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
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; } // 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 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 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"); } 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 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 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 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"); } Time Complexity: O(n)
Auxiliary Space: O(n) for Recursion Call Stack.
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