Partition Equal Subset Sum in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to calculate the total sum of the array elements.

DSA C

int totalSum = 0;
for (int i = 0; i < n; i++) {
    totalSum [1] nums[i];
}

Drag options to blanks, or click blank then click option'

A+=

B-=

C*=

D/=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '-=' instead of '+=' causes subtraction instead of addition.

Using '*=' or '/=' will not accumulate the sum correctly.

We add each element to totalSum using the '+=' operator to get the total sum.

2fill in blank

medium

Complete the code to check if the total sum is odd, which means partition is impossible.

DSA C

if (totalSum [1] 2 != 0) {
    return false;
}

Drag options to blanks, or click blank then click option'

A%

B/

C==

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '/' instead of '%' will not check odd/even correctly.

Using '==' or '!=' without modulus will cause logic errors.

The modulus operator '%' checks if totalSum is odd by checking remainder when divided by 2.

3fill in blank

hard

Fix the error in the recursive function call to check subset sum.

DSA C

bool canPartition(int* nums, int n, int sum) {
    if (sum == 0) return true;
    if (n == 0 || sum < 0) return false;
    return canPartition(nums, n - 1, sum) || canPartition(nums, n - 1, [1] - nums[n - 1]);
}

Drag options to blanks, or click blank then click option'

Asum - nums[n - 1]

Bsum + nums[n - 1]

Cn

Dsum

Attempts:

3 left

💡 Hint

Common Mistakes

Adding nums[n-1] instead of subtracting causes wrong sum calculation.

Passing 'n' instead of 'sum' causes type mismatch.

The first recursive call keeps sum unchanged, the second subtracts nums[n-1] from sum.

4fill in blank

hard

Fill both blanks to create a dynamic programming table initialization for subset sum.

DSA C

bool dp[n + 1][sum + 1];
for (int i = 0; i <= n; i++) {
    dp[i][0] = [1];
}
for (int j = 1; j <= sum; j++) {
    dp[0][j] = [2];
}

Drag options to blanks, or click blank then click option'

Atrue

Bfalse

C0

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Setting dp[i][0] to false breaks the base case.

Setting dp[0][j] to true allows impossible sums.

Sum 0 can always be formed (true), but with 0 elements no positive sum can be formed (false).

5fill in blank

hard

Fill the blank to complete the DP state update for subset sum problem.

DSA C

for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= sum; j++) {
        if (nums[i - 1] <= j) {
            dp[i][j] = dp[i - 1][j] || dp[i - 1][[1]];
        } else {
            dp[i][j] = dp[i - 1][j];
        }
    }
}

Drag options to blanks, or click blank then click option'

Ai

Bj + nums[i - 1]

Cj - nums[i - 1]

Di - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Adding nums[i-1] instead of subtracting causes wrong indexing.

Using wrong indices for dp causes logic errors.

We check if sum j can be formed without current number or by including it (sum j - nums[i-1]).

Practice

1. What does the Partition Equal Subset Sum problem ask you to find?

easy

A. If the array contains any duplicate elements

B. If the array is sorted in ascending order

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

D. If the array contains only positive numbers

Solution

1. Step 1: Understand the problem statement

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

2. Step 2: Identify the key requirement

   We need to check if there exists a subset whose sum is half of the total sum of the array.

3. Final Answer:

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

4. Quick Check:

   Partition Equal Subset Sum = Split array equally [OK]

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

Common Mistakes:

• Thinking the problem is about sorting
• Checking for duplicates instead of sums
• Assuming all numbers must be positive

2. Which of the following is the correct way to declare a boolean array in C for dynamic programming in Partition Equal Subset Sum?

easy

A. boolean dp[sum+1];

B. int dp[sum+1];

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

D. bool dp[sum+1];

Solution

1. Step 1: Identify correct C syntax for boolean arrays

   In C, bool is available via stdbool.h and arrays are declared as bool dp[size];.

2. Step 2: Check options for correctness

   bool dp[sum+1]; uses correct syntax. int dp[sum+1]; uses int, which is not boolean. bool dp = new bool[sum+1]; uses Java/C++ style dynamic allocation, not valid in C. boolean dp[sum+1]; uses 'boolean' which is not a C keyword.

3. Final Answer:

   bool dp[sum+1]; -> Option D

4. Quick Check:

   Use bool dp[size]; in C [OK]

Hint: Use stdbool.h and declare bool arrays with bool dp[size]; [OK]

Common Mistakes:

• Using 'boolean' instead of 'bool' in C
• Trying Java-style dynamic allocation in C
• Using int instead of bool for DP array

3. What will be the output of the following code snippet for array nums = {1, 5, 11, 5}?

#include 
#include 

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

int main() {
    int nums[] = {1, 5, 11, 5};
    printf("%d\n", canPartition(nums, 4));
    return 0;
}

medium

A. 1

B. 0

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 equals 11

   The DP array tracks sums achievable. The subset {11} and {1,5,5} both sum to 11, so dp[11] becomes true.

3. Final Answer:

   1 -> Option A

4. Quick Check:

   Subset sum exists = 1 [OK]

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

Common Mistakes:

• Not checking if total sum is even
• Incorrect DP initialization
• Looping forward instead of backward in DP

4. Identify the bug in this Partition Equal Subset Sum code snippet:

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

medium

A. Loop initializing dp should run till target inclusive

B. The sum calculation is incorrect

C. The inner loop should run forward, not backward

D. dp[0] should be false initially

Solution

1. Step 1: Check dp array initialization loop

   The loop runs from 0 to target-1, missing dp[target] initialization, which should be set to false.

2. Step 2: Correct the loop to include target index

   Change loop condition to i <= target to initialize all dp elements properly.

3. Final Answer:

   Loop initializing dp should run till target inclusive -> Option A

4. Quick Check:

   Initialize dp array fully including dp[target] [OK]

Hint: Initialize dp array fully including last index [OK]

Common Mistakes:

• Forgetting to initialize dp[target]
• Running inner loop forward causing wrong results
• Setting dp[0] to false instead of true

5. Given an array nums = {2, 2, 3, 5}, which of the following modifications will correctly determine if it can be partitioned into two subsets with equal sum using dynamic programming?

hard

A. Change the DP array to track counts of subsets instead of booleans

B. Use a 2D DP array where dp[i][j] means if sum j is possible with first i elements

C. Sort the array first and then apply the existing 1D DP approach

D. Only check if the largest element is less than half the total sum

Solution

1. Step 1: Understand limitations of 1D DP for some cases

   1D DP works but 2D DP is clearer and handles subset sums by tracking elements used.

2. Step 2: Use 2D DP where dp[i][j] shows if sum j is possible using first i elements

   This approach explicitly considers each element and sum, making it easier to debug and understand.

3. Step 3: Sorting or counting subsets is not necessary or sufficient

   Sorting doesn't affect subset sums. Counting subsets is more complex and not required for yes/no answer.

4. Final Answer:

   Use a 2D DP array where dp[i][j] means if sum j is possible with first i elements -> Option B

5. Quick Check:

   2D DP tracks sums per element [OK]

Hint: Use 2D DP for clear subset sum tracking [OK]

Common Mistakes:

• Sorting array expecting it to solve partition
• Counting subsets instead of checking existence
• Only checking largest element against half sum