Partition Equal Subset Sum – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

Given a non-empty array nums containing only positive integers, find if the array can be partitioned into two subsets such that the sum of elements in both subsets is equal.

Example 1:

Input: nums = [1,5,11,5]
Output: true
Explanation: The array can be partitioned as [1, 5, 5] and [11].

Example 2:

Input: nums = [1,2,3,5]
Output: false
Explanation: The array cannot be partitioned into equal sum subsets.

Constraints:

• 1 <= nums.length <= 200
• 1 <= nums[i] <= 100

C++ Partition Equal Subset Sum LeetCode Solution

class Solution {
public:
    bool canPartition(vector<int>& nums) {
        const int MAX_NUM = 100;
        const int MAX_ARRAY_SIZE = 200;
        bitset<MAX_NUM * MAX_ARRAY_SIZE / 2 + 1> bits(1);
        int sum = 0;
        for (auto n : nums) {
            sum += n;
            bits |= bits << n;
        }
        return !(sum % 2) && bits[sum / 2];
    }
};

Java Partition Equal Subset Sum LeetCode Solution

public class Solution {
    public boolean canPartition(int[] nums) {
        // check edge case
        if (nums == null || nums.length == 0) {
            return true;
        }
        // preprocess
        int volumn = 0;
        for (int num : nums) {
            volumn += num;
        }
        if (volumn % 2 != 0) {
            return false;
        }
        volumn /= 2;
        // dp def
        boolean[] dp = new boolean[volumn + 1];
        // dp init
        dp[0] = true;
        // dp transition
        for (int i = 1; i <= nums.length; i++) {
            for (int j = volumn; j >= nums[i-1]; j--) {
                dp[j] = dp[j] || dp[j - nums[i-1]];
            }
        }
        return dp[volumn];
    }
}
 

Python 3 Partition Equal Subset Sum LeetCode Solution

def canPartition(nums):
	s, n, memo = sum(nums), len(nums), {0: True}
        if s & 1: return False
        nums.sort(reverse=True)
        def dfs(i, x):
            if x not in memo:
                memo[x] = False
                if x > 0:
                    for j in range(i, n):
                        if dfs(j+1, x-nums[j]):
                            memo[x] = True
                            break
            return memo[x]
        return dfs(0, s >> 1)

Array-1180
String-562
Hash Table-412
Dynamic Programming-390
Math-368
Sorting-264

Greedy-257
Depth-First Search-256
Database-215
Breadth-First Search-200
Tree-195

Binary Search-191
Matrix-176
Binary Tree-160
Two Pointers-151
Bit Manipulation-140

Stack-133
Heap (Priority Queue)-117
Design-116
Graph-108
Simulation-103
Prefix Sum-96

Backtracking-92
Counting-86
Sliding Window-73
Linked List-69
Union Find-66

Ordered Set-48
Monotonic Stack-47
Recursion-43
Trie-41
Binary Search Tree-40
Divide and Conquer-40

Enumeration-39
Bitmask-37
Queue-33
Memoization-32
Topological Sort-31

Geometry-30
Segment Tree-27
Game Theory-24
Hash Function-24
Binary Indexed Tree-21

Interactive-18
Data Stream-17
String Matching-17
Rolling Hash-17
Shortest Path-16

Number Theory-16
Combinatorics-15
Randomized-12
Monotonic Queue-9
Iterator-9
Merge Sort-9

Concurrency-9
Doubly-Linked List-8
Brainteaser-8
Probability and Statistics-7
Quickselect-7

Bucket Sort-6
Suffix Array-6
Minimum Spanning Tree-5
Counting Sort-5
Shell-4

Line Sweep-4
Reservoir Sampling-4
Eulerian Circuit-3
Radix Sort-3

Strongly Connected Componen-t2
Rejection Sampling-2
Biconnected Component-1

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