Partition Equal Subset Sum in DSA Java - Build from Scratch

Partition Equal Subset Sum

📖 Scenario: Imagine you have a collection of weights and you want to see if you can split them into two groups that weigh the same. This is useful in packing or balancing loads evenly.

🎯 Goal: You will write a Java program that checks if an array of positive integers can be divided into two subsets with equal sums.

📋 What You'll Learn

Create an integer array called nums with the exact values {1, 5, 11, 5}.

Create an integer variable called totalSum that stores the sum of all elements in nums.

Write a method called canPartition that takes the array nums and returns true if it can be split into two equal sum subsets, otherwise false.

Print the result of calling canPartition(nums).

💡 Why This Matters

🌍 Real World

Balancing loads, dividing resources evenly, or partitioning tasks fairly in real life.

💼 Career

This problem is common in coding interviews and helps understand dynamic programming and subset problems.

1

Create the integer array nums

Create an integer array called nums with the exact values {1, 5, 11, 5}.

DSA Java

// Create the integer array nums with values {1, 5, 11, 5}
// Your code here

Hint

Use int[] nums = {1, 5, 11, 5}; to create the array.

2

Calculate the total sum of elements in nums

Create an integer variable called totalSum that stores the sum of all elements in nums using a for loop.

DSA Java

int[] nums = {1, 5, 11, 5};
// Create totalSum and calculate sum of nums elements
// Your code here

Hint

Use a for-each loop to add each number in nums to totalSum.

3

Write the canPartition method

Write a method called canPartition that takes an integer array nums and returns true if the array can be split into two subsets with equal sums, otherwise false. Use the following logic:
1. Calculate the total sum of nums. If it is odd, return false.
2. Use a boolean array dp of size sum/2 + 1 to track possible sums.
3. Initialize dp[0] to true.
4. For each number in nums, update dp from right to left.
5. Return dp[sum/2].

DSA Java

int[] nums = {1, 5, 11, 5};
int totalSum = 0;
for (int num : nums) {
    totalSum += num;
}

// Write the canPartition method below
// Your code here

Hint

Use dynamic programming with a boolean array to track sums. Remember to check if total sum is odd first.

4

Print the result of canPartition(nums)

Print the result of calling canPartition(nums) using System.out.println.

DSA Java

int[] nums = {1, 5, 11, 5};
int totalSum = 0;
for (int num : nums) {
    totalSum += num;
}

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];
}

// Print the result of canPartition(nums)
// Your code here

Hint

Use System.out.println(canPartition(nums)); inside the main method to print the result.

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