Optimal Merge Pattern

The optimal merge pattern refers to a way of scheduling merge passes in merge sort to minimize the number of merge passes required. It exploits the fact that merged runs can be merged independently.

The optimal pattern merges in powers of two:

  1. Merge sublists of size 1 into sorted sublists of size 2
  2. Merge sublists of size 2 into sorted sublists of size 4
  3. Merge sublists of size 4 into sorted sublists of size 8
  4. Continue until only 1 sorted sublist remains

This takes log N merge passes for a list of length N.

Java example:

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void mergeSort(int[] arr) {
  for (int size = 1; size < arr.length; size *= 2) {
    for (int left = 0; left < arr.length; left += 2*size) {
      int mid = left + size - 1;
      int right = Math.min(left + 2*size - 1, arr.length-1);

      merge(arr, left, mid, right); 
    }
  }
}

void merge(int[] arr, int left, int mid, int right) {
  // Merge sorted ranges arr[left..mid] and arr[mid+1..right] 
}

C++ example:

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void mergeSort(vector<int>& arr) {
  for (int size = 1; size < arr.size(); size *= 2) {
    for (int left = 0; left < arr.size(); left += 2*size) {
      int mid = left + size - 1;
      int right = min(left + 2*size - 1, (int)arr.size()-1);
      
      merge(arr, left, mid, right);
    }
  }
} 

void merge(vector<int>& arr, int left, int mid, int right) {
  // Merging logic
}

Python example:

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def merge_sort(arr):
  size = 1
  while size < len(arr):
    left = 0
    while left < len(arr):
      mid = left + size - 1
      right = min(left + 2*size - 1, len(arr)-1)

      merge(arr, left, mid, right)  
      left += 2*size

    size *= 2

def merge(arr, left, mid, right):
  # Merge logic

The optimal merge pattern minimizes expensive merge passes in merge sort.

The optimal merge pattern is an algorithm design technique used in divide-and-conquer algorithms. It works by finding the optimal substructure of a problem and using recursion to repeatedly merge solutions to subproblems.

Key aspects:

  • Break problem into non-overlapping subproblems
  • Solve subproblems recursively
  • Find optimal way to merge solutions to subproblems
  • Continue merging optimally until the whole problem is solved

This pattern is commonly applied in algorithms like merge sort, binary search, Strassen’s matrix multiplication etc.

Example in Java:

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// Merge sort using optimal merge pattern

void mergeSort(int[] arr) {
  if (arr.length <= 1) return;

  // Divide
  int[] left = Arrays.copyOfRange(arr, 0, arr.length/2);
  int[] right = Arrays.copyOfRange(arr, arr.length/2, arr.length);

  // Conquer
  mergeSort(left); 
  mergeSort(right);

  // Merge results optimally
  merge(left, right, arr); 
}

void merge(int[] left, int[] right, int[] result) {
  // TODO: Merge left and right optimally into result 
}

This decomposes the problem into smaller subproblems, solves them recursively, and merges the solutions optimally.

The optimal merge pattern can be similarly applied in C++, Python for divide-and-conquer algorithms.

The Optimal Merge Pattern problem is commonly encountered in scenarios where we have multiple sorted lists, and we need to merge them into one sorted list. The catch is that merging lists of different sizes costs differently. The cost of merging two lists is the sum of all elements in those lists. The objective is to minimize the cost of merging all the lists.

For example, consider you have three sorted arrays [1, 2], [2, 3], and [4]. The cost of merging the first two arrays is (1+2+2+3) = 8. Now we merge [8] and [4] for a cost of 8+4=12. The total cost will be 8+12=20.

Example Code

Java

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import java.util.PriorityQueue;

public class OptimalMergePattern {
    public static int findMinimumCost(int[] arr) {
        PriorityQueue<Integer> pq = new PriorityQueue<>();
        for (int x : arr) {
            pq.add(x);
        }

        int res = 0;
        while (pq.size() > 1) {
            int first = pq.poll();
            int second = pq.poll();
            int temp = first + second;
            res += temp;
            pq.add(temp);
        }
        return res;
    }

    public static void main(String[] args) {
        int[] arr = {1, 2, 3, 4, 5};
        System.out.println("Minimum merge cost is: " + findMinimumCost(arr));
    }
}

C++

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#include <iostream>
#include <queue>
#include <vector>

int findMinimumCost(std::vector<int> arr) {
    std::priority_queue<int, std::vector<int>, std::greater<int>> pq(arr.begin(), arr.end());
    int res = 0;
    while (pq.size() > 1) {
        int first = pq.top(); pq.pop();
        int second = pq.top(); pq.pop();
        int temp = first + second;
        res += temp;
        pq.push(temp);
    }
    return res;
}

int main() {
    std::vector<int> arr = {1, 2, 3, 4, 5};
    std::cout << "Minimum merge cost is: " << findMinimumCost(arr) << std::endl;
    return 0;
}

Python

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import heapq

def find_minimum_cost(arr):
    heapq.heapify(arr)
    res = 0
    while len(arr) > 1:
        first = heapq.heappop(arr)
        second = heapq.heappop(arr)
        temp = first + second
        res += temp
        heapq.heappush(arr, temp)
    return res

if __name__ == "__main__":
    arr = [1, 2, 3, 4, 5]
    print("Minimum merge cost is:", find_minimum_cost(arr))

In all three languages, we use a min-heap to keep track of the smallest elements. We pop two smallest elements, merge them, and push the sum back into the heap until only one element remains. The cost of all merges is calculated and returned.