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];
}

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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;
}

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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]);
}

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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];
}

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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];
        }
    }
}

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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]).