Meet In the Middle approach is particularly useful when the problem space is too large for a brute force solution but can be efficiently narrowed down by dividing it into smaller, more manageable parts.

The core idea behind the "Meet In the Middle" pattern is to split the problem into two halves and solve each half independently. The solutions of these halves are then combined or "met in the middle" to produce the final solution. By breaking down the problem, the pattern significantly reduces the time complexity compared to a full brute-force search.

Let's illustrate the "Meet In the Middle" approach with the following problem:

Problem Statement

Given a set of n integers where n <= 40, and each integer is at most 10¹², determine the maximum sum subset with a sum less than or equal to S, where S <= 10¹².

Why Not Brute Force?

• A brute-force solution would involve generating all possible subsets and checking their sums. However, with n as large as 40, the number of possible subsets would be 2ⁿ, which is approximately 10¹². This is computationally infeasible.

Step-by-Step Explanation of "Meet In the Middle" with Code Example

Let's walk through the "Meet In the Middle" algorithm using the example array int[] nums = {3, 15, 14, 9, 6, 2} with long S = 10. We'll explain each step of the algorithm and show how it is applied to this example.

Step 1: Splitting the Array into Two Halves

The first step in the "Meet In the Middle" approach is to divide the original array into two halves.

Code:

int[] nums = {3, 15, 14, 9, 6, 2};
int n = nums.length;
int mid = n / 2;

int[] left = Arrays.copyOfRange(nums, 0, mid);
int[] right = Arrays.copyOfRange(nums, mid, n);

Step 2: Generate All Possible Subset Sums for Both Halves

Next, we generate all possible subset sums for both the left and right halves.

Code:

List<Long> sumLeft = generateSubsetSums(left);  // Subsets of {3, 15, 14}
List<Long> sumRight = generateSubsetSums(right);  // Subsets of {9, 6, 2}

Step 3: Sort the Subset Sums of the Right Half

To facilitate efficient searching, we sort the sumRight list.

Code:

Collections.sort(sumRight);

• Sorted sumRight: {0, 2, 6, 8, 9, 11, 15, 17}

Explanation: Sorting sumRight allows us to efficiently search for the best complementary sum using binary search.

Step 4: Find the Maximum Subset Sum Less than or Equal to S

For each sum in sumLeft, we find the best complementary sum in sumRight such that their combined sum is less than or equal to S.

Code:

long maxSum = 0;

for (long sLeft : sumLeft) {
    if (sLeft > S) continue; // Skip if sLeft alone exceeds S

    long remaining = S - sLeft;

    int idx = upperBound(sumRight, remaining);  // Binary search in sumRight

    if (idx >= 0) {
        long sRight = sumRight.get(idx);
        long total = sLeft + sRight;
        if (total > maxSum) {
            maxSum = total;
        }
    }
}

• For each sum in sumLeft, calculate the remaining value that can be added from sumRight without exceeding S.
• Use binary search to find the largest value in sumRight that is less than or equal to the remaining value.
• Add this value to sLeft and keep track of the maximum sum found.

Code

Complexity Analysis

Time Complexity

• **Splitting the Array: O(n) for splitting the array into two halves.

• Generating Subset Sums (generateSubsetSums): O(2^(n/2) * n) for generating all subset sums for both halves.

• Sorting sumRight: O(2^(n/2) * (n/2)) simplifies to O(2^(n/2) * n) for sorting the subset sums.

• Searching for Best Complementary Sum: O(2^(n/2) * n) for binary search on sumRight for each element in sumLeft.

• Overall Time Complexity: O(2^(n/2) * n).

Space Complexity

• Storing Subset Sums: O(2^(n/2)) for storing subset sums for both halves.

• Overall Space Complexity: O(2^(n/2)).

Why Use This Technique?

• Traditional Approaches and Their Limitations

  • Brute force methods require checking all possible combinations, which can be extremely slow and inefficient, especially for problems with large input sizes.
  • In many cases like the Subset Sum problem, brute force algorithms exhibit exponential time complexity, making them infeasible for real-world applications when the input size grows beyond a certain point.

• Optimality of "Meet In the Middle"

  • "Meet In the Middle" reduces the problem size by dividing it into two smaller, more manageable parts. Instead of solving the entire problem in one go, this technique focuses on solving each half separately and then combining the results.
  • By doing so, the overall time complexity is often reduced to O(2^(n/2)), which is a significant improvement over the O(2⁽ⁿ⁾) complexity of brute force methods.
  • This approach effectively balances between brute force and efficiency, allowing for a solution that is both practical and feasible for large problems.

Scenarios Where "Meet In the Middle" Shines

• Subset Sum Problems: When dealing with large sets where the goal is to find a subset that meets a specific criterion (e.g., a subset that sums to a target value), traditional methods may be too slow. "Meet In the Middle" can cut down the problem size, making it solvable in a reasonable time.
• Combination and Permutation Problems: In cases where you need to explore combinations or permutations of elements from a large set, this technique offers a way to explore the search space more efficiently.
• Optimization Problems: For problems where you're trying to optimize a certain metric (like maximizing or minimizing a value), "Meet In the Middle" can help reduce the computational load.

Limitations

Scenarios Where MITM Might Not Be Best

• Small Input Size: For very small inputs, simpler methods like brute force or direct algorithms might be more efficient.
• High Memory Usage: If the problem requires storing a large number of subset sums, MITM may not be practical due to high memory consumption.

Potential Challenges

• Combining Results: Efficiently combining results from the two halves can be complex and may require careful implementation.
• Space Complexity: While MITM reduces time complexity, it may increase space complexity due to storage of subset sums.

Now, let's start solving problems of Meet In the Middle pattern.