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| 1 | +# Partition Equal Subset Sum |
| 2 | + |
| 3 | +## Problem Statement |
| 4 | +Given an integer array `nums`, return `true` if you can partition the array into two subsets such that the sum of the elements in both subsets is equal, otherwise return `false`. |
| 5 | + |
| 6 | +## Examples |
| 7 | + |
| 8 | +### Example 1: |
| 9 | +**Input:** |
| 10 | +```plaintext |
| 11 | +nums = [1,5,11,5] |
| 12 | +``` |
| 13 | +**Output:** |
| 14 | +```plaintext |
| 15 | +true |
| 16 | +``` |
| 17 | +**Explanation:** |
| 18 | +The array can be partitioned as `[1, 5, 5]` and `[11]`. |
| 19 | + |
| 20 | +### Example 2: |
| 21 | +**Input:** |
| 22 | +```plaintext |
| 23 | +nums = [1,2,3,5] |
| 24 | +``` |
| 25 | +**Output:** |
| 26 | +```plaintext |
| 27 | +false |
| 28 | +``` |
| 29 | +**Explanation:** |
| 30 | +The array cannot be partitioned into equal sum subsets. |
| 31 | + |
| 32 | +## Constraints |
| 33 | +- `1 <= nums.length <= 200` |
| 34 | +- `1 <= nums[i] <= 100` |
| 35 | + |
| 36 | +## Approach |
| 37 | +The problem can be solved using **Dynamic Programming (Subset Sum Problem)**: |
| 38 | +1. Compute the total sum of the array. |
| 39 | +2. If the total sum is odd, return `false` (since it's impossible to split into two equal parts). |
| 40 | +3. Use a **0/1 Knapsack** approach to determine if a subset with sum `total_sum/2` exists. |
| 41 | +4. Use a **bottom-up DP approach** to check if we can form the required sum. |
| 42 | + |
| 43 | +## Solution |
| 44 | +The solution uses a **boolean DP array** `dp[i]`, where `dp[i]` is `true` if a subset with sum `i` can be formed. |
| 45 | +- Initialize `dp[0] = true` (sum `0` is always possible). |
| 46 | +- Iterate over each number in `nums`, updating the `dp` array in reverse order. |
| 47 | +- If `dp[target]` is `true`, return `true`, otherwise return `false`. |
| 48 | +```python |
| 49 | +class Solution: |
| 50 | + def canPartition(self, nums: List[int]) -> bool: |
| 51 | + m, mod = divmod(sum(nums), 2) |
| 52 | + if mod: |
| 53 | + return False |
| 54 | + n = len(nums) |
| 55 | + f = [[False] * (m + 1) for _ in range(n + 1)] |
| 56 | + f[0][0] = True |
| 57 | + for i, x in enumerate(nums, 1): |
| 58 | + for j in range(m + 1): |
| 59 | + f[i][j] = f[i - 1][j] or (j >= x and f[i - 1][j - x]) |
| 60 | + return f[n][m] |
| 61 | +``` |
| 62 | + |
| 63 | +## Complexity Analysis |
| 64 | +- **Time Complexity:** `O(n * sum/2)`, where `n` is the number of elements. |
| 65 | +- **Space Complexity:** `O(sum/2)`, optimized using a 1D DP array. |
| 66 | + |
| 67 | +## Usage |
| 68 | +To run the solution, use the following function: |
| 69 | +```python |
| 70 | +from solution import canPartition |
| 71 | +print(canPartition([1,5,11,5])) # Output: True |
| 72 | +print(canPartition([1,2,3,5])) # Output: False |
| 73 | +``` |
| 74 | + |
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