Partition Equal Subset Sum in DSA Go - Time & Space Complexity
We want to understand 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 work change when the number of elements or their sums increase?
Analyze the time complexity of the following code snippet.
func canPartition(nums []int) bool {
total := 0
for _, num := range nums {
total += num
}
if total%2 != 0 {
return false
}
target := total / 2
dp := make([]bool, target+1)
dp[0] = true
for _, num := range nums {
for j := target; j >= num; j-- {
dp[j] = dp[j] || dp[j-num]
}
}
return dp[target]
}
This code checks if the array can be split into two parts with equal sums using a dynamic programming approach.
Identify the loops, recursion, array traversals that repeat.
- Primary operation: Nested loops where the outer loop goes through each number and the inner loop goes through possible sums up to the target.
- How many times: Outer loop runs once per element (n times), inner loop runs up to target sum (S/2) times.
The work grows with both the number of elements and the size of the target sum.
| 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 operations grow roughly by multiplying the number of elements by the half of the total sum, so it grows faster if numbers or their sums get bigger.
Time Complexity: O(n * S)
This means the time needed grows with the number of elements and the total sum of the array.
[X] Wrong: "The time depends only on the number of elements, so it is O(n)."
[OK] Correct: The time also depends on the sum of the numbers because the algorithm checks sums up to half the total sum, not just the count of elements.
Understanding this time complexity helps you explain how dynamic programming handles problems involving sums and subsets, a common pattern in coding challenges.
"What if we used recursion with memoization instead of this bottom-up approach? How would the time complexity change?"
Practice
Partition Equal Subset Sum problem ask you to find?Solution
Step 1: Understand the problem goal
The problem asks if we can divide the array into two parts where the sum of numbers in both parts is the same.Step 2: Clarify what is being checked
We are not checking duplicates, sorting, or positivity, only if such a partition exists.Final Answer:
If the array can be split into two subsets with equal sums -> Option BQuick Check:
Partition Equal Subset Sum = split array equally [OK]
- Confusing partition with sorting
- Checking for duplicates instead of sums
- Assuming all numbers must be positive
Solution
Step 1: Identify correct parameter type
The function should accept a slice of integers, so[]intis correct.Step 2: Identify correct return type
The function returns true or false, so return type must bebool.Final Answer:
func canPartition(nums []int) bool -> Option AQuick Check:
Function takes []int and returns bool [OK]
- Using int instead of []int as parameter
- Returning int instead of bool
- Using wrong slice type like []string
package main
import "fmt"
func canPartition(nums []int) bool {
total := 0
for _, num := range nums {
total += num
}
if total%2 != 0 {
return false
}
target := total / 2
dp := make([]bool, target+1)
dp[0] = true
for _, num := range nums {
for j := target; j >= num; j-- {
dp[j] = dp[j] || dp[j-num]
}
}
return dp[target]
}
func main() {
fmt.Println(canPartition([]int{1, 5, 11, 5}))
fmt.Println(canPartition([]int{1, 2, 3, 5}))
}Solution
Step 1: Calculate total sums and check parity
For [1,5,11,5], sum=22 (even), for [1,2,3,5], sum=11 (odd). Odd sum means no equal partition.Step 2: Use dynamic programming to check subset sum
For first array, dp[target] becomes true, so returns true. For second, returns false.Final Answer:
true false -> Option CQuick Check:
Even sum array returns true, odd sum returns false [OK]
- Ignoring sum parity check
- Incorrect DP update order
- Returning wrong boolean values
func canPartition(nums []int) bool {
total := 0
for _, num := range nums {
total += num
}
if total%2 == 1 {
return false
}
target := total / 2
dp := make([]bool, target+1)
dp[0] = true
for _, num := range nums {
for j := num; j <= target; j++ {
dp[j] = dp[j] || dp[j-num]
}
}
return dp[target]
}Solution
Step 1: Analyze the inner loop direction
The inner loop runs from num to target increasing, which causes dp to use updated values in the same iteration, leading to incorrect results.Step 2: Correct loop direction for subset sum DP
The loop should run backwards from target down to num to avoid using updated dp values prematurely.Final Answer:
The inner loop should run backwards from target to num -> Option AQuick Check:
Loop backwards to avoid dp overwrite [OK]
- Looping forwards in DP causing wrong results
- Incorrect parity check condition
- Wrong dp slice size
- Wrong dp[0] initialization
[2, 2, 3, 5], which of the following modifications to the Partition Equal Subset Sum algorithm will correctly identify if the array can be partitioned into two subsets with equal sum, considering that the array may contain zero or negative numbers?Solution
Step 1: Understand the limitation of standard DP
The standard DP approach assumes all numbers are positive, so negative numbers break the indexing logic.Step 2: Modify algorithm for negative numbers
Offset sums to make all numbers non-negative and use a 2D DP array to track possible sums including negatives.Step 3: Check if partition is possible with adjusted DP
This approach correctly handles zero and negative values, unlike ignoring or sorting.Final Answer:
Adjust the algorithm to handle negative numbers by offsetting sums and using a 2D DP array -> Option DQuick Check:
Handle negatives with offset and 2D DP [OK]
- Ignoring negative numbers causes wrong results
- Sorting does not fix negative sum indexing
- Greedy approach fails for equal partition
Ready to prove it?
Pass the test and this pattern moves unknown
