (sum % 2 == 0) return false;
    int target = sum / 2;
    boolean[] dp = new boolean[target + 1];
    dp[0] = true;
    for (int num : nums) {
        for (int i = target; i >= num; i--) {
            dp[i] = dp[i] || dp[i - num];
        }
    }
    return dp[target];
}

medium

A. The dp array size is incorrect

B. The sum parity check condition is reversed

C. The inner loop should increment i instead of decrement

D. The method should return int instead of boolean

Solution

1. Step 1: Check sum parity condition

   The code returns false if sum % 2 == 0, but it should return false if sum is odd (sum % 2 != 0).

2. Step 2: Verify other parts

   dp array size and loops are correct; only the parity check is wrong.

3. Final Answer:

   The sum parity check condition is reversed -> Option B

4. Quick Check:

   Return false if sum is odd, not even [OK]

Hint: Remember: odd sum means no equal partition [OK]

Common Mistakes:

• Returning false when sum is even instead of odd
• Changing dp array size incorrectly
• Looping forward instead of backward causing wrong dp updates

5. Given the array nums = {2, 2, 3, 5}, which of the following modifications to the standard Partition Equal Subset Sum approach will correctly determine if the array can be partitioned into two subsets with equal sum, considering that elements can be used multiple times?

hard

A. Use a recursive approach without memoization

B. Keep the inner loop backward but initialize dp[1] = true

C. Double the target sum before processing

D. Change the inner loop to iterate from num to target (forward) instead of backward

Solution

1. Step 1: Understand the problem change

   Allowing elements multiple times means this is an unbounded subset sum problem, requiring forward iteration in dp update.

2. Step 2: Modify dp iteration accordingly

   Changing the inner loop to go from num to target (forward) allows reusing elements multiple times correctly.

3. Step 3: Verify other options

   Initializing dp[1] is incorrect; recursion without memo is inefficient; doubling target sum is wrong.

4. Final Answer:

   Change the inner loop to iterate from num to target (forward) instead of backward -> Option D

5. Quick Check:

   Unbounded subset sum needs forward dp loop [OK]

Hint: For repeated elements, loop forward in dp array [OK]

Common Mistakes:

• Using backward loop for repeated elements
• Incorrect dp initialization
• Ignoring efficiency by using recursion without memo