Partition Equal Subset Sum in DSA Python - Time & Space Complexity

Time Complexity: Partition Equal Subset Sum

O(n * sum)

Understanding Time Complexity

We want to know how the time needed to check if an array can be split into two equal-sum parts grows as the array gets bigger.

How does the number of steps change when the array size or sum increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

def can_partition(nums):
    total = sum(nums)
    if total % 2 != 0:
        return False
    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]
    return dp[target]
    

This code checks if the list can be split into two parts with equal sums using a dynamic programming approach.

Identify Repeating Operations

• Primary operation: Nested loops - outer loop over each number, inner loop over possible sums up to target.
• How many times: Outer loop runs n times (n = number of elements), inner loop runs up to target sum times.

How Execution Grows With Input

As the number of elements (n) or the target sum (half of total) grows, the steps increase roughly by multiplying these two.

Input Size (n)	Approx. Operations (n * target)
10 (sum ~ 100)	10 * 50 = 500
100 (sum ~ 1000)	100 * 500 = 50,000
1000 (sum ~ 10,000)	1000 * 5000 = 5,000,000

Pattern observation: The time grows roughly with the product of the number of elements and the target sum, so doubling either roughly doubles the work.

Final Time Complexity

Time Complexity: O(n * sum)

This means the time needed grows with both the number of elements and the total sum of the array.

Common Mistake

[X] Wrong: "The time depends only on the number of elements (n)."

[OK] Correct: The sum value also affects the time because the inner loop runs up to the target sum, so bigger sums mean more steps.

Interview Connect

Understanding this time complexity helps you explain how dynamic programming solves subset problems efficiently compared to checking all subsets.

Self-Check

"What if we used recursion with memoization instead of this bottom-up approach? How would the time complexity change?"

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