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; } |
Java
// 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 |
Python3
# 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 |
C#
// 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
<?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 ?> |
Javascript
<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> |
Output
-50
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; } |
Java
// 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 |
Python3
# 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 |
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
<?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 ?> |
Javascript
<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> |
Output
45
Related article:
Program to find largest element in an array
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