Check if Array is Sorted

Easy Arrays

Check whether an array is sorted in non-decreasing order.

Sample input

[1, 2, 3, 4, 5]

Sample output

true

Solution

arr = [1, 2, 3, 4, 5]
is_sorted = all(arr[i] <= arr[i + 1] for i in range(len(arr) - 1))
print(is_sorted)

const arr = [1, 2, 3, 4, 5];
let sorted = true;
for (let i = 0; i < arr.length - 1; i++) {
  if (arr[i] > arr[i + 1]) {
    sorted = false;
    break;
  }
}
console.log(sorted);

public class Main {
    public static void main(String[] args) {
        int[] arr = {1, 2, 3, 4, 5};
        boolean sorted = true;
        for (int i = 0; i < arr.length - 1; i++) {
            if (arr[i] > arr[i + 1]) {
                sorted = false;
                break;
            }
        }
        System.out.println(sorted);
    }
}

fun main() {
    val arr = intArrayOf(1, 2, 3, 4, 5)
    var sorted = true
    for (i in 0 until arr.size - 1) {
        if (arr[i] > arr[i + 1]) {
            sorted = false
            break
        }
    }
    println(sorted)
}

let arr = [1, 2, 3, 4, 5]
var sorted = true
for i in 0..<(arr.count - 1) {
    if arr[i] > arr[i + 1] {
        sorted = false
        break
    }
}
print(sorted)

void main() {
  List<int> arr = [1, 2, 3, 4, 5];
  bool sorted = true;
  for (int i = 0; i < arr.length - 1; i++) {
    if (arr[i] > arr[i + 1]) {
      sorted = false;
      break;
    }
  }
  print(sorted);
}

#include <iostream>
using namespace std;

int main() {
    int arr[] = {1, 2, 3, 4, 5};
    int n = 5;
    bool sorted = true;
    for (int i = 0; i < n - 1; i++) {
        if (arr[i] > arr[i + 1]) {
            sorted = false;
            break;
        }
    }
    cout << (sorted ? "true" : "false") << endl;
    return 0;
}

#include <stdio.h>

int main() {
    int arr[] = {1, 2, 3, 4, 5};
    int n = 5, sorted = 1;
    for (int i = 0; i < n - 1; i++) {
        if (arr[i] > arr[i + 1]) {
            sorted = 0;
            break;
        }
    }
    printf("%s\n", sorted ? "true" : "false");
    return 0;
}