# Maximum and minimum of an array using minimum number of comparisons

Last Updated : 15 Apr, 2024

Given an array of size N. The task is to find the maximum and the minimum element of the array using the minimum number of comparisons.

Examples:

Input: arr[] = {3, 5, 4, 1, 9}
Output: Minimum element is: 1
Maximum element is: 9

Input: arr[] = {22, 14, 8, 17, 35, 3}
Output:  Minimum element is: 3
Maximum element is: 35

Recommended Practice

## Maximum and minimum of an array :

To solve the problem of finding the minimum and maximum elements in an array, you can follow these steps:

Step 1: Write functions to find the minimum (setmini) and maximum (setmaxi) values in the array.

Step 2: In the setmini function:

• Initialize a variable (mini) to INT_MAX.
• Iterate through the array, and if an element is smaller than the current mini, update mini to that element.
• Return the final value of mini.

Step 3: In the setmaxi function:

• Initialize a variable (maxi) to INT_MIN.
• Iterate through the array, and if an element is larger than the current maxi, update maxi to that element.
• Return the final value of maxi.

Step 4: In the main function:

• Declare an array and its size.
• Print the result ,Call the setminimum and setmaxi functions to find the minimum and maximum values.

Below is the implementation of the above approach:

C++ ```#include <iostream> #include <limits.h> using namespace std; int setmini(int A[], int N) { int mini = INT_MAX; for (int i = 0; i < N; i++) { if (A[i] < mini) { mini = A[i]; } } return mini; } int setmaxi(int A[], int N) { int maxi = INT_MIN; for (int i = 0; i < N; i++) { if (A[i] > maxi) { maxi = A[i]; } } return maxi; } int main() { int A[] = { 4, 9, 6, 5, 2, 3 }; int N = 6; cout <<"Minimum element is: " <<setmini(A, N) << endl; cout <<"Miximum element is: "<< setmaxi(A, N) << endl; } ``` Java ```import java.util.Arrays; public class MinMaxFinder { /** * Finds the minimum element in an array of integers. * * @param A The input array. * @param N The size of the array. * @return The minimum element in the array. */ public static int setmini(int[] A, int N) { int mini = Integer.MAX_VALUE; for (int i = 0; i < N; i++) { if (A[i] < mini) { mini = A[i]; } } return mini; } /** * Finds the maximum element in an array of integers. * * @param A The input array. * @param N The size of the array. * @return The maximum element in the array. */ public static int setmaxi(int[] A, int N) { int maxi = Integer.MIN_VALUE; for (int i = 0; i < N; i++) { if (A[i] > maxi) { maxi = A[i]; } } return maxi; } public static void main(String[] args) { int[] A = { 4, 9, 6, 5, 2, 3 }; int N = A.length; System.out.println("Minimum element is: " + setmini(A, N)); System.out.println("Maximum element is: " + setmaxi(A, N)); } } ``` Python3 ```# Python program to find Maximum and minimum of an array using minimum number of comparisons def set_min(A): mini = float('inf') # Initialize mini as positive infinity for num in A: if num < mini: mini = num return mini def set_max(A): maxi = float('-inf') # Initialize maxi as negative infinity for num in A: if num > maxi: maxi = num return maxi # Driver code if __name__ == "__main__": A = [4, 9, 6, 5, 2, 3] N = len(A) print("Minimum element is:", set_min(A)) print("Maximum element is:", set_max(A)) ``` JavaScript ```// JavaScript code to find the minimum and maximum elements in an array function setMini(A) { let mini = Infinity; for (let i = 0; i < A.length; i++) { if (A[i] < mini) { mini = A[i]; } } return mini; } function setMaxi(A) { let maxi = -Infinity; for (let i = 0; i < A.length; i++) { if (A[i] > maxi) { maxi = A[i]; } } return maxi; } function main() { const A = [4, 9, 6, 5, 2, 3]; console.log("Minimum element is: " + setMini(A)); console.log("Maximum element is: " + setMaxi(A)); } main(); //This code is contributed by Utkarsh. ```

Output
```Minimum element is: 2
Miximum  element is: 9

```

Time Complexity: O(N)

Auxiliary Space: O(1)

## Maximum and minimum of an array using Sorting:

One approach to find the maximum and minimum element in an array is to first sort the array in ascending order. Once the array is sorted, the first element of the array will be the minimum element and the last element of the array will be the maximum element.

Step-by-step approach:

• Initialize an array.
• Sort the array in ascending order.
• The first element of the array will be the minimum element.
• The last element of the array will be the maximum element.
• Print the minimum and maximum element.

Below is the implementation of the above approach:

C++ ```// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std; struct Pair { int min; int max; }; Pair getMinMax(int arr[], int n) { Pair minmax; sort(arr, arr + n); minmax.min = arr[0]; minmax.max = arr[n - 1]; return minmax; } int main() { int arr[] = { 1000, 11, 445, 1, 330, 3000 }; int arr_size = sizeof(arr) / sizeof(arr[0]); Pair minmax = getMinMax(arr, arr_size); cout << "Minimum element is " << minmax.min << endl; cout << "Maximum element is " << minmax.max << endl; return 0; } // This code is contributed by Tapesh(tapeshdua420) ``` Java ```import java.io.*; import java.util.*; class Pair { public int min; public int max; } class Main { static Pair getMinMax(int arr[], int n) { Pair minmax = new Pair(); Arrays.sort(arr); minmax.min = arr[0]; minmax.max = arr[n - 1]; return minmax; } public static void main(String[] args) { int arr[] = { 1000, 11, 445, 1, 330, 3000 }; int arr_size = arr.length; Pair minmax = getMinMax(arr, arr_size); System.out.println("Minimum element is " + minmax.min); System.out.println("Maximum element is " + minmax.max); } } ``` Python ```def getMinMax(arr): arr.sort() minmax = {"min": arr[0], "max": arr[-1]} return minmax arr = [1000, 11, 445, 1, 330, 3000] minmax = getMinMax(arr) print("Minimum element is", minmax["min"]) print("Maximum element is", minmax["max"]) ``` C# ```using System; public struct Pair { public int min; public int max; } class Program { // Function to find the minimum and maximum elements in // an array static Pair GetMinMax(int[] arr, int n) { Pair minmax = new Pair(); // Sort the array in ascending order Array.Sort(arr); // The first element after sorting is the minimum minmax.min = arr[0]; // The last element after sorting is the maximum minmax.max = arr[n - 1]; return minmax; } static void Main() { int[] arr = { 1000, 11, 445, 1, 330, 3000 }; int arrSize = arr.Length; // Find the minimum and maximum elements in the // array Pair minmax = GetMinMax(arr, arrSize); // Print the minimum and maximum elements Console.WriteLine("Minimum element is " + minmax.min); Console.WriteLine("Maximum element is " + minmax.max); } } ``` Javascript ```// Function to find the minimum and maximum elements in an array function getMinMax(arr) { // Create an object to store the minimum and maximum values const minmax = {}; // Sort the array in ascending order arr.sort((a, b) => a - b); // Store the minimum element as the first element of the sorted array minmax.min = arr[0]; // Store the maximum element as the last element of the sorted array minmax.max = arr[arr.length - 1]; // Return the object containing the minimum and maximum values return minmax; } // Main function function main() { // Given array const arr = [1000, 11, 445, 1, 330, 3000]; // Call the getMinMax function to find the minimum and maximum values const minmax = getMinMax(arr); // Print the minimum element console.log("Minimum element is " + minmax.min); // Print the maximum element console.log("Maximum element is " + minmax.max); } // Call the main function to start the program main(); ```

Output
```Minimum element is 1
Maximum element is 3000

```

Time complexity: O(n log n), where n is the number of elements in the array, as we are using a sorting algorithm.
Auxilary Space: is O(1), as we are not using any extra space.

### Number of Comparisons:

The number of comparisons made to find the minimum and maximum elements is equal to the number of comparisons made during the sorting process. For any comparison-based sorting algorithm, the minimum number of comparisons required to sort an array of n elements is O(n log n). Hence, the number of comparisons made in this approach is also O(n log n).

## Maximum and minimum of an array using Linear search:

Initialize values of min and max as minimum and maximum of the first two elements respectively. Starting from 3rd, compare each element with max and min, and change max and min accordingly (i.e., if the element is smaller than min then change min, else if the element is greater than max then change max, else ignore the element)

Below is the implementation of the above approach:

C++ ```// C++ program of above implementation #include<bits/stdc++.h> using namespace std; // Pair struct is used to return // two values from getMinMax() struct Pair { int min; int max; }; Pair getMinMax(int arr[], int n) { struct Pair minmax; int i; // If there is only one element // then return it as min and max both if (n == 1) { minmax.max = arr[0]; minmax.min = arr[0]; return minmax; } // If there are more than one elements, // then initialize min and max if (arr[0] > arr[1]) { minmax.max = arr[0]; minmax.min = arr[1]; } else { minmax.max = arr[1]; minmax.min = arr[0]; } for(i = 2; i < n; i++) { if (arr[i] > minmax.max) minmax.max = arr[i]; else if (arr[i] < minmax.min) minmax.min = arr[i]; } return minmax; } // Driver code int main() { int arr[] = { 1000, 11, 445, 1, 330, 3000 }; int arr_size = 6; struct Pair minmax = getMinMax(arr, arr_size); cout << "Minimum element is " << minmax.min << endl; cout << "Maximum element is " << minmax.max; return 0; } // This code is contributed by nik_3112 ``` C ```/* structure is used to return two values from minMax() */ #include<stdio.h> struct pair { int min; int max; }; struct pair getMinMax(int arr[], int n) { struct pair minmax; int i; /*If there is only one element then return it as min and max both*/ if (n == 1) { minmax.max = arr[0]; minmax.min = arr[0]; return minmax; } /* If there are more than one elements, then initialize min and max*/ if (arr[0] > arr[1]) { minmax.max = arr[0]; minmax.min = arr[1]; } else { minmax.max = arr[1]; minmax.min = arr[0]; } for (i = 2; i<n; i++) { if (arr[i] > minmax.max) minmax.max = arr[i]; else if (arr[i] < minmax.min) minmax.min = arr[i]; } return minmax; } /* Driver program to test above function */ int main() { int arr[] = {1000, 11, 445, 1, 330, 3000}; int arr_size = 6; struct pair minmax = getMinMax (arr, arr_size); printf("nMinimum element is %d", minmax.min); printf("nMaximum element is %d", minmax.max); getchar(); } ``` Java ```import java.io.*; import java.util.*; // Java program of above implementation public class GFG { /* Class Pair is used to return two values from getMinMax() */ static class Pair { int min; int max; } static Pair getMinMax(int arr[], int n) { Pair minmax = new Pair(); int i; /*If there is only one element then return it as min and max both*/ if (n == 1) { minmax.max = arr[0]; minmax.min = arr[0]; return minmax; } /* If there are more than one elements, then initialize min and max*/ if (arr[0] > arr[1]) { minmax.max = arr[0]; minmax.min = arr[1]; } else { minmax.max = arr[1]; minmax.min = arr[0]; } for (i = 2; i < n; i++) { if (arr[i] > minmax.max) { minmax.max = arr[i]; } else if (arr[i] < minmax.min) { minmax.min = arr[i]; } } return minmax; } /* Driver program to test above function */ public static void main(String args[]) { int arr[] = {1000, 11, 445, 1, 330, 3000}; int arr_size = 6; Pair minmax = getMinMax(arr, arr_size); System.out.printf("\nMinimum element is %d", minmax.min); System.out.printf("\nMaximum element is %d", minmax.max); } } ``` Python3 ```# Python program of above implementation # structure is used to return two values from minMax() class pair: def __init__(self): self.min = 0 self.max = 0 def getMinMax(arr: list, n: int) -> pair: minmax = pair() # If there is only one element then return it as min and max both if n == 1: minmax.max = arr[0] minmax.min = arr[0] return minmax # If there are more than one elements, then initialize min # and max if arr[0] > arr[1]: minmax.max = arr[0] minmax.min = arr[1] else: minmax.max = arr[1] minmax.min = arr[0] for i in range(2, n): if arr[i] > minmax.max: minmax.max = arr[i] elif arr[i] < minmax.min: minmax.min = arr[i] return minmax # Driver Code if __name__ == "__main__": arr = [1000, 11, 445, 1, 330, 3000] arr_size = 6 minmax = getMinMax(arr, arr_size) print("Minimum element is", minmax.min) print("Maximum element is", minmax.max) # This code is contributed by # sanjeev2552 ``` C# ```// C# program of above implementation using System; class GFG { /* Class Pair is used to return two values from getMinMax() */ class Pair { public int min; public int max; } static Pair getMinMax(int []arr, int n) { Pair minmax = new Pair(); int i; /* If there is only one element then return it as min and max both*/ if (n == 1) { minmax.max = arr[0]; minmax.min = arr[0]; return minmax; } /* If there are more than one elements, then initialize min and max*/ if (arr[0] > arr[1]) { minmax.max = arr[0]; minmax.min = arr[1]; } else { minmax.max = arr[1]; minmax.min = arr[0]; } for (i = 2; i < n; i++) { if (arr[i] > minmax.max) { minmax.max = arr[i]; } else if (arr[i] < minmax.min) { minmax.min = arr[i]; } } return minmax; } // Driver Code public static void Main(String []args) { int []arr = {1000, 11, 445, 1, 330, 3000}; int arr_size = 6; Pair minmax = getMinMax(arr, arr_size); Console.Write("Minimum element is {0}", minmax.min); Console.Write("\nMaximum element is {0}", minmax.max); } } // This code is contributed by PrinciRaj1992 ``` Javascript ```<script> // JavaScript program of above implementation /* Class Pair is used to return two values from getMinMax() */ function getMinMax(arr, n) { minmax = new Array(); var i; var min; var max; /*If there is only one element then return it as min and max both*/ if (n == 1) { minmax.max = arr[0]; minmax.min = arr[0]; return minmax; } /* If there are more than one elements, then initialize min and max*/ if (arr[0] > arr[1]) { minmax.max = arr[0]; minmax.min = arr[1]; } else { minmax.max = arr[1]; minmax.min = arr[0]; } for (i = 2; i < n; i++) { if (arr[i] > minmax.max) { minmax.max = arr[i]; } else if (arr[i] < minmax.min) { minmax.min = arr[i]; } } return minmax; } /* Driver program to test above function */ var arr = [1000, 11, 445, 1, 330, 3000]; var arr_size = 6; minmax = getMinMax(arr, arr_size); document.write("\nMinimum element is " ,minmax.min +"<br>"); document.write("\nMaximum element is " , minmax.max); // This code is contributed by shivanisinghss2110 </script> ```

Output
```Minimum element is 1
Maximum element is 3000

```

Time Complexity: O(n)
Auxiliary Space: O(1) as no extra space was needed.

### Number of Comparisons:

In this method, the total number of comparisons is 1 + 2*(n-2) in the worst case and 1 + (n-2) in the best case.
In the above implementation, the worst case occurs when elements are sorted in descending order and the best case occurs when elements are sorted in ascending order.

## Maximum and minimum of an array using the Tournament Method:

The idea is to divide the array into two parts and compare the maximums and minimums of the two parts to get the maximum and the minimum of the whole array.

Below is the implementation of the above approach:

C++ ```// C++ program of above implementation #include <bits/stdc++.h> using namespace std; // structure is used to return // two values from minMax() struct Pair { int min; int max; }; struct Pair getMinMax(int arr[], int low, int high) { struct Pair minmax, mml, mmr; int mid; // If there is only one element if (low == high) { minmax.max = arr[low]; minmax.min = arr[low]; return minmax; } // If there are two elements if (high == low + 1) { if (arr[low] > arr[high]) { minmax.max = arr[low]; minmax.min = arr[high]; } else { minmax.max = arr[high]; minmax.min = arr[low]; } return minmax; } // If there are more than 2 elements mid = (low + high) / 2; mml = getMinMax(arr, low, mid); mmr = getMinMax(arr, mid + 1, high); // Compare minimums of two parts if (mml.min < mmr.min) minmax.min = mml.min; else minmax.min = mmr.min; // Compare maximums of two parts if (mml.max > mmr.max) minmax.max = mml.max; else minmax.max = mmr.max; return minmax; } // Driver code int main() { int arr[] = { 1000, 11, 445, 1, 330, 3000 }; int arr_size = 6; struct Pair minmax = getMinMax(arr, 0, arr_size - 1); cout << "Minimum element is " << minmax.min << endl; cout << "Maximum element is " << minmax.max; return 0; } // This code is contributed by nik_3112 ``` C ```/* structure is used to return two values from minMax() */ #include <stdio.h> struct pair { int min; int max; }; struct pair getMinMax(int arr[], int low, int high) { struct pair minmax, mml, mmr; int mid; // If there is only one element if (low == high) { minmax.max = arr[low]; minmax.min = arr[low]; return minmax; } /* If there are two elements */ if (high == low + 1) { if (arr[low] > arr[high]) { minmax.max = arr[low]; minmax.min = arr[high]; } else { minmax.max = arr[high]; minmax.min = arr[low]; } return minmax; } /* If there are more than 2 elements */ mid = (low + high) / 2; mml = getMinMax(arr, low, mid); mmr = getMinMax(arr, mid + 1, high); /* compare minimums of two parts*/ if (mml.min < mmr.min) minmax.min = mml.min; else minmax.min = mmr.min; /* compare maximums of two parts*/ if (mml.max > mmr.max) minmax.max = mml.max; else minmax.max = mmr.max; return minmax; } /* Driver program to test above function */ int main() { int arr[] = { 1000, 11, 445, 1, 330, 3000 }; int arr_size = 6; struct pair minmax = getMinMax(arr, 0, arr_size - 1); printf("nMinimum element is %d", minmax.min); printf("nMaximum element is %d", minmax.max); getchar(); } ``` Java ```import java.io.*; import java.util.*; // Java program of above implementation public class GFG { /* Class Pair is used to return two values from * getMinMax() */ static class Pair { int min; int max; } static Pair getMinMax(int arr[], int low, int high) { Pair minmax = new Pair(); Pair mml = new Pair(); Pair mmr = new Pair(); int mid; // If there is only one element if (low == high) { minmax.max = arr[low]; minmax.min = arr[low]; return minmax; } /* If there are two elements */ if (high == low + 1) { if (arr[low] > arr[high]) { minmax.max = arr[low]; minmax.min = arr[high]; } else { minmax.max = arr[high]; minmax.min = arr[low]; } return minmax; } /* If there are more than 2 elements */ mid = (low + high) / 2; mml = getMinMax(arr, low, mid); mmr = getMinMax(arr, mid + 1, high); /* compare minimums of two parts*/ if (mml.min < mmr.min) { minmax.min = mml.min; } else { minmax.min = mmr.min; } /* compare maximums of two parts*/ if (mml.max > mmr.max) { minmax.max = mml.max; } else { minmax.max = mmr.max; } return minmax; } /* Driver program to test above function */ public static void main(String args[]) { int arr[] = { 1000, 11, 445, 1, 330, 3000 }; int arr_size = 6; Pair minmax = getMinMax(arr, 0, arr_size - 1); System.out.printf("\nMinimum element is %d", minmax.min); System.out.printf("\nMaximum element is %d", minmax.max); } } ``` Python3 ```# Python program of above implementation def getMinMax(low, high, arr): arr_max = arr[low] arr_min = arr[low] # If there is only one element if low == high: arr_max = arr[low] arr_min = arr[low] return (arr_max, arr_min) # If there is only two element elif high == low + 1: if arr[low] > arr[high]: arr_max = arr[low] arr_min = arr[high] else: arr_max = arr[high] arr_min = arr[low] return (arr_max, arr_min) else: # If there are more than 2 elements mid = int((low + high) / 2) arr_max1, arr_min1 = getMinMax(low, mid, arr) arr_max2, arr_min2 = getMinMax(mid + 1, high, arr) return (max(arr_max1, arr_max2), min(arr_min1, arr_min2)) # Driver code arr = [1000, 11, 445, 1, 330, 3000] high = len(arr) - 1 low = 0 arr_max, arr_min = getMinMax(low, high, arr) print('Minimum element is ', arr_min) print('nMaximum element is ', arr_max) # This code is contributed by DeepakChhitarka ``` C# ```// C# implementation of the approach using System; public class GFG { /* Class Pair is used to return two values from * getMinMax() */ public class Pair { public int min; public int max; } static Pair getMinMax(int[] arr, int low, int high) { Pair minmax = new Pair(); Pair mml = new Pair(); Pair mmr = new Pair(); int mid; // If there is only one element if (low == high) { minmax.max = arr[low]; minmax.min = arr[low]; return minmax; } /* If there are two elements */ if (high == low + 1) { if (arr[low] > arr[high]) { minmax.max = arr[low]; minmax.min = arr[high]; } else { minmax.max = arr[high]; minmax.min = arr[low]; } return minmax; } /* If there are more than 2 elements */ mid = (low + high) / 2; mml = getMinMax(arr, low, mid); mmr = getMinMax(arr, mid + 1, high); /* compare minimums of two parts*/ if (mml.min < mmr.min) { minmax.min = mml.min; } else { minmax.min = mmr.min; } /* compare maximums of two parts*/ if (mml.max > mmr.max) { minmax.max = mml.max; } else { minmax.max = mmr.max; } return minmax; } /* Driver program to test above function */ public static void Main(String[] args) { int[] arr = { 1000, 11, 445, 1, 330, 3000 }; int arr_size = 6; Pair minmax = getMinMax(arr, 0, arr_size - 1); Console.Write("\nMinimum element is {0}", minmax.min); Console.Write("\nMaximum element is {0}", minmax.max); } } // This code contributed by Rajput-Ji ``` Javascript ```<script> // Javascript program of above implementation /* Class Pair is used to return two values from getMinMax() */ class Pair { constructor(){ this.min = -1; this.max = 10000000; } } function getMinMax(arr , low , high) { var minmax = new Pair(); var mml = new Pair(); var mmr = new Pair(); var mid; // If there is only one element if (low == high) { minmax.max = arr[low]; minmax.min = arr[low]; return minmax; } /* If there are two elements */ if (high == low + 1) { if (arr[low] > arr[high]) { minmax.max = arr[low]; minmax.min = arr[high]; } else { minmax.max = arr[high]; minmax.min = arr[low]; } return minmax; } /* If there are more than 2 elements */ mid = parseInt((low + high) / 2); mml = getMinMax(arr, low, mid); mmr = getMinMax(arr, mid + 1, high); /* compare minimums of two parts */ if (mml.min < mmr.min) { minmax.min = mml.min; } else { minmax.min = mmr.min; } /* compare maximums of two parts */ if (mml.max > mmr.max) { minmax.max = mml.max; } else { minmax.max = mmr.max; } return minmax; } /* Driver program to test above function */ var arr = [ 1000, 11, 445, 1, 330, 3000 ]; var arr_size = 6; var minmax = getMinMax(arr, 0, arr_size - 1); document.write("\nMinimum element is ", minmax.min); document.write("<br/>Maximum element is ", minmax.max); // This code is contributed by Rajput-Ji </script> ```

Output
```Minimum element is 1
Maximum element is 3000

```

Time Complexity: O(n)
Auxiliary Space: O(log n) as the stack space will be filled for the maximum height of the tree formed during recursive calls same as a binary tree.

### Number of Comparisons:

Let the number of comparisons be T(n). T(n) can be written as follows:

T(n) = T(floor(n/2)) + T(ceil(n/2)) + 2
T(2) = 1
T(1) = 0

If n is a power of 2, then we can write T(n) as:

T(n) = 2T(n/2) + 2

After solving the above recursion, we get

T(n) = 3n/2 -2

Thus, the approach does 3n/2 -2 comparisons if n is a power of 2. And it does more than 3n/2 -2 comparisons if n is not a power of 2.

## Maximum and minimum of an array by comparing in pairs:

The idea is that when n is odd then initialize min and max as the first element. If n is even then initialize min and max as minimum and maximum of the first two elements respectively. For the rest of the elements, pick them in pairs and compare their maximum and minimum with max and min respectively.

Below is the implementation of the above approach:

C++ ```// C++ program of above implementation #include<bits/stdc++.h> using namespace std; // Structure is used to return // two values from minMax() struct Pair { int min; int max; }; struct Pair getMinMax(int arr[], int n) { struct Pair minmax; int i; // If array has even number of elements // then initialize the first two elements // as minimum and maximum if (n % 2 == 0) { if (arr[0] > arr[1]) { minmax.max = arr[0]; minmax.min = arr[1]; } else { minmax.min = arr[0]; minmax.max = arr[1]; } // Set the starting index for loop i = 2; } // If array has odd number of elements // then initialize the first element as // minimum and maximum else { minmax.min = arr[0]; minmax.max = arr[0]; // Set the starting index for loop i = 1; } // In the while loop, pick elements in // pair and compare the pair with max // and min so far while (i < n - 1) { if (arr[i] > arr[i + 1]) { if(arr[i] > minmax.max) minmax.max = arr[i]; if(arr[i + 1] < minmax.min) minmax.min = arr[i + 1]; } else { if (arr[i + 1] > minmax.max) minmax.max = arr[i + 1]; if (arr[i] < minmax.min) minmax.min = arr[i]; } // Increment the index by 2 as // two elements are processed in loop i += 2; } return minmax; } // Driver code int main() { int arr[] = { 1000, 11, 445, 1, 330, 3000 }; int arr_size = 6; Pair minmax = getMinMax(arr, arr_size); cout << "Minimum element is " << minmax.min << endl; cout << "Maximum element is " << minmax.max; return 0; } // This code is contributed by nik_3112 ``` C ```#include<stdio.h> /* structure is used to return two values from minMax() */ struct pair { int min; int max; }; struct pair getMinMax(int arr[], int n) { struct pair minmax; int i; /* If array has even number of elements then initialize the first two elements as minimum and maximum */ if (n%2 == 0) { if (arr[0] > arr[1]) { minmax.max = arr[0]; minmax.min = arr[1]; } else { minmax.min = arr[0]; minmax.max = arr[1]; } i = 2; /* set the starting index for loop */ } /* If array has odd number of elements then initialize the first element as minimum and maximum */ else { minmax.min = arr[0]; minmax.max = arr[0]; i = 1; /* set the starting index for loop */ } /* In the while loop, pick elements in pair and compare the pair with max and min so far */ while (i < n-1) { if (arr[i] > arr[i+1]) { if(arr[i] > minmax.max) minmax.max = arr[i]; if(arr[i+1] < minmax.min) minmax.min = arr[i+1]; } else { if (arr[i+1] > minmax.max) minmax.max = arr[i+1]; if (arr[i] < minmax.min) minmax.min = arr[i]; } i += 2; /* Increment the index by 2 as two elements are processed in loop */ } return minmax; } /* Driver program to test above function */ int main() { int arr[] = {1000, 11, 445, 1, 330, 3000}; int arr_size = 6; struct pair minmax = getMinMax (arr, arr_size); printf("Minimum element is %d", minmax.min); printf("\nMaximum element is %d", minmax.max); getchar(); } ``` Java ```import java.io.*; import java.util.*; // Java program of above implementation public class GFG { /* Class Pair is used to return two values from getMinMax() */ static class Pair { int min; int max; } static Pair getMinMax(int arr[], int n) { Pair minmax = new Pair(); int i; /* If array has even number of elements then initialize the first two elements as minimum and maximum */ if (n % 2 == 0) { if (arr[0] > arr[1]) { minmax.max = arr[0]; minmax.min = arr[1]; } else { minmax.min = arr[0]; minmax.max = arr[1]; } i = 2; /* set the starting index for loop */ } /* If array has odd number of elements then initialize the first element as minimum and maximum */ else { minmax.min = arr[0]; minmax.max = arr[0]; i = 1; /* set the starting index for loop */ } /* In the while loop, pick elements in pair and compare the pair with max and min so far */ while (i < n - 1) { if (arr[i] > arr[i + 1]) { if (arr[i] > minmax.max) { minmax.max = arr[i]; } if (arr[i + 1] < minmax.min) { minmax.min = arr[i + 1]; } } else { if (arr[i + 1] > minmax.max) { minmax.max = arr[i + 1]; } if (arr[i] < minmax.min) { minmax.min = arr[i]; } } i += 2; /* Increment the index by 2 as two elements are processed in loop */ } return minmax; } /* Driver program to test above function */ public static void main(String args[]) { int arr[] = {1000, 11, 445, 1, 330, 3000}; int arr_size = 6; Pair minmax = getMinMax(arr, arr_size); System.out.printf("Minimum element is %d", minmax.min); System.out.printf("\nMaximum element is %d", minmax.max); } } ``` Python3 ```# Python3 program of above implementation def getMinMax(arr): n = len(arr) # If array has even number of elements then # initialize the first two elements as minimum # and maximum if(n % 2 == 0): if arr[0] < arr[1]: mn = arr[0] mx = arr[1] else: mn = arr[1] mx = arr[0] # set the starting index for loop i = 2 # If array has odd number of elements then # initialize the first element as minimum # and maximum else: mx = mn = arr[0] # set the starting index for loop i = 1 # In the while loop, pick elements in pair and # compare the pair with max and min so far while(i < n - 1): if arr[i] < arr[i + 1]: mx = max(mx, arr[i + 1]) mn = min(mn, arr[i]) else: mx = max(mx, arr[i]) mn = min(mn, arr[i + 1]) # Increment the index by 2 as two # elements are processed in loop i += 2 return (mx, mn) # Driver Code if __name__ =='__main__': arr = [1000, 11, 445, 1, 330, 3000] mx, mn = getMinMax(arr) print("Minimum element is", mn) print("Maximum element is", mx) # This code is contributed by Kaustav ``` C# ```// C# program of above implementation using System; class GFG { /* Class Pair is used to return two values from getMinMax() */ public class Pair { public int min; public int max; } static Pair getMinMax(int []arr, int n) { Pair minmax = new Pair(); int i; /* If array has even number of elements then initialize the first two elements as minimum and maximum */ if (n % 2 == 0) { if (arr[0] > arr[1]) { minmax.max = arr[0]; minmax.min = arr[1]; } else { minmax.min = arr[0]; minmax.max = arr[1]; } i = 2; } /* set the starting index for loop */ /* If array has odd number of elements then initialize the first element as minimum and maximum */ else { minmax.min = arr[0]; minmax.max = arr[0]; i = 1; /* set the starting index for loop */ } /* In the while loop, pick elements in pair and compare the pair with max and min so far */ while (i < n - 1) { if (arr[i] > arr[i + 1]) { if (arr[i] > minmax.max) { minmax.max = arr[i]; } if (arr[i + 1] < minmax.min) { minmax.min = arr[i + 1]; } } else { if (arr[i + 1] > minmax.max) { minmax.max = arr[i + 1]; } if (arr[i] < minmax.min) { minmax.min = arr[i]; } } i += 2; /* Increment the index by 2 as two elements are processed in loop */ } return minmax; } // Driver Code public static void Main(String []args) { int []arr = {1000, 11, 445, 1, 330, 3000}; int arr_size = 6; Pair minmax = getMinMax(arr, arr_size); Console.Write("Minimum element is {0}", minmax.min); Console.Write("\nMaximum element is {0}", minmax.max); } } // This code is contributed by 29AjayKumar ``` Javascript ```<script> // JavaScript program of above implementation function getMinMax(arr){ let n = arr.length let mx,mn,i // If array has even number of elements then // initialize the first two elements as minimum // and maximum if(n % 2 == 0){ mx = Math.max(arr[0], arr[1]) mn = Math.min(arr[0], arr[1]) // set the starting index for loop i = 2 } // If array has odd number of elements then // initialize the first element as minimum // and maximum else{ mx = mn = arr[0] // set the starting index for loop i = 1 } // In the while loop, pick elements in pair and // compare the pair with max and min so far while(i < n - 1){ if(arr[i] < arr[i + 1]){ mx = Math.max(mx, arr[i + 1]) mn = Math.min(mn, arr[i]) } else{ mx = Math.max(mx, arr[i]) mn = Math.min(mn, arr[i + 1]) } // Increment the index by 2 as two // elements are processed in loop i += 2 } return [mx, mn] } // Driver Code let arr = [1000, 11, 445, 1, 330, 3000] let mx = getMinMax(arr)[0] let mn = getMinMax(arr)[1] document.write("Minimum element is", mn,"</br>") document.write("Maximum element is", mx,"</br>") // This code is contributed by shinjanpatra </script> ```

Output
```Minimum element is 1
Maximum element is 3000

```

Time Complexity: O(n)
Auxiliary Space: O(1) as no extra space was needed.

### Number of Comparisons:

The total number of comparisons: Different for even and odd n, see below:

If n is odd: 3*(n-1)/2

If n is even: 1 + 3*(n-2)/2 = 3n/2-2, 1 comparison for initializing min and max,
and 3(n-2)/2 comparisons for rest of the elements

The third and fourth approaches make an equal number of comparisons when n is a power of 2.
In general, method 4 seems to be the best.
Please write comments if you find any bug in the above programs/algorithms or a better way to solve the same problem.

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