Sorted Array Checker in Java
Code
import java.util.*;
public class SortedArray {
public static boolean SortedArray(int [ ] arr){
for(int i=1;i<arr.length;i++){
if( arr[i]>arr[i-1]){
}
else return false;
}
return true;
}
public static void main (String [] arg){
Scanner sc = new Scanner(System.in);
int [ ] ar={1,2,3,4,5};
if( SortedArray(ar)){
System.out.println("true");
}else {System.out.println("false");}
}
}
Overview
This Java program checks whether an array is sorted in ascending order. It traverses the array and compares each element with its previous one to determine if the array is sorted.
Functionality
Input:
- The program checks a predefined array of integers.
Output:
- It prints
true
if the array is sorted in ascending order. - It prints
false
if the array is not sorted.
Methods
1. SortedArray(int[] arr)
This method checks if the array is sorted by iterating through it.
Steps:
- The method starts from the second element (index
1
) and compares it with the previous element. - If at any point, the current element is less than or equal to the previous one, the method returns
false
indicating the array is not sorted. - If the loop completes without returning
false
, the method returnstrue
indicating the array is sorted.
Time Complexity:
- Iterating through the array has a time complexity of O(n), where
n
is the number of elements.
Main Method
- The program defines an array of integers to check.
- It calls the
SortedArray
method to check if the array is sorted. - Depending on the result, it prints
true
orfalse
.
Example Run:
Array: [1, 2, 3, 4, 5]
true
Time Complexity:
Method | Time Complexity |
---|---|
SortedArray() | O(n) |
Conclusion
- This program provides a simple and efficient method to check whether an array is sorted.
- With a time complexity of O(n), the solution is optimal for checking sorting in a single pass through the array.