Max-Min Problem
”;
Let us consider a simple problem that can be solved by divide and conquer technique.
Max-Min Problem
The Max-Min Problem in algorithm analysis is finding the maximum and minimum value in an array.
Solution
To find the maximum and minimum numbers in a given array numbers[] of size n, the following algorithm can be used. First we are representing the naive method and then we will present divide and conquer approach.
Naive Method
Naive method is a basic method to solve any problem. In this method, the maximum and minimum number can be found separately. To find the maximum and minimum numbers, the following straightforward algorithm can be used.
Algorithm: Max-Min-Element (numbers[])
max := numbers[1]
min := numbers[1]
for i = 2 to n do
if numbers[i] > max then
max := numbers[i]
if numbers[i] < min then
min := numbers[i]
return (max, min)
Example
Following are the implementations of the above approach in various programming languages −
#include <stdio.h>
struct Pair {
int max;
int min;
};
// Function to find maximum and minimum using the naive algorithm
struct Pair maxMinNaive(int arr[], int n) {
struct Pair result;
result.max = arr[0];
result.min = arr[0];
// Loop through the array to find the maximum and minimum values
for (int i = 1; i < n; i++) {
if (arr[i] > result.max) {
result.max = arr[i]; // Update the maximum value if a larger element is found
}
if (arr[i] < result.min) {
result.min = arr[i]; // Update the minimum value if a smaller element is found
}
}
return result; // Return the pair of maximum and minimum values
}
int main() {
int arr[] = {6, 4, 26, 14, 33, 64, 46};
int n = sizeof(arr) / sizeof(arr[0]);
struct Pair result = maxMinNaive(arr, n);
printf("Maximum element is: %dn", result.max);
printf("Minimum element is: %dn", result.min);
return 0;
}
Output
Maximum element is: 64 Minimum element is: 4
#include <iostream>
using namespace std;
struct Pair {
int max;
int min;
};
// Function to find maximum and minimum using the naive algorithm
Pair maxMinNaive(int arr[], int n) {
Pair result;
result.max = arr[0];
result.min = arr[0];
// Loop through the array to find the maximum and minimum values
for (int i = 1; i < n; i++) {
if (arr[i] > result.max) {
result.max = arr[i]; // Update the maximum value if a larger element is found
}
if (arr[i] < result.min) {
result.min = arr[i]; // Update the minimum value if a smaller element is found
}
}
return result; // Return the pair of maximum and minimum values
}
int main() {
int arr[] = {6, 4, 26, 14, 33, 64, 46};
int n = sizeof(arr) / sizeof(arr[0]);
Pair result = maxMinNaive(arr, n);
cout << "Maximum element is: " << result.max << endl;
cout << "Minimum element is: " << result.min << endl;
return 0;
}
Output
Maximum element is: 64 Minimum element is: 4
public class MaxMinNaive {
static class Pair {
int max;
int min;
}
// Function to find maximum and minimum using the naive algorithm
static Pair maxMinNaive(int[] arr) {
Pair result = new Pair();
result.max = arr[0];
result.min = arr[0];
// Loop through the array to find the maximum and minimum values
for (int i = 1; i < arr.length; i++) {
if (arr[i] > result.max) {
result.max = arr[i]; // Update the maximum value if a larger element is found
}
if (arr[i] < result.min) {
result.min = arr[i]; // Update the minimum value if a smaller element is found
}
}
return result; // Return the pair of maximum and minimum values
}
public static void main(String[] args) {
int[] arr = {6, 4, 26, 14, 33, 64, 46};
Pair result = maxMinNaive(arr);
System.out.println("Maximum element is: " + result.max);
System.out.println("Minimum element is: " + result.min);
}
}
Output
Maximum element is: 64 Minimum element is: 4
def max_min_naive(arr):
max_val = arr[0]
min_val = arr[0]
# Loop through the array to find the maximum and minimum values
for i in range(1, len(arr)):
if arr[i] > max_val:
max_val = arr[i] # Update the maximum value if a larger element is found
if arr[i] < min_val:
min_val = arr[i] # Update the minimum value if a smaller element is found
return max_val, min_val # Return the pair of maximum and minimum values
arr = [6, 4, 26, 14, 33, 64, 46]
max_val, min_val = max_min_naive(arr)
print("Maximum element is:", max_val)
print("Minimum element is:", min_val)
Output
Maximum element is: 64 Minimum element is: 4
Analysis
The number of comparison in Naive method is 2n – 2.
The number of comparisons can be reduced using the divide and conquer approach. Following is the technique.
Divide and Conquer Approach
In this approach, the array is divided into two halves. Then using recursive approach maximum and minimum numbers in each halves are found. Later, return the maximum of two maxima of each half and the minimum of two minima of each half.
In this given problem, the number of elements in an array is $y – x + 1$, where y is greater than or equal to x.
$mathbf{mathit{Max – Min(x, y)}}$ will return the maximum and minimum values of an array $mathbf{mathit{numbers[x…y]}}$.
Algorithm: Max - Min(x, y) if y – x ≤ 1 then return (max(numbers[x],numbers[y]),min((numbers[x],numbers[y])) else (max1, min1):= maxmin(x, ⌊((x + y)/2)⌋) (max2, min2):= maxmin(⌊((x + y)/2) + 1)⌋,y) return (max(max1, max2), min(min1, min2))
Example
Following are implementations of the above approach in various programming languages −
#include <stdio.h>
// Structure to store both maximum and minimum elements
struct Pair {
int max;
int min;
};
struct Pair maxMinDivideConquer(int arr[], int low, int high) {
struct Pair result;
struct Pair left;
struct Pair right;
int mid;
// If only one element in the array
if (low == high) {
result.max = arr[low];
result.min = arr[low];
return result;
}
// If there are two elements in the array
if (high == low + 1) {
if (arr[low] < arr[high]) {
result.min = arr[low];
result.max = arr[high];
} else {
result.min = arr[high];
result.max = arr[low];
}
return result;
}
// If there are more than two elements in the array
mid = (low + high) / 2;
left = maxMinDivideConquer(arr, low, mid);
right = maxMinDivideConquer(arr, mid + 1, high);
// Compare and get the maximum of both parts
result.max = (left.max > right.max) ? left.max : right.max;
// Compare and get the minimum of both parts
result.min = (left.min < right.min) ? left.min : right.min;
return result;
}
int main() {
int arr[] = {6, 4, 26, 14, 33, 64, 46};
int n = sizeof(arr) / sizeof(arr[0]);
struct Pair result = maxMinDivideConquer(arr, 0, n - 1);
printf("Maximum element is: %dn", result.max);
printf("Minimum element is: %dn", result.min);
return 0;
}
Output
Maximum element is: 64 Minimum element is: 4
#include <iostream>
using namespace std;
// Structure to store both maximum and minimum elements
struct Pair {
int max;
int min;
};
Pair maxMinDivideConquer(int arr[], int low, int high) {
Pair result, left, right;
int mid;
// If only one element in the array
if (low == high) {
result.max = arr[low];
result.min = arr[low];
return result;
}
// If there are two elements in the array
if (high == low + 1) {
if (arr[low] < arr[high]) {
result.min = arr[low];
result.max = arr[high];
} else {
result.min = arr[high];
result.max = arr[low];
}
return result;
}
// If there are more than two elements in the array
mid = (low + high) / 2;
left = maxMinDivideConquer(arr, low, mid);
right = maxMinDivideConquer(arr, mid + 1, high);
// Compare and get the maximum of both parts
result.max = (left.max > right.max) ? left.max : right.max;
// Compare and get the minimum of both parts
result.min = (left.min < right.min) ? left.min : right.min;
return result;
}
int main() {
int arr[] = {6, 4, 26, 14, 33, 64, 46};
int n = sizeof(arr) / sizeof(arr[0]);
Pair result = maxMinDivideConquer(arr, 0, n - 1);
cout << "Maximum element is: " << result.max << endl;
cout << "Minimum element is: " << result.min << endl;
return 0;
}
Output
Maximum element is: 64 Minimum element is: 4
public class MaxMinDivideConquer {
// Class to store both maximum and minimum elements
static class Pair {
int max;
int min;
}
static Pair maxMinDivideConquer(int[] arr, int low, int high) {
Pair result = new Pair();
Pair left, right;
int mid;
// If only one element in the array
if (low == high) {
result.max = arr[low];
result.min = arr[low];
return result;
}
// If there are two elements in the array
if (high == low + 1) {
if (arr[low] < arr[high]) {
result.min = arr[low];
result.max = arr[high];
} else {
result.min = arr[high];
result.max = arr[low];
}
return result;
}
// If there are more than two elements in the array
mid = (low + high) / 2;
left = maxMinDivideConquer(arr, low, mid);
right = maxMinDivideConquer(arr, mid + 1, high);
// Compare and get the maximum of both parts
result.max = Math.max(left.max, right.max);
// Compare and get the minimum of both parts
result.min = Math.min(left.min, right.min);
return result;
}
public static void main(String[] args) {
int[] arr = {6, 4, 26, 14, 33, 64, 46};
Pair result = maxMinDivideConquer(arr, 0, arr.length - 1);
System.out.println("Maximum element is: " + result.max);
System.out.println("Minimum element is: " + result.min);
}
}
Output
Maximum element is: 64 Minimum element is: 4
def max_min_divide_conquer(arr, low, high):
# Structure to store both maximum and minimum elements
class Pair:
def __init__(self):
self.max = 0
self.min = 0
result = Pair()
# If only one element in the array
if low == high:
result.max = arr[low]
result.min = arr[low]
return result
# If there are two elements in the array
if high == low + 1:
if arr[low] < arr[high]:
result.min = arr[low]
result.max = arr[high]
else:
result.min = arr[high]
result.max = arr[low]
return result
# If there are more than two elements in the array
mid = (low + high) // 2
left = max_min_divide_conquer(arr, low, mid)
right = max_min_divide_conquer(arr, mid + 1, high)
# Compare and get the maximum of both parts
result.max = max(left.max, right.max)
# Compare and get the minimum of both parts
result.min = min(left.min, right.min)
return result
arr = [6, 4, 26, 14, 33, 64, 46]
result = max_min_divide_conquer(arr, 0, len(arr) - 1)
print("Maximum element is:", result.max)
print("Minimum element is:", result.min)
Output
Maximum element is: 64 Minimum element is: 4
Analysis
Let T(n) be the number of comparisons made by $mathbf{mathit{Max – Min(x, y)}}$, where the number of elements $n = y – x + 1$.
If T(n) represents the numbers, then the recurrence relation can be represented as
$$T(n) = begin{cases}Tleft(lfloorfrac{n}{2}rfloorright)+Tleft(lceilfrac{n}{2}rceilright)+2 & for: n>2\1 & for:n = 2 \0 & for:n = 1end{cases}$$
Let us assume that n is in the form of power of 2. Hence, n = 2k where k is height of the recursion tree.
So,
$$T(n) = 2.T (frac{n}{2}) + 2 = 2.left(begin{array}{c}2.T(frac{n}{4}) + 2end{array}right) + 2 ….. = frac{3n}{2} – 2$$
Compared to Naïve method, in divide and conquer approach, the number of comparisons is less. However, using the asymptotic notation both of the approaches are represented by O(n).
Advertisements
”;