Partition Equal Subset Sum in DSA C# - Execution Trace

Concept Flow - Partition Equal Subset Sum

Calculate total sum of array

↓

Is total sum even?

No→Return False: cannot partition

Yes↓

Set target = total sum / 2

↓

Use DP to check if subset sums to target

↓

If subset found

→Return True

No↓

Return False

Check if array can be split into two parts with equal sum by finding a subset with sum equal to half of total.

Execution Sample

DSA C#

int[] nums = {1, 5, 11, 5};
int total = 0;
foreach (int num in nums) total += num;
int target = total / 2;
bool canPartition = SubsetSum(nums, target);

Check if array {1,5,11,5} can be partitioned into two subsets with equal sum 11.

Execution Table

Step	Operation	Subset Sum Target	DP Array State	Result
1	Calculate total sum	22	N/A	Total sum = 22
2	Check if total sum is even	22	N/A	Yes, even
3	Set target = total / 2	11	N/A	Target = 11
4	Initialize DP array	11	[True, False, False, False, False, False, False, False, False, False, False, False]	DP[0] = True
5	Process num = 1	11	[True, True, False, False, False, False, False, False, False, False, False, False]	DP[1] = True
6	Process num = 5	11	[True, True, False, False, False, True, True, False, False, False, False, False]	DP[5] and DP[6] = True
7	Process num = 11	11	[True, True, False, False, False, True, True, False, False, False, False, True]	DP[11] = True (subset found)
8	Result	11	N/A	Return True: partition possible

💡 DP[11] is True, meaning subset with sum 11 exists, so partition is possible

Variable Tracker

Variable	Start	After Step 4	After Step 5	After Step 6	After Step 7	Final
total	0	22	22	22	22	22
target	0	11	11	11	11	11
DP array	N/A	[True, False x11]	[True, True, False x10]	[True, True, False x3, True, True, False x5]	[True, True, False x3, True, True, False x4, True]	[True, True, False x3, True, True, False x4, True]
num	N/A	N/A	1	5	11	N/A

Key Moments - 3 Insights

Why do we check if total sum is even before proceeding?

Because if total sum is odd (see Step 2), it's impossible to split into two equal parts, so we return False immediately.

What does DP array represent in this problem?

DP array at each index i shows if a subset sum of i is possible with processed numbers (see Steps 4-7). True means subset sum i can be formed.

Why does DP[11] being True mean partition is possible?

Because target is half of total sum (11), DP[11] True means a subset sums to 11, so the rest also sums to 11, allowing equal partition (Step 7).

Visual Quiz - 3 Questions

Look at the execution table at Step 5, what does DP[1] = True mean?

A Partition is already found

B Total sum is 1

C A subset with sum 1 is possible using processed numbers

D DP array is reset

💡 Hint

Refer to DP Array State column at Step 5 showing DP[1] = True means subset sum 1 is achievable

At which step does the algorithm find a subset equal to the target sum?

A Step 7

B Step 4

C Step 6

D Step 8

💡 Hint

Check Result column; Step 7 shows DP[11] = True meaning subset found

If the total sum was 23 instead of 22, what would happen at Step 2?

A Algorithm proceeds to DP

B Returns False immediately

C Sets target to 11

D Throws error

💡 Hint

Step 2 checks if total sum is even; odd sum means immediate False

Concept Snapshot

Partition Equal Subset Sum:
- Find if array can split into two equal sum subsets
- Calculate total sum; if odd, return False
- Use DP to check subset sums to total/2
- DP[i] = True means subset sum i possible
- If DP[target] True, partition exists

Full Transcript

Partition Equal Subset Sum checks if an array can be split into two subsets with equal sums. First, it calculates the total sum of the array. If the sum is odd, it returns False immediately because equal partition is impossible. If even, it sets the target as half the total sum. Then it uses a dynamic programming array where each index represents if a subset sum is possible. It initializes DP[0] as True because sum zero is always possible. It processes each number, updating the DP array to mark sums achievable with the current numbers. If at any point DP[target] becomes True, it means a subset with sum equal to half the total exists, so the array can be partitioned equally. Otherwise, it returns False. This approach efficiently solves the problem by reducing it to a subset sum check.

Practice

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

easy

A. If the array is sorted in ascending order

B. If the array contains any duplicate elements

C. If the array contains only positive numbers

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

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: Clarify what is being checked

   It is not about duplicates, sorting, or positivity, but about equal sum partitioning.

3. Final Answer:

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

4. Quick Check:

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

Hint: Focus on splitting array into two equal sum parts [OK]

Common Mistakes:

• Confusing with checking duplicates
• Thinking about sorting order
• Assuming all numbers must be positive

2. Which of the following C# code snippets correctly initializes a boolean array named dp of size sum + 1 for the Partition Equal Subset Sum problem?

easy

A. bool[] dp = new bool[sum];

B. bool[] dp = new bool(sum + 1);

C. bool[] dp = new bool[sum + 1];

D. bool dp = new bool[sum + 1];

Solution

1. Step 1: Understand array initialization syntax in C#

   Arrays are initialized with square brackets and the size inside new[]. For size sum+1, use new bool[sum + 1].

2. Step 2: Check each option

   bool[] dp = new bool[sum]; misses +1, bool dp = new bool[sum + 1]; uses wrong type declaration, bool[] dp = new bool(sum + 1); uses parentheses which is invalid.

3. Final Answer:

   bool[] dp = new bool[sum + 1]; -> Option C

4. Quick Check:

   Correct C# array syntax = bool[] dp = new bool[sum + 1]; [OK]

Hint: Use square brackets and new keyword for arrays in C# [OK]

Common Mistakes:

• Forgetting +1 for array size
• Using parentheses instead of square brackets
• Declaring array as single bool variable

3. What will be the output of the following C# code snippet for the input array nums = new int[] {1, 5, 11, 5};?

bool CanPartition(int[] nums) {
    int sum = 0;
    foreach (int num in nums) sum += num;
    if (sum % 2 != 0) return false;
    int target = sum / 2;
    bool[] dp = new bool[target + 1];
    dp[0] = true;
    foreach (int num in nums) {
        for (int i = target; i >= num; i--) {
            dp[i] = dp[i] || dp[i - num];
        }
    }
    return dp[target];
}

Console.WriteLine(CanPartition(nums));

medium

A. true

B. false

C. Compile-time error

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

   Target = 11. The dp array tracks if sum i is achievable. The code finds that subset {11} or {5,5,1} sums to 11, so dp[11] = true.

3. Final Answer:

   true -> Option A

4. Quick Check:

   Sum even and subset sum found = true [OK]

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

Common Mistakes:

• Not checking if sum is even
• Incorrect dp array update direction
• Returning false prematurely

4. Identify the bug in the following C# code snippet for Partition Equal Subset Sum and select the correct fix:

bool CanPartition(int[] nums) {
    int sum = 0;
    foreach (int num in nums) sum += num;
    if (sum % 2 != 0) return false;
    int target = sum / 2;
    bool[] dp = new bool[target];
    dp[0] = true;
    foreach (int num in nums) {
        for (int i = target; i >= num; i--) {
            dp[i] = dp[i] || dp[i - num];
        }
    }
    return dp[target];
}

medium

A. Initialize dp array with false values explicitly

B. Change dp array size to target + 1

C. Change loop to iterate from 0 to target

D. Remove the dp[0] = true assignment

Solution

1. Step 1: Check dp array size and indexing

   dp array is declared with size target, but dp[target] is accessed, causing out-of-range error.

2. Step 2: Fix array size to include target index

   Change dp size to target + 1 to safely access dp[target].

3. Final Answer:

   Change dp array size to target + 1 -> Option B

4. Quick Check:

   Array size must cover all indices used [OK]

Hint: Array size must be one more than max index accessed [OK]

Common Mistakes:

• Ignoring array index out of range
• Changing loop direction unnecessarily
• Removing dp[0] initialization

5. Given the array nums = new int[] {2, 2, 3, 5}, which of the following statements is true regarding the Partition Equal Subset Sum problem?

hard

A. The array cannot be partitioned into two subsets with equal sum

B. The array can be partitioned into two subsets with equal sum: {2,3} and {2,5}

C. The array can be partitioned into two subsets with equal sum: {2,2,3} and {5}

D. The array can be partitioned into two subsets with equal sum: {3,5} and {2,2}

Solution

1. Step 1: Calculate total sum of array elements

   Sum = 2 + 2 + 3 + 5 = 12, which is even, so partition might be possible.

2. Step 2: Check possible subsets for half sum (6)

   Half sum = 6. Possible subsets: {2,2,2} no (only two 2s), {3,3} no (only one 3), {5,1} no 1 missing. No subset sums to 6 exactly.

3. Step 3: Verify given options

   The array can be partitioned into two subsets with equal sum: {2,3} and {2,5} sums: {2,3}=5 and {2,5}=7 (not equal). The array can be partitioned into two subsets with equal sum: {2,2,3} and {5} sums: {2,2,3}=7 and {5}=5 (not equal). The array can be partitioned into two subsets with equal sum: {3,5} and {2,2} sums: {3,5}=8 and {2,2}=4 (not equal).

4. Final Answer:

   The array cannot be partitioned into two subsets with equal sum -> Option A

5. Quick Check:

   Sum even but no subset sums to half sum = false [OK]

Hint: Check if any subset sums exactly to half total sum [OK]

Common Mistakes:

• Assuming even sum means partition always possible
• Incorrectly summing subsets
• Ignoring subset element counts