Partition Equal Subset Sum in DSA Java - Time & Space Complexity

Time Complexity: Partition Equal Subset Sum

O(n * sum)

Understanding Time Complexity

We want to know how the time needed to check if an array can be split into two equal sum parts grows as the array size grows.

How does the algorithm's work increase when the input numbers get bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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 j = target; j >= num; j--) {
            dp[j] = dp[j] || dp[j - num];
        }
    }
    return dp[target];
}
    

This code checks if the array can be split into two parts with equal sums using a dynamic programming approach.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops where the outer loop goes through each number and the inner loop goes through possible sums up to half the total sum.
• How many times: Outer loop runs once per number (n times), inner loop runs up to target sum (sum/2 times) for each number.

How Execution Grows With Input

As the number of elements or the target sum grows, the work grows roughly by multiplying these two factors.

Input Size (n)	Approx. Operations
10 (sum ~ 100)	10 * 50 = 500
100 (sum ~ 1000)	100 * 500 = 50,000
1000 (sum ~ 10,000)	1000 * 5000 = 5,000,000

Pattern observation: The operations grow roughly with the product of the number of elements and the half sum value.

Final Time Complexity

Time Complexity: O(n * sum)

This means the time needed grows with both how many numbers there are and how large their total sum is.

Common Mistake

[X] Wrong: "The time depends only on the number of elements, so it is O(n)."

[OK] Correct: The time also depends on the sum of the numbers because the inner loop runs up to half the sum, not just the count of numbers.

Interview Connect

Understanding this time complexity helps you explain how dynamic programming balances between input size and value range, a key skill in problem solving.

Self-Check

"What if we used recursion with memoization instead of this bottom-up approach? How would the time complexity change?"

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