Determine in JavaScript whether an array is sorted

Question

Determine in JavaScript whether an array is sorted

Looking to develop a program that can determine if the elements in an array are sorted, I am working with three sets of input data:

1,2,3,4,5

1,2,8,9,9

1,2,2,3,2

Here is the code snippet I have come up with:

let sorts = +gets(); // 3
let list = [];

for (let i = 0; i < sorts; i++) {
    list[i] = gets().split(',').map(Number); // The Array will be: [ [ 1, 2, 3, 4, 5 ], [ 1, 2, 8, 9, 9 ], [ 1, 2, 2, 3, 2 ] ]
}

for (let i = 0; i < list[i][i].length; i++){
    if (list[i][i] < list[i][i +1]) {
        print('true');
    } else {
        print('false');
    }
}

The objective is to output true or false on individual lines for each list. Thus, the expected output for the given example should be as follows:

true

false

I would greatly appreciate any guidance on how to proceed with this task.

javascript arrays sorting

Answer 1

Answer №1

If you want to determine if each value in an array is greater than the previous one, you can use the array#every method.

const isSorted = arr => arr.every((v,i,a) => !i || a[i-1] <= v);
console.log(isSorted([1,2,3,4,5]));
console.log(isSorted([1,2,8,9,9])); 
console.log(isSorted([1,2,2,3,2]));

Answer 2

If you want to determine if each value in an array is greater than the previous one, you can use the array#every method.

const isSorted = arr => arr.every((v,i,a) => !i || a[i-1] <= v);
console.log(isSorted([1,2,3,4,5]));
console.log(isSorted([1,2,8,9,9])); 
console.log(isSorted([1,2,2,3,2]));

Answer 3

Answer №2

Consider the following code snippet:

const result1 = ![1,2,3,4,5].reduce((acc, item) => acc !== false && item >= acc)
// true

const result2 = ![1,2,8,9,9].reduce((acc, item) => acc !== false && item >= acc)
// true 

const result3 = ![1,2,2,3,2].reduce((acc, item) => acc !== false && item >= acc)
// false

The reduce method simplifies the array into a single value - in this case, a boolean.

Within each iteration, we call a function with the signature (acc, item), checking if acc is not false, if item is greater than or equal to acc. This process continues for every element in the array. Finally, we use ! twice to convert the final outcome into a pure boolean.

Answer 4

Consider the following code snippet:

const result1 = ![1,2,3,4,5].reduce((acc, item) => acc !== false && item >= acc)
// true

const result2 = ![1,2,8,9,9].reduce((acc, item) => acc !== false && item >= acc)
// true 

const result3 = ![1,2,2,3,2].reduce((acc, item) => acc !== false && item >= acc)
// false

The reduce method simplifies the array into a single value - in this case, a boolean.

Within each iteration, we call a function with the signature (acc, item), checking if acc is not false, if item is greater than or equal to acc. This process continues for every element in the array. Finally, we use ! twice to convert the final outcome into a pure boolean.

Answer 5

Answer №3

Here's a simple method using the `slice` method: It checks if the previous element is less than the next element. If this condition holds true for every element, it will return true; otherwise, it will return false.

arr.slice(1).every((item, i) => arr[i] <= item);

Take a look at the following demo:

var arr = [[1,2,3,4,5],[1,2,8,9,9],[1,2,2,3,2],[0,1,2,3,4,5]];

function isArraySorted (arr) {
  return arr.slice(1).every((item, i) => arr[i] <= item)
}

var result= [];
for (var i = 0; i < arr.length; i++){
result.push(isArraySorted(arr[i]))
}
console.log(result);

Answer 6

Here's a simple method using the `slice` method: It checks if the previous element is less than the next element. If this condition holds true for every element, it will return true; otherwise, it will return false.

arr.slice(1).every((item, i) => arr[i] <= item);

Take a look at the following demo:

var arr = [[1,2,3,4,5],[1,2,8,9,9],[1,2,2,3,2],[0,1,2,3,4,5]];

function isArraySorted (arr) {
  return arr.slice(1).every((item, i) => arr[i] <= item)
}

var result= [];
for (var i = 0; i < arr.length; i++){
result.push(isArraySorted(arr[i]))
}
console.log(result);

Answer 7

Answer №4

let numbers = ["9,8,7,6,5", "1,2,3,4,5", "9,2,3,4,9"];

for (let numList of numbers){
    let array = numList.split(',').map(Number);
    console.log(array);
    let isSorted = true;
    for(let index = 0; index < array.length - 1; index++){
        if(array[index] > array[index+1]) {
            isSorted = false;
            break;
        }
    }
    console.log(isSorted);
}

Answer 8

let numbers = ["9,8,7,6,5", "1,2,3,4,5", "9,2,3,4,9"];

for (let numList of numbers){
    let array = numList.split(',').map(Number);
    console.log(array);
    let isSorted = true;
    for(let index = 0; index < array.length - 1; index++){
        if(array[index] > array[index+1]) {
            isSorted = false;
            break;
        }
    }
    console.log(isSorted);
}

Answer 9

Answer №5

Arranged Lists of Numbers

Covering Negative Values, Null Elements, and Consecutive Repeats

Utilize the every() function to check if the numbers are correctly ordered. The logic is:

(num <= arr[idx + 1]) || (idx === arr.length - 1)

if the current number is less than or equal to the next number...

OR...
if the current index equals the last index...
```
 return 1 (true)
```

Example

var list0 = [1, 2, 3, 4, 5];
var list1 = [1, 2, 8, 9, 9];
var list2 = [1, 2, 2, 3, 2];
var list3 = [0, 0, 0, 1, 3];
var list4 = [-3, 0, 1, 3, 3];
var list5 = [-4, -2, 0, 0, -4];

function arranged(array) {
  return array.every(function(num, idx, arr) {
    return (num <= arr[idx + 1]) || (idx === arr.length - 1) ? 1 : 0;
  });
}

console.log(list0 +' | '+arranged(list0));
console.log(list1 +' | '+arranged(list1));
console.log(list2 +' | '+arranged(list2));
console.log(list3 +' | '+arranged(list3));
console.log(list4 +' | '+arranged(list4));
console.log(list5 +' | '+arranged(list5));

Answer 10

Arranged Lists of Numbers

Covering Negative Values, Null Elements, and Consecutive Repeats

Utilize the every() function to check if the numbers are correctly ordered. The logic is:

(num <= arr[idx + 1]) || (idx === arr.length - 1)

if the current number is less than or equal to the next number...

OR...
if the current index equals the last index...
```
 return 1 (true)
```

Example

var list0 = [1, 2, 3, 4, 5];
var list1 = [1, 2, 8, 9, 9];
var list2 = [1, 2, 2, 3, 2];
var list3 = [0, 0, 0, 1, 3];
var list4 = [-3, 0, 1, 3, 3];
var list5 = [-4, -2, 0, 0, -4];

function arranged(array) {
  return array.every(function(num, idx, arr) {
    return (num <= arr[idx + 1]) || (idx === arr.length - 1) ? 1 : 0;
  });
}

console.log(list0 +' | '+arranged(list0));
console.log(list1 +' | '+arranged(list1));
console.log(list2 +' | '+arranged(list2));
console.log(list3 +' | '+arranged(list3));
console.log(list4 +' | '+arranged(list4));
console.log(list5 +' | '+arranged(list5));

Answer 11

Answer №6

There are numerous ways to achieve this task. Let me share one of my methods.

const checkArraySorted = arr =>
  arr
  .slice(0) // create a copy of the array
  .sort((a, b) => a - b) // sort the array in ascending order
  .every((el, i) => el === arr[i]) // compare sorted array with original)

Answer 12

There are numerous ways to achieve this task. Let me share one of my methods.

const checkArraySorted = arr =>
  arr
  .slice(0) // create a copy of the array
  .sort((a, b) => a - b) // sort the array in ascending order
  .every((el, i) => el === arr[i]) // compare sorted array with original)

Answer 13

Answer №7

Here is a handy method you can use to verify if a list is sorted correctly:

    var list1 = [4, 7, 9, 12, 15];
    var list2 = [10, 8, 6];

console.log(checkSorted(list1));
console.log(checkSorted(list2));

    function checkSorted(list) {
        for (var i = 0; i < list.length; i++) {
            if (list[i + 1]) {
                if (list[i] > list[i + 1]) {
                    return false;
                }
            }

        }
        return true;
    }

Answer 14

Here is a handy method you can use to verify if a list is sorted correctly:

    var list1 = [4, 7, 9, 12, 15];
    var list2 = [10, 8, 6];

console.log(checkSorted(list1));
console.log(checkSorted(list2));

    function checkSorted(list) {
        for (var i = 0; i < list.length; i++) {
            if (list[i + 1]) {
                if (list[i] > list[i + 1]) {
                    return false;
                }
            }

        }
        return true;
    }

Answer 15

Answer №8

To determine if the sorted and stringified version of an array matches the original array, you can use a simple approach without worrying about efficiency or elegance.

const arraysToVerify = [
  [3, 6, 8, 2, 4],
  [2, 4, 9, 10, 11],
  [5, 3, 7]
]

const isEqual = arraysToVerify.map(
  arr => JSON.stringify([...arr].sort((a, b) => a - b)) === JSON.stringify(arr)
);

console.log(isEqual);

Answer 16

To determine if the sorted and stringified version of an array matches the original array, you can use a simple approach without worrying about efficiency or elegance.

const arraysToVerify = [
  [3, 6, 8, 2, 4],
  [2, 4, 9, 10, 11],
  [5, 3, 7]
]

const isEqual = arraysToVerify.map(
  arr => JSON.stringify([...arr].sort((a, b) => a - b)) === JSON.stringify(arr)
);

console.log(isEqual);

Answer 17

Answer №9

It seems like you're trying to determine whether an array is sorted or not. Here's a solution for your query. Take a look at the code snippet provided below.

var myArray=[1,4,3,6];

if(isSorted(myArray)){

    console.log("List is sorted");
}else{
    console.log("List is not sorted");
}

function isSorted(X){

var sorted=false;

for(var i=0;i<X.length;i++){

        var next=i+1;

    if (next<=X.length-1){

        if(X[i]>X[next]){
            sorted=false;
            break;
        }else{
            sorted=true;

        }
    }

}


return sorted;

}

Answer 18

It seems like you're trying to determine whether an array is sorted or not. Here's a solution for your query. Take a look at the code snippet provided below.

var myArray=[1,4,3,6];

if(isSorted(myArray)){

    console.log("List is sorted");
}else{
    console.log("List is not sorted");
}

function isSorted(X){

var sorted=false;

for(var i=0;i<X.length;i++){

        var next=i+1;

    if (next<=X.length-1){

        if(X[i]>X[next]){
            sorted=false;
            break;
        }else{
            sorted=true;

        }
    }

}


return sorted;

}

Determine in JavaScript whether an array is sorted

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Arranged Lists of Numbers

Covering Negative Values, Null Elements, and Consecutive Repeats

Example

Answer №6

Answer №7

Answer №8

Answer №9

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