416. Partition Equal Subset Sum

Difficulty: Medium

Topics: Array, Dynamic Programming

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 or false otherwise.

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

Solution:

We need to determine if a given integer array can be partitioned into two subsets such that the sum of the elements in both subsets is equal. This can be efficiently achieved using a dynamic programming approach.

Approach

  1. Check Total Sum: First, calculate the total sum of the array. If the sum is odd, it's impossible to split it into two equal parts, so return false immediately.
  2. Target Subset Sum: If the total sum is even, each subset must sum to half of the total sum. We need to check if there exists a subset of the array that sums up to this target value.
  3. Dynamic Programming: Use a dynamic programming (DP) approach to determine if a subset with the target sum exists. We maintain a boolean array dp where dp[j] indicates whether a subset with sum j can be formed. Initialize dp[0] to true because a sum of 0 is always possible. For each number in the array, update the dp array from the target sum down to the number's value to avoid reusing the same element multiple times.

Let's implement this solution in PHP: 416. Partition Equal Subset Sum

/**
 * @param Integer[] $nums
 * @return Boolean
 */
function canPartition($nums) {
    ...
    ...
    ...
    /**
     * go to ./solution.php
     */
}

// Example Test Cases
var_dump(canPartition([1, 5, 11, 5])); // true
var_dump(canPartition([1, 2, 3, 5]));  // false
?>

Explanation:

  1. Initial Check: The total sum of the array is calculated. If it's odd, return false immediately as it's impossible to split an odd sum into two equal parts.
  2. Target Calculation: If the sum is even, compute the target sum as half of the total sum.
  3. DP Array Initialization: Initialize a boolean array dp where dp[j] indicates if sum j can be formed. Initially, only dp[0] is true because a sum of 0 is always possible.
  4. Updating DP Array: For each number in the array, update the dp array from the target sum down to the number's value. This ensures that each number is only considered once per iteration, preventing reuse in the same subset.
  5. Result Check: Finally, check if dp[target] is true, indicating that a subset with the target sum exists.

This approach efficiently checks for the possibility of partitioning the array using dynamic programming, ensuring optimal time and space complexity.

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