Partition Equal Subset Sum in DSA C++ - 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's impossible to split the array into two equal sum subsets.

intermediate

What technique is commonly used to solve the Partition Equal Subset Sum problem efficiently?

Dynamic programming, specifically a subset sum approach to check if a subset with sum equal to half of total sum exists.

intermediate

In the dynamic programming solution, what does the boolean DP table represent?

DP[i][j] is true if a subset of the first i numbers has a sum equal to j, otherwise false.

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 before applying dynamic programming in Partition Equal Subset Sum?

A Reverse the array

B Sort the array

C Check if total sum is even

D Find the maximum element

Checking if the total sum is even is essential because if it's odd, partitioning into equal subsets is impossible.

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

A 5

B 10

C 20

D 15

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

What does DP[i][j] = true mean in the DP table?

A Subset of first i elements has sum j

B Sum j is impossible with first i elements

C Subset of first j elements has sum i

D Sum i is impossible with first j elements

DP[i][j] = true means a subset of the first i elements can form sum j.

Which of these is NOT a valid approach to solve Partition Equal Subset Sum?

A Greedy sorting and splitting

B Backtracking with memoization

C Dynamic programming subset sum

D Bitset optimization for subset sums

Greedy sorting and splitting does not guarantee correct partitioning for this problem.

What is the space complexity of the classic DP solution for Partition Equal Subset Sum?

A O(n)

B O(1)

C O(sum)

D O(n * sum/2)

The DP table stores boolean values for n elements and sums up to sum/2, so space is O(n * sum/2).

Explain how dynamic programming helps solve the Partition Equal Subset Sum problem.

Think about how to find if a subset with a specific sum exists.

Describe the steps to implement the Partition Equal Subset Sum solution using dynamic programming.

Start from sum calculation and end with DP table lookup.

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 into equal sum subsets [OK]

Hint: Check if total sum is even to start [OK]

Common Mistakes:

• Thinking it checks for duplicates
• Assuming sorting is required
• Confusing with subset sum existence

2. Which of the following C++ code snippets correctly initializes a boolean DP array for the Partition Equal Subset Sum problem with size (sum/2 + 1) and sets the base case?

easy

A. bool dp[sum / 2 + 1]; std::fill(dp, dp + sum / 2 + 1, false); dp[0] = true;

B. std::vector<int> dp(sum / 2 + 1, 0); dp[0] = 1;

C. std::vector<bool> dp(sum / 2 + 1, false); dp[0] = true;

D. std::vector<bool> dp(sum + 1, false); dp[sum] = true;

Solution

1. Step 1: Choose correct data type and size

   We need a boolean vector of size (sum/2 + 1) to track subset sums. std::vector<bool> dp(sum / 2 + 1, false); dp[0] = true; uses correct size and type.

2. Step 2: Set base case correctly

   dp[0] = true means sum zero is always possible. std::vector<bool> dp(sum / 2 + 1, false); dp[0] = true; does this correctly. std::vector<int> dp(sum / 2 + 1, 0); dp[0] = 1; uses int vector, which is less appropriate. std::vector<bool> dp(sum + 1, false); dp[sum] = true; uses wrong size and sets dp[sum] instead of dp[0].

3. Final Answer:

   std::vector<bool> dp(sum / 2 + 1, false); dp[0] = true; -> Option C

4. Quick Check:

   Boolean vector with dp[0] true = Correct initialization [OK]

Hint: Use bool vector and set dp[0] = true for base case [OK]

Common Mistakes:

• Using int vector instead of bool
• Incorrect dp array size
• Setting wrong base case index

3. What is the output of the following C++ code snippet for the input array {1, 5, 11, 5}?

bool canPartition(std::vector<int>& nums) {
    int sum = 0;
    for (int num : nums) sum += num;
    if (sum % 2 != 0) return false;
    int target = sum / 2;
    std::vector<bool> dp(target + 1, false);
    dp[0] = true;
    for (int num : nums) {
        for (int j = target; j >= num; --j) {
            dp[j] = dp[j] || dp[j - num];
        }
    }
    return dp[target];
}

int main() {
    std::vector<int> nums = {1, 5, 11, 5};
    std::cout << (canPartition(nums) ? "true" : "false") << std::endl;
    return 0;
}

medium

A. true

B. false

C. Compilation error

D. Runtime error

Solution

1. Step 1: Calculate total sum and check parity

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

2. Step 2: Use DP to check subset sum of 11

   The code uses a boolean dp array to check if subset sum 11 is possible. The subsets {11} and {1,5,5} both sum to 11, so dp[11] becomes true.

3. Final Answer:

   true -> Option A

4. Quick Check:

   Subset sum 11 exists = true [OK]

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

Common Mistakes:

• Assuming sum odd means true
• Not iterating dp backwards
• Confusing output with subset elements

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

bool canPartition(std::vector<int>& nums) {
    int sum = 0;
    for (int num : nums) sum += num;
    if (sum % 2 != 0) return false;
    int target = sum / 2;
    std::vector<bool> dp(target + 1, false);
    dp[0] = true;
    for (int num : nums) {
        for (int j = num; j <= target; ++j) {
            dp[j] = dp[j] || dp[j - num];
        }
    }
    return dp[target];
}

medium

A. Change sum check to if (sum % 2 == 0) return false

B. Initialize dp vector with true instead of false

C. Remove dp[0] = true assignment

D. Change inner loop to iterate backwards: for (int j = target; j >= num; --j)

Solution

1. Step 1: Analyze inner loop direction

   The inner loop iterates forwards, which causes dp to use updated values in the same iteration, leading to incorrect results.

2. Step 2: Correct loop direction

   Iterating backwards ensures dp[j - num] uses values from previous iteration, preserving correctness.

3. Final Answer:

   Change inner loop to iterate backwards: for (int j = target; j >= num; --j) -> Option D

4. Quick Check:

   Backward iteration in DP loop fixes state update [OK]

Hint: Iterate dp array backwards to avoid overwriting states [OK]

Common Mistakes:

• Iterating dp forwards causing wrong results
• Incorrect dp initialization
• Wrong sum parity check

5. Given the array {2, 2, 3, 5}, which of the following statements is true regarding the Partition Equal Subset Sum problem?

hard

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

B. The array cannot be partitioned into two subsets with equal sum because no subset sums to half the total

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

D. The array cannot be partitioned into two subsets with equal sum because total sum is odd

Solution

1. Step 1: Calculate total sum

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

2. Step 2: Check for subset with sum 6

   Half sum = 6. Possible subsets: - {2,2,3} sum = 7 (too high) - {2,3} sum = 5 (too low) - {5} sum = 5 (too low) - {2,2} sum = 4 (too low) No subset sums exactly to 6.

3. Final Answer:

   The array cannot be partitioned into two subsets with equal sum because no subset sums to half the total -> Option B

4. Quick Check:

   Half sum subset missing = No partition [OK]

Hint: Check if half sum subset exists before concluding [OK]

Common Mistakes:

• Assuming total sum even means partition possible
• Incorrect subset sums calculation
• Confusing subsets with unequal sums