Check If An Array is Sorted

Input: arr[] = {18, 25, 27}
Output: Yes

Input: arr[] = {18, 30, 30, 99}
Output: Yes

Input: arr[] = {100}
Output: Yes

Input: arr[] = {100, 20, 200}
Output: No

Note
- We are only considering array wheree array are in assending order
- It there are duplicate then is also true. Ex. 2nd example above

Naive Approach

package main

import (
    "fmt"
)

func main() {
    arr := []int{18, 25, 27} // Yes
    // arr := []int{18, 30, 30, 99} // Yes
    // arr := []int{100} // Yes
    // arr := []int{100, 20, 200} // No
    result := isArraySorted(arr)
    if result {
        fmt.Println("Yes")
    } else {
        fmt.Println("No")
    }

}

func isArraySorted(arr []int) bool {
    for i := 0; i < len(arr); i++ {
        for j := i + 1; j < len(arr); j++ {
            if arr[i] > arr[j] {
                return false
            }
        }
    }
    return true
}

Time Complexity: O(n2)

Efficient Approach

package main

import (
    "fmt"
)

func main() {
    // arr := []int{18, 25, 27} // Yes
    // arr := []int{18, 30, 30, 99} // Yes
    // arr := []int{100} // Yes
    arr := []int{100, 20, 200} // No
    result := isArraySorted(arr)
    if result {
        fmt.Println("Yes")
    } else {
        fmt.Println("No")
    }

}

func isArraySorted(arr []int) bool {
    for i := 0; i < len(arr)-2; i++ {
        if arr[i] > arr[i+1] {
            return false
        }
    }
    return true
}

Time Complexity: O(n)