divide and conquer based algorithm to find maximum and minimum of an array

A divide and conquer algorithm to find the maximum and minimum of an array in C++ can be implemented using the following steps:

Step 1: Define the function To begin, define a function that takes in the array and the indices for the start and end of the array as parameters. This function will be responsible for finding the maximum and minimum within the specified range.

Step 2: Check the base case Next, check if the range specified by the start and end indices contains only one element. If it does, then the maximum and minimum values are the same, which should be returned.

Step 3: Divide the array If the range contains more than one element, divide the array into two halves. Find the mid index by calculating the average of the start and end indices. This will give you the mid index that separates the array into two halves.

Step 4: Recursively find the maximum and minimum Recursively call the function on the two halves of the array, passing in the appropriate start and end indices. This will find the maximum and minimum values for each half of the array.

Step 5: Compare the maximum and minimum values After the recursive calls, you will have the maximum and minimum values for each half of the array. Compare these values to find the overall maximum and minimum for the entire array.

Step 6: Return the maximum and minimum Finally, return the maximum and minimum values found in the array.

Here is an example implementation of the divide and conquer algorithm to find the maximum and minimum of an array in C++:

#include <iostream>
#include <climits>
using namespace std;

pair<int, int> findMinMax(int arr[], int start, int end) {
    pair<int, int> minMax;
    // Base case: if there is only one element in the range
    if (start == end) {
        minMax.first = arr[start];
        minMax.second = arr[start];
        return minMax;
    }
    // Divide the array into two halves
    int mid = (start + end) / 2;
    // Recursively find the maximum and minimum for each half
    pair<int, int> leftMinMax = findMinMax(arr, start, mid);
    pair<int, int> rightMinMax = findMinMax(arr, mid + 1, end);
    // Compare the maximum and minimum values
    minMax.first = min(leftMinMax.first, rightMinMax.first);
    minMax.second = max(leftMinMax.second, rightMinMax.second);
    // Return the maximum and minimum values
    return minMax;
}

int main() {
    int arr[] = { 1, 8, 3, 9, 2, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    pair<int, int> minMax = findMinMax(arr, 0, n - 1);
    cout << "Minimum: " << minMax.first << endl;
    cout << "Maximum: " << minMax.second << endl;
    return 0;
}

This implementation divides the array into halves at each step, recursively finds the maximum and minimum for each half, and then compares the results to find the overall maximum and minimum for the entire array. The time complexity of this algorithm is O(n), where n is the number of elements in the array, as each element is compared only once.