Recursive Programs to find Minimum and Maximum elements of array
Last Updated :
19 Sep, 2023
Given an array of integers arr, the task is to find the minimum and maximum element of that array using recursion.
Examples :
Input: arr = {1, 4, 3, -5, -4, 8, 6};
Output: min = -5, max = 8
Input: arr = {1, 4, 45, 6, 10, -8};
Output: min = -8, max = 45
Recursive approach to find the Minimum element in the array
Approach:
- Get the array for which the minimum is to be found
- Recursively find the minimum according to the following:
- Recursively traverse the array from the end
- Base case: If the remaining array is of length 1, return the only present element i.e. arr[0]
if(n == 1)
return arr[0];
- Recursive call: If the base case is not met, then call the function by passing the array of one size less from the end, i.e. from arr[0] to arr[n-1].
- Return statement: At each recursive call (except for the base case), return the minimum of the last element of the current array (i.e. arr[n-1]) and the element returned from the previous recursive call.
return min(arr[n-1], recursive_function(arr, n-1));
- Print the returned element from the recursive function as the minimum element
Pseudocode for Recursive function:
If there is single element, return it.
Else return minimum of following.
a) Last Element
b) Value returned by recursive call
for n-1 elements.
Below is the implementation of the above approach:
C++
// Recursive C++ program to find minimum
#include <iostream>
using namespace std;
// function to print Minimum element using recursion
int findMinRec(int A[], int n)
{
// if size = 0 means whole array has been traversed
if (n == 1)
return A[0];
return min(A[n-1], findMinRec(A, n-1));
}
// driver code to test above function
int main()
{
int A[] = {1, 4, 45, 6, -50, 10, 2};
int n = sizeof(A)/sizeof(A[0]);
cout << findMinRec(A, n);
return 0;
}
// Recursive Java program to find minimum
import java.util.*;
class GFG {
// function to return minimum element using recursion
public static int findMinRec(int A[], int n)
{
// if size = 0 means whole array
// has been traversed
if(n == 1)
return A[0];
return Math.min(A[n-1], findMinRec(A, n-1));
}
// Driver code
public static void main(String args[])
{
int A[] = {1, 4, 45, 6, -50, 10, 2};
int n = A.length;
// Function calling
System.out.println(findMinRec(A, n));
}
}
//This code is contributed by Niraj_Pandey
# Recursive python 3 program to
# find minimum
# function to print Minimum element
# using recursion
def findMinRec(A, n):
# if size = 0 means whole array
# has been traversed
if (n == 1):
return A[0]
return min(A[n - 1], findMinRec(A, n - 1))
# Driver Code
if __name__ == '__main__':
A = [1, 4, 45, 6, -50, 10, 2]
n = len(A)
print(findMinRec(A, n))
# This code is contributed by
# Shashank_Sharma
// Recursive C# program to find minimum
using System;
class GFG
{
// function to return minimum
// element using recursion
public static int findMinRec(int []A,
int n)
{
// if size = 0 means whole array
// has been traversed
if(n == 1)
return A[0];
return Math.Min(A[n - 1],
findMinRec(A, n - 1));
}
// Driver code
static public void Main ()
{
int []A = {1, 4, 45, 6, -50, 10, 2};
int n = A.Length;
// Function calling
Console.WriteLine(findMinRec(A, n));
}
}
// This code is contributed by Sachin
<?php
// Recursive PHP program to find minimum
// function to print Minimum
// element using recursion
function findMinRec($A, $n)
{
// if size = 0 means whole
// array has been traversed
if ($n == 1)
return $A[0];
return min($A[$n - 1], findMinRec($A, $n - 1));
}
// Driver Code
$A = array (1, 4, 45, 6, -50, 10, 2);
$n = sizeof($A);
echo findMinRec($A, $n);
// This code is contributed by akt
?>
<script>
// Javascript program to find minimum
// Function to print Minimum
// element using recursion
function findMinRec(A, n)
{
// If size = 0 means whole
// array has been traversed
if (n == 1)
return A[0];
return Math.min(A[n - 1],
findMinRec(A, n - 1));
}
// Driver Code
let A = [ 1, 4, 45, 6, -50, 10, 2 ];
let n = A.length;
document.write( findMinRec(A, n));
// This code is contributed by sravan kumar G
</script>
Recursive approach to find the Maximum element in the array
Approach:
- Get the array for which the maximum is to be found
- Recursively find the maximum according to the following:
- Recursively traverse the array from the end
- Base case: If the remaining array is of length 1, return the only present element i.e. arr[0]
if(n == 1)
return arr[0];
- Recursive call: If the base case is not met, then call the function by passing the array of one size less from the end, i.e. from arr[0] to arr[n-1].
- Return statement: At each recursive call (except for the base case), return the maximum of the last element of the current array (i.e. arr[n-1]) and the element returned from the previous recursive call.
return max(arr[n-1], recursive_function(arr, n-1));
- Print the returned element from the recursive function as the maximum element
Pseudocode for Recursive function:
If there is single element, return it.
Else return maximum of following.
a) Last Element
b) Value returned by recursive call
for n-1 elements.
Below is the implementation of the above approach:
C++
// Recursive C++ program to find maximum
#include <iostream>
using namespace std;
// function to return maximum element using recursion
int findMaxRec(int A[], int n)
{
// if n = 0 means whole array has been traversed
if (n == 1)
return A[0];
return max(A[n-1], findMaxRec(A, n-1));
}
// driver code to test above function
int main()
{
int A[] = {1, 4, 45, 6, -50, 10, 2};
int n = sizeof(A)/sizeof(A[0]);
cout << findMaxRec(A, n);
return 0;
}
// Recursive Java program to find maximum
import java.util.*;
class GFG {
// function to return maximum element using recursion
public static int findMaxRec(int A[], int n)
{
// if size = 0 means whole array
// has been traversed
if(n == 1)
return A[0];
return Math.max(A[n-1], findMaxRec(A, n-1));
}
// Driver code
public static void main(String args[])
{
int A[] = {1, 4, 45, 6, -50, 10, 2};
int n = A.length;
// Function calling
System.out.println(findMaxRec(A, n));
}
}
//This code is contributed by Niraj_Pandey
# Recursive Python 3 program to
# find maximum
# function to return maximum element
# using recursion
def findMaxRec(A, n):
# if n = 0 means whole array
# has been traversed
if (n == 1):
return A[0]
return max(A[n - 1], findMaxRec(A, n - 1))
# Driver Code
if __name__ == "__main__":
A = [1, 4, 45, 6, -50, 10, 2]
n = len(A)
print(findMaxRec(A, n))
# This code is contributed by ita_c
// Recursive C# program to find maximum
using System;
class GFG
{
// function to return maximum
// element using recursion
public static int findMaxRec(int []A,
int n)
{
// if size = 0 means whole array
// has been traversed
if(n == 1)
return A[0];
return Math.Max(A[n - 1],
findMaxRec(A, n - 1));
}
// Driver code
static public void Main ()
{
int []A = {1, 4, 45, 6, -50, 10, 2};
int n = A.Length;
// Function calling
Console.WriteLine(findMaxRec(A, n));
}
}
// This code is contributed by Sach_Code
<?php
// Recursive PHP program to find maximum
// function to return maximum
// element using recursion
function findMaxRec($A, $n)
{
// if n = 0 means whole array
// has been traversed
if ($n == 1)
return $A[0];
return max($A[$n - 1],
findMaxRec($A, $n - 1));
}
// Driver Code
$A = array(1, 4, 45, 6, -50, 10, 2);
$n = sizeof($A);
echo findMaxRec($A, $n);
// This code is contributed
// by Akanksha Rai
?>
<script>
// Recursive Java program to find maximum
// Function to return maximum element
// using recursion
function findMaxRec(A, n)
{
// If size = 0 means whole array
// has been traversed
if (n == 1)
return A[0];
return Math.max(A[n - 1],
findMaxRec(A, n - 1));
}
// Driver code
let A = [ 1, 4, 45, 6, -50, 10, 2 ];
let n = A.length;
// Function calling
document.write(findMaxRec(A, n));
// This code is contributed by sravan kumar G
</script>
Related article:
Program to find largest element in an array
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