I am working on a function that takes an array of numbers as input. The aim is to determine if the array is in ascending order and return true if it is, otherwise return false.
arr = [2, 10, 99, 150]
function checkOrder(arr) {
let sortedArray = arr.sort((a, b) => (a - b));
if (arr === sortedArray) {
return true
}
}
To determine if each value in an array is greater than the previous one, you can iterate through the array and compare each value with the preceding one. Here's an example of how you could achieve this:
const numbers1 = [2, 10, 99, 150];
const numbers2 = [2, 1, 3, 4];
const isAscendingOrder = (arr) => {
for (let i = 0; i < arr.length; i++) {
if (i > 0 && arr[i - 1] > arr[i]) {
return false;
}
}
return true;
}
console.log(numbers1, isAscendingOrder(numbers1));
console.log(numbers2, isAscendingOrder(numbers2));If you want to check for ascending order in an array, the .every() method can be used:
const numbers1 = [2, 10, 99, 150];
const numbers2 = [10, 2, 99, 150];
const isAscendingOrder = arr => arr.every((v, i, a) => (i == 0 || v >= a[i - 1]));
console.log(isAscendingOrder(numbers1));
console.log(isAscendingOrder(numbers2));If you want to check if an array is ordered, one way is to stringify the array and compare it with a sorted clone. To avoid mutating the original array while sorting, create a new clone using [...arr]
function areOrdered(arr) {
let sortedArray = [...arr].sort((a, b) => a - b);
return JSON.stringify(arr) === JSON.stringify(sortedArray)
}
console.log(areOrdered([2, 10, 99, 150]))
console.log(areOrdered([2, 100, 99]))Alternatively, you can use a simple for loop to iterate through the array and check if each element is less than or equal to the next one.
function areOrdered(arr) {
for (let i = 0; i < arr.length - 1; i++) {
if (arr[i] > arr[i + 1]) {
return false;
}
}
return true
}
console.log(areOrdered([2, 10, 99, 150]))
console.log(areOrdered([2, 100, 99]))I believe this approach should be successful (utilizing the .some function to verify if any two consecutive elements break the rule of ascending order) -:
const checkIfAscendingArr = (arr) => !arr.some((item,i) => {
if(i > 0){
return (arr[i-1] > item)
}
})
function checkAscendingOrder( arr ) {
for( let i = 1; i < arr.length; i++ ){
if( arr[i-1] > arr[i] )
return false;
}
return true;
}
// sample tests
let testArr1 = [1,2,5,5,8]; // true
let testArr2 = [1,2,7,5,8]; // false
let testArr3 = [-10]; // true
const res1 = checkAscendingOrder( testArr1 );
const res2 = checkAscendingOrder( testArr2 );
const res3 = checkAscendingOrder( testArr3 );
console.log( res1 );
console.log( res2 );
console.log( res3 );In my opinion, the most efficient way to solve this problem is by utilizing the Array.every method. This method allows you to compare each element in an array with the next one using their indexes. And in case we reach the end of the array, we can simply compare it with x+1.
let isOrdered = arr => arr.every((x,i,a) => x < (a[i+1] || x+1))
console.log(isOrdered([2, 10, 99, 150]))
console.log(isOrdered([21, 10, 99, 150]))Check out this cool single line code snippet using Array.reduce. I understand that single liners can sometimes be hard to read, but this one is just for fun!
var ascending = [1, 2, 3, 4].reduce((prev, curr, index, arr) => {return prev === false ? false : curr > prev ? index === arr.length - 1 ? true :curr : false}, 0);
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