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Program to check if an array is sorted or not (Iterative and Recursive)
  • Difficulty Level : Easy
  • Last Updated : 22 Dec, 2020

Given an array of size n, write a program to check if it is sorted in ascending order or not. Equal values are allowed in an array and two consecutive equal values are considered sorted.

Examples: 

Input : 20 21 45 89 89 90
Output : Yes

Input : 20 20 45 89 89 90
Output : Yes

Input : 20 20 78 98 99 97
Output : No

Recursive approach:
The basic idea for the recursive approach:  

1: If size of array is zero or one, return true.
2: Check last two elements of array, if they are
   sorted, perform a recursive call with n-1
   else, return false.
If all the elements will be found sorted, n will
eventually fall to one, satisfying Step 1.

Below is the implementation using recursion:

C++




// Recursive approach to check if an
// Array is sorted or not
#include <bits/stdc++.h>
using namespace std;
 
// Function that returns 0 if a pair
// is found unsorted
int arraySortedOrNot(int arr[], int n)
{
    // Array has one or no element or the
    // rest are already checked and approved.
    if (n == 1 || n == 0)
        return 1;
 
    // Unsorted pair found (Equal values allowed)
    if (arr[n - 1] < arr[n - 2])
        return 0;
 
    // Last pair was sorted
    // Keep on checking
    return arraySortedOrNot(arr, n - 1);
}
 
// Driver code
int main()
{
    int arr[] = { 20, 23, 23, 45, 78, 88 };
    int n = sizeof(arr) / sizeof(arr[0]);
    if (arraySortedOrNot(arr, n))
        cout << "Yes\n";
    else
        cout << "No\n";
}


Java




// Recursive approach to check if an
// Array is sorted or not
 
class CkeckSorted {
    // Function that returns 0 if a pair
    // is found unsorted
    static int arraySortedOrNot(int arr[], int n)
    {
        // Array has one or no element or the
        // rest are already checked and approved.
        if (n == 1 || n == 0)
            return 1;
 
        // Unsorted pair found (Equal values allowed)
        if (arr[n - 1] < arr[n - 2])
            return 0;
 
        // Last pair was sorted
        // Keep on checking
        return arraySortedOrNot(arr, n - 1);
    }
 
    // main function
    public static void main(String[] args)
    {
        int arr[] = { 20, 23, 23, 45, 78, 88 };
        int n = arr.length;
        if (arraySortedOrNot(arr, n) != 0)
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}


Python3




# Recursive approach to check if an
# Array is sorted or not
 
# Function that returns 0 if a pair
# is found unsorted
 
 
def arraySortedOrNot(arr):
 
    # Calculating length
    n = len(arr)
 
    # Array has one or no element or the
    # rest are already checked and approved.
    if n == 1 or n == 0:
        return True
 
    # Recursion applied till last element
    return arr[0] <= arr[1] and arraySortedOrNot(arr[1:])
 
 
arr = [20, 23, 23, 45, 78, 88]
 
# Displaying result
if arraySortedOrNot(arr):
    print("Yes")
else:
    print("No")


C#




// Recursive approach to check if an
// Array is sorted or not
using System;
 
class CkeckSorted {
    // Function that returns 0 if a pair
    // is found unsorted
    static int arraySortedOrNot(int[] arr, int n)
    {
        // Array has one or no element or the
        // rest are already checked and approved.
        if (n == 1 || n == 0)
            return 1;
 
        // Unsorted pair found (Equal values allowed)
        if (arr[n - 1] < arr[n - 2])
            return 0;
 
        // Last pair was sorted
        // Keep on checking
        return arraySortedOrNot(arr, n - 1);
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        int[] arr = { 20, 23, 23, 45, 78, 88 };
        int n = arr.Length;
        if (arraySortedOrNot(arr, n) != 0)
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
 
// This code is contributed by 29AjayKumar


Output:



Yes

Time Complexity: O(n) 
Auxiliary Space: O(n) for Recursion Call Stack.

Another Recursive approach:

C++




// C++ Recursive approach to check if an
// Array is sorted or not
#include <iostream>
using namespace std;
 
// Function that returns true if array is
// sorted in non-decreasing order.
bool arraySortedOrNot(int a[], int n)
{
     
    // Base case
    if (n == 1 || n == 0)
    {
        return true;
    }
     
    // Check if present index and index
    // previous to it are in correct order
    // and rest of the array is also sorted
    // if true then return true else return
    // false
    return a[n - 1] >= a[n - 2] &&
     arraySortedOrNot(a, n - 1);
}
 
// Driver code
int main()
{
    int arr[] = { 20, 23, 23, 45, 78, 88 };
    int n = sizeof(arr) / sizeof(arr[0]);
     
    // Function Call
    if (arraySortedOrNot(arr, n))
    {
        cout << "Yes" << endl;
    }
    else
    {
        cout << "No" << endl;
    }
     
    return 0;
}
 
// This code is contributed by avanitrachhadiya2155


Java




// Java Recursive approach to check if an
// Array is sorted or not
class GFG {
 
    // Function that returns true if array is
    // sorted in non-decreasing order.
    static boolean arraySortedOrNot(int a[], int n)
    {
          // base case
        if (n == 1 || n == 0)
            return true;
     
          // check if present index and index
        // previous to it are in correct order
        // and rest of the array is also sorted
        // if true then return true else return
        // false
        return a[n - 1] >= a[n - 2]
            && arraySortedOrNot(a, n - 1);
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        int arr[] = { 20, 23, 23, 45, 78, 88 };
        int n = arr.length;
         
          // Function Call
        if (arraySortedOrNot(arr, n))
            System.out.print("Yes");
        else
            System.out.print("No");
    }
}
 
// This code is contributed by Durgesh N. Birmiwal.


Python3




# Python3 recursive program to check
# if an Array is sorted or not
 
# Function that returns true if array
# is sorted in non-decreasing order.
def arraySortedOrNot(arr, n):
 
    # Base case
    if (n == 0 or n == 1):
        return True
         
    # Check if present index and index
    # previous to it are in correct order
    # and rest of the array is also sorted
    # if true then return true else return
    # false
    return (arr[n - 1] >= arr[n - 2] and
            arraySortedOrNot(arr, n - 1))
 
# Driver code
arr = [ 20, 23, 23, 45, 78, 88 ]
n = len(arr)
 
# Function Call
if (arraySortedOrNot(arr, n)):
    print("Yes")
else:
    print("No")
     
# This code is contributed by Virusbuddah


C#




// C# recursive approach to check if an
// Array is sorted or not
using System;
 
class GFG{
     
// Function that returns true if array is
// sorted in non-decreasing order.
static bool arraySortedOrNot(int[] a, int n)
{
     
    // Base case
    if (n == 1 || n == 0)
    {
        return true;
    }
     
    // Check if present index and index
    // previous to it are in correct order
    // and rest of the array is also sorted
    // if true then return true else return
    // false
    return a[n - 1] >= a[n - 2] &&
     arraySortedOrNot(a, n - 1);
}
 
// Driver code
static public void Main()
{
    int[] arr = { 20, 23, 23, 45, 78, 88 };
    int n = arr.Length;
     
    // Function Call
    if (arraySortedOrNot(arr, n))
    {
        Console.WriteLine("Yes");
    }
    else
    {
        Console.WriteLine("No");
    }
}
}
 
// This code is contributed by rag2127


Output

Yes

Time Complexity: O(n) 
Auxiliary Space: O(n) for Recursion Call Stack.

Iterative approach: The idea is pretty much the same. The benefit of the iterative approach is it avoids the usage of recursion stack space and recursion overhead.

Below is the implementation using iteration: 

C++




// C++ program to check if an
// Array is sorted or not
#include <bits/stdc++.h>
using namespace std;
 
// Function that returns true if array is
// sorted in non-decreasing order.
bool arraySortedOrNot(int arr[], int n)
{
    // Array has one or no element
    if (n == 0 || n == 1)
        return true;
 
    for (int i = 1; i < n; i++)
 
        // Unsorted pair found
        if (arr[i - 1] > arr[i])
            return false;
 
    // No unsorted pair found
    return true;
}
 
// Driver code
int main()
{
    int arr[] = { 20, 23, 23, 45, 78, 88 };
    int n = sizeof(arr) / sizeof(arr[0]);
    if (arraySortedOrNot(arr, n))
        cout << "Yes\n";
    else
        cout << "No\n";
}


Java




// Recursive approach to check if an
// Array is sorted or not
class GFG {
 
    // Function that returns true if array is
    // sorted in non-decreasing order.
    static boolean arraySortedOrNot(int arr[], int n)
    {
 
        // Array has one or no element
        if (n == 0 || n == 1)
            return true;
 
        for (int i = 1; i < n; i++)
 
            // Unsorted pair found
            if (arr[i - 1] > arr[i])
                return false;
 
        // No unsorted pair found
        return true;
    }
 
    // driver code
    public static void main(String[] args)
    {
 
        int arr[] = { 20, 23, 23, 45, 78, 88 };
        int n = arr.length;
 
        if (arraySortedOrNot(arr, n))
            System.out.print("Yes\n");
        else
            System.out.print("No\n");
    }
}
 
// This code is contributed by Anant Agarwal.


Python3




# Python3 program to check if an
# Array is sorted or not
 
# Function that returns true if array is
# sorted in non-decreasing order.
def arraySortedOrNot(arr, n):
 
    # Array has one or no element
    if (n == 0 or n == 1):
        return True
 
    for i in range(1, n):
 
        # Unsorted pair found
        if (arr[i-1] > arr[i]):
            return False
 
    # No unsorted pair found
    return True
 
 
# Driver code
arr = [20, 23, 23, 45, 78, 88]
n = len(arr)
if (arraySortedOrNot(arr, n)):
    print("Yes")
else:
    print("No")
     
# This code is contributed by Anant Agarwal.


C#




// Recursive approach to check if an
// Array is sorted or not
using System;
 
class GFG
{
 
    // Function that returns true if array is
    // sorted in non-decreasing order.
    static bool arraySortedOrNot(int []arr, int n)
    {
 
        // Array has one or no element
        if (n == 0 || n == 1)
            return true;
 
        for (int i = 1; i < n; i++)
 
            // Unsorted pair found
            if (arr[i - 1] > arr[i])
                return false;
 
        // No unsorted pair found
        return true;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int []arr = { 20, 23, 23, 45, 78, 88 };
        int n = arr.Length;
 
        if (arraySortedOrNot(arr, n))
            Console.Write("Yes\n");
        else
            Console.Write("No\n");
    }
}
 
// This code is contributed by PrinciRaj1992


Output: 

Yes

Time Complexity: O(n) 
Auxiliary Space: O(1)

This article is contributed by Rohit Thapliyal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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