Partition Equal Subset Sum in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the total sum is even.

DSA Python

total_sum = sum(nums)
if total_sum [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 '%' to check evenness.

Using '//' which is floor division, not modulo.

We use the modulo operator '%' to check if total_sum is divisible by 2 (even). If total_sum % 2 != 0, it means it's odd, so return False.

2fill in blank

medium

Complete the code to initialize the target sum for subset.

DSA Python

target = total_sum [1] 2

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

A-

B+

C//

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using '/' which returns float instead of integer.

Using '+' or '-' which are incorrect operations here.

We use floor division '//' to get the integer half of total_sum as the target sum for the subset.

3fill in blank

hard

Fix the error in the loop to iterate over nums correctly.

DSA Python

for num in [1]:
    for i in range(target, num - 1, -1):
        dp[i] = dp[i] or dp[i - num]

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

Anums

Brange(nums)

Ctarget

Ddp

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'range(nums)' which is invalid because nums is a list.

Using 'target' or 'dp' which are not iterable lists of numbers.

We need to iterate over the list 'nums' directly to access each number.

4fill in blank

hard

Fill both blanks to initialize the dp array and set the base case.

DSA Python

dp = [[1]] * (target + 1)
dp[[2]] = True

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

AFalse

BTrue

C0

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Setting dp[0] to False which breaks the base case.

Initializing dp with True which is incorrect.

We initialize dp with False values and set dp[0] = True because sum 0 is always possible with empty subset.

5fill in blank

hard

Fill all three blanks to complete the function that returns if partition is possible.

DSA Python

def canPartition(nums):
    total_sum = sum(nums)
    if total_sum [1] 2 != 0:
        return False
    target = total_sum [2] 2
    dp = [False] * (target + 1)
    dp[0] = True
    for num in nums:
        for i in range(target, num - 1, -1):
            dp[i] = dp[i] or dp[i - num]
    return dp[[3]]

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

A%

B//

Ctarget

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using '/' instead of '//' for integer division.

Returning dp[0] instead of dp[target].

Using '==' instead of '%' for odd check.

We check if total_sum % 2 != 0 to return False, use floor division '//' to get target, and return dp[target] to check if subset sum is possible.

Practice

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

easy

A. If the list contains any duplicate numbers

B. If the list can be split into two subsets with equal sums

C. If the list is sorted in ascending order

D. If the list contains only even numbers

Solution

1. Step 1: Understand the problem statement

   The problem asks if the input list can be divided into two parts where the sum of numbers in both parts is the same.

2. Step 2: Identify the key condition

   This means the total sum of the list must be even and there must be a subset with sum equal to half of the total sum.

3. Final Answer:

   If the list can be split into two subsets with equal sums -> Option B

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 list must be sorted
• Confusing with even number check

2. Which of the following Python code snippets correctly initializes a boolean list dp of size target + 1 with false values, except dp[0] set to true?

easy

A. dp = [True for _ in range(target + 1)]; dp[0] = True

B. dp = [True] * (target + 1); dp[0] = False

C. dp = [False for _ in range(target)]; dp[0] = True

D. dp = [False] * (target + 1); dp[0] = True

Solution

1. Step 1: Understand list initialization

   We want a list of size target+1 with all False except dp[0] = True.

2. Step 2: Check each option

   dp = [False] * (target + 1); dp[0] = True correctly creates a list of False of length target+1 and sets dp[0] to True.

3. Final Answer:

   dp = [False] * (target + 1); dp[0] = True -> Option D

4. Quick Check:

   Initialize dp with False, set dp[0] True [OK]

Hint: Use [False] * size and set dp[0] True manually [OK]

Common Mistakes:

• Using wrong list size (target instead of target+1)
• Setting dp[0] to False instead of True
• Using list comprehension with wrong range

3. What will be the output of the following Python code snippet?

nums = [1, 5, 11, 5]
total = sum(nums)
target = total // 2

dp = [False] * (target + 1)
dp[0] = True

for num in nums:
    for j in range(target, num - 1, -1):
        dp[j] = dp[j] or dp[j - num]

print(dp[target])

medium

A. True

B. False

C. IndexError

D. SyntaxError

Solution

1. Step 1: Calculate total and target

   Total sum = 1+5+11+5 = 22, target = 11.

2. Step 2: Trace dp updates

   dp array tracks if sum j is possible. After processing all nums, dp[11] becomes True because subset [11] or [5,5,1] sums to 11.

3. Final Answer:

   True -> Option A

4. Quick Check:

   Subset sum 11 possible = True [OK]

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

Common Mistakes:

• Forgetting to iterate backwards in inner loop
• Miscomputing total sum
• Confusing dp meaning (True means sum possible)

4. The following code attempts to solve the Partition Equal Subset Sum problem but has a bug. What is the bug?

def can_partition(nums):
    total = sum(nums)
    if total % 2 != 0:
        return False
    target = total // 2
    dp = [False] * target
    dp[0] = True
    for num in nums:
        for j in range(target, num - 1, -1):
            dp[j] = dp[j] or dp[j - num]
    return dp[target]

medium

A. dp[0] should be False initially

B. The total sum check should be for odd numbers

C. dp list size should be target + 1, not target

D. The inner loop should iterate forwards, not backwards

Solution

1. Step 1: Check dp list size

   dp list is created with size target, but indices go up to target, so size must be target + 1.

2. Step 2: Identify impact of bug

   Accessing dp[target] causes IndexError because dp size is too small.

3. Final Answer:

   dp list size should be target + 1, not target -> Option C

4. Quick Check:

   dp size = target + 1 needed [OK]

Hint: dp size must cover index target inclusive [OK]

Common Mistakes:

• Using dp size = target causes index error
• Changing loop direction incorrectly
• Setting dp[0] to False breaks base case

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

hard

A. It's not possible to partition into two subsets with equal sum

B. It's possible to partition into two subsets with equal sum: [2,2,3] and [5]

C. It's possible to partition into two subsets with equal sum: [2,3] and [2,5]

D. The problem requires subsets to have different sums

Solution

1. Step 1: Calculate total sum and target

   Total sum = 2+2+3+5 = 12, target = 6.

2. Step 2: Check possible subsets summing to 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 partition

   Since no subset sums to 6, partition into equal sum subsets is not possible.

4. Final Answer:

   It's not possible to partition into two subsets with equal sum -> Option A

5. Quick Check:

   No subset sums to half total sum = no partition [OK]

Hint: Sum list, check if half sum subset exists [OK]

Common Mistakes:

• Assuming subsets like [2,3] and [2,5] sum equally
• Ignoring that subsets must sum exactly half
• Miscounting elements in subsets