Check if an Array is Sorted

Last Updated : 13 May, 2025

Given an array of size n, the task is 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: arr[] = [20, 21, 45, 89, 89, 90]
Output: Yes

Input: arr[] = [20, 20, 45, 89, 89, 90]
Output: Yes

Input: arr[] = [20, 20, 78, 98, 99, 97]
Output: No

Table of Content

Iterative approach - O(n) Time and O(1) Space
Recursive approach - O(n) Time and O(n) Space

Iterative approach - O(n) Time and O(1) Space

The idea is simple. We start traversing from the second element. For every element we check if it is smaller than or equal to previous element or not. At any point if we find previous element greater, we return false.

For example [10, 20, 30, 5, 6[

i = 1 : (10 <= 20), continue
i = 2 : (20 <= 30), continue
i = 3 : (30 > 5), return false.

C++

#include <bits/stdc++.h>
using namespace std;

// Function that returns true if vector is
// sorted in non-decreasing order.
bool isSorted(const vector<int>& arr)
{
    // For an array to be sorted, every
    // element must be greater than the 
    // previous element
    for (int i = 1; i < arr.size(); i++)
        if (arr[i - 1] > arr[i])
            return false;

    return true;
}

// Driver code
int main()
{
    vector<int> arr = { 20, 23, 23, 45, 78, 88 };
    cout << (isSorted(arr) ? "Yes\n" : "No\n");
    return 0;
}

// C program to check if an
// Array is sorted or not

#include <stdio.h>

// Function that returns true if array is
// sorted in non-decreasing order.
int arraySortedOrNot(int arr[], int n)
{
    // Array has one or no element
    if (n == 0 || n == 1)
        return 1;

    for (int i = 1; i < n; i++) {
        // Unsorted pair found
        if (arr[i - 1] > arr[i])
            return 0;
    }

    // No unsorted pair found
    return 1;
}

// Driver code
int main()
{
    int arr[] = { 20, 23, 23, 45, 78, 88 };
    int n = sizeof(arr) / sizeof(arr[0]);

    if (arraySortedOrNot(arr, n))
        printf("Yes\n");
    else
        printf("No\n");

    return 0;
}

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.

Python

# 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.

// 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

JavaScript

// JavaScript program Iterative approach to check if an
// Array is sorted or not

// Function that returns true if array is
// sorted in non-decreasing order.
function arraySortedOrNot(arr, n){

    // Array has one or no element
    if (n === 0 || n === 1) {
        return true;
    }

    for (let i = 1; i < n; i++) {
        // Unsorted pair found
        if (arr[i - 1] > arr[i]) {
            return false;
        }
    }

    // No unsorted pair found
    return true;
}

// Driver Code
let arr = [20, 23, 23, 45, 78, 88];
let n = arr.length;

if (arraySortedOrNot(arr, n)) {
    console.log("Yes");
} else {
    console.log("No");
}

Output

Yes

Recursive approach - O(n) Time and O(n) Space

The basic idea for the recursive approach:

If size of array is zero or one, return true.
Check last two elements of array, if they are sorted, perform a recursive call with n-1 else, return false.

C++

#include <iostream>
#include <vector>
using namespace std;

// Recursive function that returns true if vector is
// sorted in non-decreasing order.
bool isSorted(const vector<int>& arr, int n)
{
    // Base case
    if (n == 1 || n == 0)
        return true;

    // Check if current and previous elements are in order
    // and recursively check the rest of the array
    return arr[n - 1] >= arr[n - 2] && isSorted(arr, n - 1);
}

// Driver code
int main()
{
    vector<int> arr = { 20, 23, 23, 45, 78, 88 };
    cout << (isSorted(arr, arr.size()) ? "Yes\n" : "No\n");
    return 0;
}

// C Recursive approach to check if an
// Array is sorted or not

#include <stdio.h>

// Function that returns true if array is
// sorted in non-decreasing order.
int arraySortedOrNot(int a[], int n)
{
    // Base case
    if (n == 1 || n == 0) {
        return 1;
    }

    // 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)) {
        printf("Yes\n");
    }
    else {
        printf("No\n");
    }

    return 0;
}

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.

Python

# 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# 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

JavaScript

// JavaScript Recursive approach to check if an
// Array is sorted or not

// Function that returns true if array is
// sorted in non-decreasing order.
function arraySortedOrNot(a, 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
let arr = [20, 23, 23, 45, 78, 88];
let n = arr.length;

// Function Call
if (arraySortedOrNot(arr, n)) {
    console.log("Yes");
} else {
    console.log("No");
}

Output

Yes

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

Check if an array is Wave Array

Rohit Thapliyal

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Check if an Array is Sorted

Iterative approach - O(n) Time and O(1) Space

Recursive approach - O(n) Time and O(n) Space

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