# 1. Introduction

In this blog post, we explore the Equal Subset Sum Partition problem, which is a common problem in dynamic programming. The challenge is to determine whether a given array of positive integers can be partitioned into two subsets with equal sums.

## Problem

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.```

# 2. Solution Steps

1. Calculate the total sum of the array. If the sum is odd, return false, as it cannot be partitioned into two equal subsets.

2. Find if there is a subset with a sum equal to half of the total sum.

3. Use dynamic programming to solve the subset sum problem.

4. Create a boolean 2D array dp where dp[i][j] is true if there is a subset of the first i numbers that adds up to j.

5. Fill the dp table using the given set of numbers.

6. Return the value in dp[n][half of sum].

# 3. Code Program

``````public class Solution {

// Main method for testing
public static void main(String[] args) {
int[] nums1 = {1, 5, 11, 5};
System.out.println("Can partition: " + canPartition(nums1));

int[] nums2 = {1, 2, 3, 5};
System.out.println("Can partition: " + canPartition(nums2));
}

// Method to check if the array can be partitioned into two subsets with equal sum
public static boolean canPartition(int[] nums) {
int totalSum = 0;
for (int num : nums) {
totalSum += num;
}

// If totalSum is odd, cannot partition
if (totalSum % 2 != 0) return false;

int subsetSum = totalSum / 2;
boolean[][] dp = new boolean[nums.length + 1][subsetSum + 1];
dp[0][0] = true;

for (int i = 1; i <= nums.length; i++) {
int curr = nums[i - 1];
for (int j = 0; j <= subsetSum; j++) {
if (j < curr) {
dp[i][j] = dp[i - 1][j];
} else {
dp[i][j] = dp[i - 1][j] || dp[i - 1][j - curr];
}
}
}
return dp[nums.length][subsetSum];
}
}
``````

### Output:

```Can partition: true
Can partition: false
```

### Explanation:

In the first example, the array [1, 5, 11, 5] can be partitioned into two subsets [1, 5, 5] and [11], both of which sum up to 11.

The canPartition method first checks if the total sum is even. It then uses dynamic programming to find if there is a subset with a sum equal to half of the total sum.

The second example returns false as the array [1, 2, 3, 5] cannot be partitioned into subsets with equal sums.