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
Approach 1: 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 <stdio.h>// Function that returns 0 if a pair// is found unsortedint 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 codeint 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;} |
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 unsortedint 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 codeint 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 notclass 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 unsorteddef 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 resultif arraySortedOrNot(arr): print("Yes")else: print("No") |
C#
// Recursive approach to check if an// Array is sorted or notusing 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 |
Javascript
<script>// Recursive approach to check if an// Array is sorted or not// Function that returns 0 if a pair// is found unsortedfunction arraySortedOrNot(arr, 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 codelet arr = [ 20, 23, 23, 45, 78, 88 ];let n = arr.length;if (arraySortedOrNot(arr, n) != 0) document.write("Yes");else document.write("No");// This code is contributed by sravan kumar G</script> |
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 <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 codeint 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;} |
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 codeint 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 notclass 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 codearr = [ 20, 23, 23, 45, 78, 88 ]n = len(arr)# Function Callif (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 notusing 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 codestatic 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
<script>// 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 codelet arr = [ 20, 23, 23, 45, 78, 88 ];let n = arr.length;// Function Callif (arraySortedOrNot(arr, n)){ document.write("Yes" + "<br>");}else{ document.write("No" + "<br>");}// This code is contributed by Surbhi Tyagi.</script> |
Output
Yes
Time Complexity: O(n)
Auxiliary Space: O(n) for Recursion Call Stack.
Approach 2: 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 <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 codeint 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;} |
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 codeint 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 notclass 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 codearr = [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 notusing 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
<script>// Javascript program 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(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)) document.write("Yes\n"); else document.write("No\n"); // This code is contributed by sanjoy_62.</script> |
Output
Yes
Time Complexity: O(n)
Auxiliary Space: O(1)
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