Related
I solved this problem by iterating through the array then find the item when the sum equals to array[i] + item returning true otherwise returning false.
My Question is => How I can return the indices of those numbers that add up to sum not just true? Using the same code below:
function hasPairsWithSum(array,sum) {
for (let i = 0; i < array.length; i++) {
if (array.find((item) => {return sum === array[i] + item}
));
return true;
};
return false;
};
console.log(hasPairsWithSum([1,2,4,4],8))
Note: Time complexity must be less than O(n ^ 2).
JavaScript O(n) Solution.
function hasPairsWithSum(array, sum) {
const map = new Map ();
for(let i = 0; i < array.length; i++) {
let currVal = array[i];
if (map.has(currVal)) {
return [map.get(currVal),i]
}
// difference value = sum - current value
let diff = sum - currVal
map.set(diff,i)
}
};
console.log(hasPairsWithSum([2,2,4,4], 8))
Please refer this code.
function hasPairsWithSum(array,sum) {
let result = [];
for (let i = 0; i < array.length; i++) {
if (array.some((item, index) => {return i === index ? false : sum === array[i] + item}))
result.push(i);
};
return result;
};
console.log(hasPairsWithSum([1,2,4,4],8))
console.log(hasPairsWithSum([3,2,4],6))
console.log(hasPairsWithSum([0,4,3,0],0))
O(n) Soln ... using math concept a+b = n then if a is present in our array then need to find b = n - a is present or not ..
def hasPairsWithSum(array,sum):
d = {}
for i in range(len(array)):
if(array[i] in d):
d[array[i]].append(i)
else:
d[array[i]] = [i]
ans = []
for i in range(len(array)):
val = sum - array[i]
if(val in d):
if(d[val][0] == i):
if(len(d[val]) > 1):
ans.append((i,d[val][1]))
break
else:
continue
else:
ans.append((i,d[val][0]))
break
return ans
print(hasPairsWithSum([4, 4, 4, 4], 8))
O(nlogn) soln ....just store the index with elements .. then sort it by their values .. next step run a loop with complexity of O(n) [concept : Two pointers]
def hasPairsWithSum(array,sum):
arr = []
for i in range(len(array)):
arr.append((array[i],i))
arr.sort()
i = 0
j = len(array)-1
ans = []
while(i<j):
tmp_sum = arr[i][0] + arr[j][0]
if(tmp_sum == sum):
ans.append((arr[i][1] , arr[j][1]))
#add your logic if you want to find all possible indexes instead of break
break
elif(tmp_sum < sum):
i = i + 1
elif(tmp_sum > sum):
j = j - 1
return ans
print(hasPairsWithSum([1,2,4,4],8))
note : if you want to find all possible soln then these approaches will not work either add you own logic in while loop or another approach is use binary search with traversal on every element and store the indexes in set (worst case this will be O(n^2) as we have to find all possible values) Eg: [4,4,4,4,4,4] , sum = 8 and you want to print all possible indexes then we end up running it upto n^2 (why? reason: total possible solns. are 5+4+3+2+1 = n*(n-1)/2 ≈ n^2)
You have to iterate over the array elements checking at every iteration for every element of the array (except the last one) all the elements at the right of it like below:
function findIndexes(array, sum) {
const result = [];
for (let i = 0; i < array.length -1; ++i) {
for (let j = i + 1; j < array.length; ++j) {
if ((array[i] + array[j]) === sum) {
result.push([i, j]);
}
}
}
return result;
}
console.log(findIndexes([1, 2, 4, 4], 8));
console.log(findIndexes([3, 2, 4], 6));
Update:
It is possible to obtain a linear O(n) complexity using an auxiliary Map structure associating an integer value as key with as a value the list containing all the indexes of the elements in the array equal to the integer key like below:
function findIndexes(array, sum) {
const map = new Map();
const result = [];
for (let i = 0; i < array.length; ++i) {
const a = array[i];
const b = sum - a;
if (map.has(b)) {
for (const index of map.get(b)) {
result.push([index, i]);
}
}
const l = map.has(a) ? map.get(a) : [];
l.push(i);
map.set(a, l);
}
return result;
}
console.log(findIndexes([1, 2, 4, 4], 8));
console.log(findIndexes([3, 2, 4], 6));
console.log(findIndexes([1, 1, 1], 2));
For some reason, the manipulated doubleArray below is not shown in the console. Any variables that I declare after the for loop won't show to the console on both cases. Consider that in the first algorithm, there is only one for loop with x being incremented everytime. Whereas, in the second algorithm, it's a nested for loop. Can someone help me fix my error in both algorithms?
First Algorithm:
var isDuplicate = function() {
var helloWorld = [1,2,3,4,3];
var doubleValue = [];
var x = 0;
for (i = 0; i < helloWorld.length; i++) {
x = x + 1;
if (helloWorld[i] === helloWorld[x] && i !== x) {
doubleValue.push(helloWorld[i])
console.log(helloWorld[i]);
} else {
continue;
}
}
console.log(doubleValue);
};
The second Algorithm:
var isDuplicate = function() {
var helloWorld = [1,2,3,4,3];
var doubleValue = [];
for (i = 0; i < helloWorld.length; i++) {
for (x = 1; x < helloWorld.length; i++) {
if (helloWorld[i] === helloWorld[x] && i !== x) {
doubleValue.push(helloWorld[x]);
}
}
}
console.log(doubleValue);
};
In first algorithm, you are only checking if the number at current index is equal to the number at the next index, meaning you are only comparing numbers at consecutive indexes. First algorithm will work only if you have duplicate numbers on consecutive indexes.
In second algorithm, you are incrementing i in both loops, increment x in nested loop, change x = 1 to x = i + 1 and your error will be fixed.
Here's the fixed second code snippet
var isDuplicate = function() {
var helloWorld = [1,2,3,4,3, 1, 2];
var doubleValue = [];
for (let i = 0; i < helloWorld.length; i++) {
for (let x = i + 1; x < helloWorld.length; x++) {
if (helloWorld[i] === helloWorld[x] && i !== x) {
doubleValue.push(helloWorld[x]);
}
}
}
console.log(doubleValue);
};
isDuplicate();
Heres's another way to find the duplicates in an array, using an object. Loop over the array, if current number is present as key in the object, push the current number in the doubleValue array otherwise add the current number as key-value pair in the object.
const isDuplicate = function() {
const helloWorld = [1,2,3,4,3, 1, 2];
const doubleValue = [];
const obj = {};
helloWorld.forEach(n => obj[n] ? doubleValue.push(n): obj[n] = n);
console.log(doubleValue);
};
isDuplicate();
Not entirely sure what you are trying to do. If you are only looking for a method to remove duplicates you can do the following:
const hello_world = [1, 2, 2, 3, 4, 5, 5];
const duplicates_removed = Array.from(new Set(hello_world));
A set is a data object that only allows you to store unique values so, when converting an array to a set it will automatically remove all duplicate values. In the example above we are creating a set from hello_world and converting it back to an array.
If you are looking for a function that can identify all the duplicates in an array you can try the following:
const hello_world = [1, 2, 2, 3, 4, 5, 5];
const duplicates_found = hello_world.filter((item, index) => hello_world.indexOf(item) != index);
The main problem by finding duplicates is to have nested loop to compare each element of the array with any other element exept the element at the same position.
By using the second algorithm, you can iterate from the known position to reduce the iteration count.
var isDuplicate = function(array) {
var doubleValue = [];
outer: for (var i = 0; i < array.length - 1; i++) { // add label,
// declare variable i
// no need to check last element
for (var j = i + 1; j < array.length; j++) { // start from i + 1,
// increment j
if (array[i] === array[j]) { // compare values, not indices
doubleValue.push(array[i]);
continue outer; // prevent looping
}
}
}
return doubleValue;
};
console.log(isDuplicate([1, 2, 3, 4, 3])); // [3]
You could take an object for storing seen values and use a single loop for getting duplicate values.
const
getDuplicates = array => {
const
seen = {}
duplicates = [];
for (let value of array) {
if (seen[value]) duplicates.push(value);
else seen[value] = true;
}
return duplicates;
};
console.log(getDuplicates([1, 2, 3, 4, 3])); // [3]
Your first algorithm doesn't work because it only looks for duplicates next to each other. You can fix it by first sorting the array, then finding the duplicates. You can also remove the x and replace it by ++i in the loop.
var isDuplicate = function() {
var helloWorld = [1,2,3,4,3,6];
var doubleValue = [];
helloWorld = helloWorld.sort((a, b) => { return a - b });
for (i = 0; i < helloWorld.length; i++) {
if (helloWorld[i] === helloWorld[++i]) {
doubleValue.push(helloWorld[i])
console.log(helloWorld[i]);
} else {
continue;
}
}
console.log(doubleValue);
};
isDuplicate();
For the second algorithm loop, you probably meant x++ instead of i++ in the second loop. This would fix the problem.
var isDuplicate = function() {
var helloWorld = [1,2,3,4,3,4];
var doubleValue = [];
for (i = 0; i < helloWorld.length; i++) {
for (x = i + 1; x < helloWorld.length; x++) {
if (helloWorld[i] === helloWorld[x]) {
doubleValue.push(helloWorld[x]);
}
}
}
console.log(doubleValue);
};
isDuplicate()
The first algorithm can't be fixed, it can only detect consecutive duplicates,
in the second algorithm you increment i in both loops.
To avoid the duplicates beeing listed too often, you should start the second loop with i + 1
I am trying to build logic currently with arrays and data structure. I am trying to implement the logic using for loop
function getRepeatingNumber(arr) {
for (var i = 0; i < arr.length; i++) {
for (var j = i + 1; j < arr.length; j++) {
if (arr[i] === arr[j]) {
return arr[i];
}
}
}
return undefined;
}
getRepeatingNumber([2, 3, 6, 5, 2]);
the above function takes in array and returns a repeated item in the array so in the above case it will return 2. But what if I have an array something like this arr[2,3,3,6,5,2] in this case it should return 3 but as the outer loop has index [0] which is 2 as the reference it will return 2 as the answer.
How to implement a function that returns the first occurrence of the repeated item.
Instead of iterating with j in the part after i, iterate the part before i:
function getRepeatingNumber(arr){
for (var i = 1; i < arr.length; i++) {
for (var j = 0; j < i; j++) {
if (arr[i] === arr[j]) {
return arr[i];
}
}
}
}
console.log(getRepeatingNumber([2,3,3,6,5,2]));
Note that an explicit return undefined is not needed, that is the default behaviour already.
You could also use indexOf to shorten the code a bit:
function getRepeatingNumber(arr){
for (var i = 1; i < arr.length; i++) {
if (arr.indexOf(arr[i]) < i) {
return arr[i];
}
}
}
console.log(getRepeatingNumber([2,3,3,6,5,2]));
You could even decide to make use of find -- which will return undefined in case of no match (i.e. no duplicates in our case):
function getRepeatingNumber(arr){
return arr.find((a, i) => {
if (arr.indexOf(a) < i) {
return true;
}
});
}
console.log(getRepeatingNumber([2,3,3,6,5,2]));
If you do this for huge arrays, then it would become important to have a solution that runs with linear time complexity. In that case, a Set will be useful:
function getRepeatingNumber(arr){
var set = new Set;
return arr.find(a => {
if (set.has(a)) return true;
set.add(a);
});
}
console.log(getRepeatingNumber([2,3,3,6,5,2]));
And if you are into functions of functions, and one-liners, then:
const getRepeatingNumber = r=>(t=>r.find(a=>[t.has(a),t.add(a)][0]))(new Set);
console.log(getRepeatingNumber([2,3,3,6,5,2]));
You need a data structure to keep track of first occurring index.
My recommendation is to use an array to store all the index of repeating numbers. Sort the array in ascending order and return the item at first index from the array.
function getRepeatingNumber(arr){
var resultIndexArr = [];
var count = 0;
var flag = 0;
for(var i=0;i<arr.length;i++)
{
for(var j=i+1;j<arr.length;j++)
{
if(arr[i] === arr[j])
{
flag = 1;
resultIndexArr[count++] = j;
}
}
}
resultIndexArr.sort((a, b) => a - b);
var resultIndex = resultIndexArr[0];
if(flag === 1)
return arr[resultIndex];
else
return;
}
console.log(getRepeatingNumber([2,3,6,5,2])); // test case 1
console.log(getRepeatingNumber([2,3,3,6,5,2])); // test case 2
console.log(getRepeatingNumber([2,5,3,6,5,2])); // test case 3
This will return correct result, but this is not the best solution. The best solution is to store your items in an array, check for each iteration if the item already exists in your array, if it exists then just return that item.
as a javascript dev you should be comfortable wit functional programming & higher-order functions so check the doc to get more understanding of some useful functions: like filter - find - reduce - findIndex map ...
Documentation
Now to answer your question:
at first you should think by step :
Get the occurrence of an item in an array as function:
const arr = [2, 5, 6, 2, 4, 5, 6, 8, 2, 5, 2]
const res = arr.reduce((numberofOcc, item) => {
if (item === 2)
numberofOcc++
return numberofOcc
}, 0);
console.log(`result without function ${res}`);
/* so my function will be */
const occurenceFun = (num, arr) => {
return arr.reduce((numberofOcc, item) => {
if (item === num)
numberofOcc++
return numberofOcc
}, 0);
}
console.log(`result using my function ${occurenceFun(2, arr)}`);
Now i have this function so i can use it inside another function to get the higher occurrence i have in an array
const arr = [1, 2, 5, 6, 8, 7, 2, 2, 2, 10, 10, 2]
const occurenceFun = (num, arr) => {
return arr.reduce((numberofOcc, item) => {
if (item === num)
numberofOcc++
return numberofOcc
}, 0);
}
/*let's create our function*/
const maxOccurenceFun = arr => {
let max = 0;
arr.forEach(el => {
if (max < occurenceFun(el, arr)) {
max = el
}
})
return max;
}
console.log(`the max occurence in this array is : ${maxOccurenceFun(arr)}`);
I got an array
var myArray = [5,8,1,4,2,9,3,7,6];
I want the output to be [ 9, 1, 8, 2, 7, 3, 6, 4, 5 ]. I tried the following code:
function firstAndLast(array) {
var arr= [];
array = myArray.sort().reverse();
for(var i = 0; i < array.length; i++){
var firstItem = myArray[i];
var lastItem = myArray[myArray.length - 1];
if(lastItem > firstItem){
arr.push(array[i]);
}}
var display = firstAndLast(myArray);
console.log(display);
Can anyone suggest what am I missing to achieve the targeted result?
What I want to acheive is to arrange the array in even odd indexes where odd indexes contain larger values in descending order and even indexes contain values in ascending order
Your code actually fits your description, except this part:
if(lastItem > firstItem){
arr.push(array[i]);
}
Why don't you just push both items to the array:
if(lastItem > firstItem){
arr.push(firstItem, lastItem);
}
And the lastItem should be dependent on i:
var lastItem = array[array.length - i - 1];
Them you only have to
return arr;
At the end and it should work :)
function firstAndLast(array) {
const result = [];
array = array.sort((a, b) => a - b).reverse();
for(var i = 0; i < array.length; i++){
var firstItem = array[i];
var lastItem = array[array.length - i - 1];
if(lastItem < firstItem){
result.push(firstItem, lastItem);
}
}
return result;
}
var myArray = [5,8,1,4,2,9,3,7,6];
console.log(firstAndLast(myArray));
Now this only omits the value in the middle, which you can easily add like this in the loop:
if(firstItem === lastItem) {
result.push(firstItem);
}
Apparently you want to shuffle the array?
If that is the case the simplest way of doing it is just using this
function shuffle(array) {
var m = array.length, t, i;
// While there remain elements to shuffle…
while (m) {
// Pick a remaining element…
i = Math.floor(Math.random() * m--);
// And swap it with the current element.
t = array[m];
array[m] = array[i];
array[i] = t;
}
return array;
}
It is Fisher-Yates shuffle.
More on it on the link : https://bost.ocks.org/mike/shuffle/
If that is not the case post a reply to the comment so we can find some new sorting logic!
function firstAndLast(array) { //You're declaring array here but you're using it in line 3 using the same array without, you're pasing myArray in line 12, my sugestion is to declare array inside de function
var arr = [];
array = myArray.sort().reverse();
for(var i = 0; i < array.length; i++) {
var firstItem = myArray[i];
var lastItem = myArray[myArray.length - 1]; //You're always using the last item of your array, if i'm not wrong (or confused) you want decrement the position, right? you have to use myArray.length-i or a new variable to decrement the position
if(lastItem > firstItem)
arr.push(array[i]); //Again, you're using array when you want to use myArray (it is what you're using for the position in line 5 and 6)
}
var display = firstAndLast(myArray);
console.log(display);
I want to reverse an array without using reverse() function like this:
function reverse(array){
var output = [];
for (var i = 0; i<= array.length; i++){
output.push(array.pop());
}
return output;
}
console.log(reverse([1,2,3,4,5,6,7]));
However, the it shows [7, 6, 5, 4] Can someone tell me, why my reverse function is wrong? Thanks in advance!
array.pop() removes the popped element from the array, reducing its size by one. Once you're at i === 4, your break condition no longer evaluates to true and the loop ends.
One possible solution:
function reverse(array) {
var output = [];
while (array.length) {
output.push(array.pop());
}
return output;
}
console.log(reverse([1, 2, 3, 4, 5, 6, 7]));
You can make use of Array.prototype.reduceright and reverse it
check the following snippet
var arr = ([1, 2, 3, 4, 5, 6, 7]).reduceRight(function(previous, current) {
previous.push(current);
return previous;
}, []);
console.log(arr);
In ES6 this could be written as
reverse = (array) => array.map(array.pop, [... array]);
No need to pop anything... Just iterate through the existing array in reverse order to make your new one.
function reverse(array){
var output = [];
for (var i = array.length - 1; i> -1; i--){
output.push(array[i]);
}
return output;
}
console.log(reverse([1,2,3,4,5,6,7]));
Edit after answer got accepted.
A link in a comment on your opening post made me test my way VS the accepted answer's way. I was pleased to see that my way, at least in my case, turned out to be faster every single time. By a small margin but, faster non the less.
Here's the copy/paste of what I used to test it (tested from Firefox developer scratch pad):
function reverseMyWay(array){
var output = [];
for (var i = array.length - 1; i> -1; i--){
output.push(array[i]);
}
return output;
}
function reverseTheirWay(array) {
var output = [];
while (array.length) {
output.push(array.pop());
}
return output;
}
function JustDoIt(){
console.log("their way starts")
var startOf = new Date().getTime();
for(var p = 0; p < 10000; p++)
{
console.log(reverseTheirWay([7,6,5,4,3,2,1]))
}
var endOf = new Date().getTime();
console.log("ran for " + (endOf - startOf) + " ms");
console.log("their way ends")
}
function JustDoIMyWay(){
console.log("my way starts")
var startOf = new Date().getTime();
for(var p = 0; p < 10000; p++)
{
console.log(reverseMyWay([7,6,5,4,3,2,1]))
}
var endOf = new Date().getTime();
console.log("ran for " + (endOf - startOf) + " ms");
console.log("my way ends")
}
JustDoIt();
JustDoIMyWay();
Solution to reverse an array without using built-in function and extra space.
let arr = [1, 2, 3, 4, 5, 6, 7];
let n = arr.length-1;
for(let i=0; i<=n/2; i++) {
let temp = arr[i];
arr[i] = arr[n-i];
arr[n-i] = temp;
}
console.log(arr);
Do it in a reverse way, Because when you do .pop() every time the array's length got affected.
function reverse(array){
var output = [];
for (var i = array.length; i > 0; i--){
output.push(array.pop());
}
return output;
}
console.log(reverse([1,2,3,4,5,6,7]));
Or you could cache the length of the array in a variable before popping out from the array,
function reverse(array){
var output = [];
for (var i = 0, len= array.length; i< len; i++){
output.push(array.pop());
}
return output;
}
console.log(reverse([1,2,3,4,5,6,7]));
You are modifying the existing array with your reverse function, which is affecting array.length.
Don't pop off the array, just access the item in the array and unshift the item on the new array so that the first element of the existing array becomes the last element of the new array:
function reverse(array){
var output = [],
i;
for (i = 0; i < array.length; i++){
output.unshift(array[i]);
}
return output;
}
console.log(reverse([1,2,3,4,5,6,7]));
If you'd like to modify the array in-place similar to how Array.prototype.reverse does (it's generally inadvisable to cause side-effects), you can splice the array, and unshift the item back on at the beginning:
function reverse(array) {
var i,
tmp;
for (i = 1; i < array.length; i++) {
tmp = array.splice(i, 1)[0];
array.unshift(tmp);
}
return array;
}
var a = [1, 2, 3, 4, 5];
console.log('reverse result', reverse(a));
console.log('a', a);
This piece allows to reverse the array in place, without pop, splice, or push.
var arr = [1, 2, 3, 4, 5];
function reverseArrayInPlace(arr2) {
var half = Math.floor(arr2.length / 2);
for (var i = 0; i < half; i++) {
var temp = arr2[arr2.length - 1 - i];
arr2[arr2.length - 1 - i] = arr2[i];
arr2[i] = temp;
}
return arr2;
}
As you pop items off the first array, it's length changes and your loop count is shortened. You need to cache the original length of the original array so that the loop will run the correct amount of times.
function reverse(array){
var output = [];
var len = array.length;
for (var i = 0; i< len; i++){
output.push(array.pop());
}
return output;
}
console.log(reverse([1,2,3,4,5,6,7]));
You're modifying the original array and changing it's size. instead of a for loop you could use a while
function reverse(array){
var output = [];
while(array.length){
//this removes the last element making the length smaller
output.push(array.pop());
}
return output;
}
console.log(reverse([1,2,3,4,5,6,7]));
function rvrc(arr) {
for (let i = 0; i < arr.length / 2; i++) {
const buffer = arr[i];
arr[i] = arr[arr.length - 1 - i];
arr[arr.length - 1 - i] = buffer;
}
};
const reverse = (array)=>{
var output = [];
for(let i=array.length; i>0; i--){
output.push(array.pop());
}
console.log(output);
}
reverse([1, 2, 3, 4, 5, 6, 7, 8]);
This happens because every time you do array.pop(), whilst it does return the last index in the array, it also removes it from the array. The loop recalculates the length of the array at each iteration. Because the array gets 1 index shorter at each iteration, you get a much shorter array returned from the function.
This piece of code will work without using a second array. It is using the built in method splice.
function reverse(array){
for (let i = 0; i < array.length; i++) {
array.splice(i, 0, array.splice(array.length - 1)[0]);
}
return array;
}
Here, let's define the function
function rev(arr) {
const na = [];
for (let i=0; i<arr.length; i++) {
na.push(arr[arr.length-i])
}
return na;
}
Let's say your array is defined as 'abca' and contains ['a','b','c','d','e','foo','bar']
We would do:
var reva = rev(abca)
This would make 'reva' return ['bar','foo','e','d','c','b','a'].
I hope I helped!
You can use .map as it is perfect for this situation and is only 1 line:
const reverse = a =>{ i=a.length; return a.map(_=>a[i-=1]) }
This will take the array, and for each index, change it to the length of the array - index, or the opposite side of the array.
with reverse for loop
let array = ["ahmet", "mehmet", "aslı"]
length = array.length
newArray = [];
for (let i = length-1; i >-1; i--) {
newArray.push(array[i])
}
console.log(newArray)
And this one:
function reverseArray(arr) {
let top = arr.length - 1;
let bottom = 0;
let swap = 0;
while (top - bottom >= 1) {
swap = arr[bottom];
arr[bottom] = arr[top];
arr[top] = swap;
bottom++;
top--;
}
}
function reverse(arr) {
for (let i = 0; i < arr.length - 1; i++) {
arr.splice(i, 0, arr.pop())
}
return arr;
}
console.log(reverse([1, 2, 3, 4, 5]))
//without another array
reverse=a=>a.map((x,y)=>a[a.length-1-y])
reverse=a=>a.map((x,y)=>a[a.length-1-y])
console.log(reverse(["Works","It","One","Line"]))
One of shortest:
let reverse = arr = arr.map(arr.pop, [...arr])
This is an old question, but someone may find this helpful.
There are two main ways to do it:
First, out of place, you basically push the last element to a new array, and use the new array:
function arrReverse(arr) {
let newArr = [];
for(let i = 0; i<arr.length; i++){
newArr.push(arr.length -1 -i);
}
return newArr;
}
arrReverse([0,1,2,3,4,5,6,7,8,9]);
Then there's in place. This is a bit tricky, but the way I think of it is like having four objects in front of you. You need to hold the first in your hand, then move the last item to the first place, and then place the item in your hand in the last place.
Afterwards, you increase the leftmost side by one and decrease the rightmost side by one:
function reverseArr(arr) {
let lh;
for(let i = 0; i<arr.length/2; i++){
lh = arr[i];
arr[i] = arr[arr.length -i -1];
arr[arr.length -i -1] = lh;
}
return arr;
}
reverseArr([0,1,2,3,4,5,6,7,8,9]);
Like so. I even named my variable lh for "left hand" to help the idea along.
Understanding arrays is massively important, and figuring out how they work will not only save you from unnecessarily long and tedious ways of solving this, but will also help you grasp certain data concepts way better!
I found a way of reversing the array this way:
function reverse(arr){
for (let i = arr.length-1; i >= 0; i--){
arr.splice(i, 0, arr.shift());
}
return arr;
}
Without Using any Pre-define function
const reverseArray = (array) => {
for (let i = 0; i < Math.floor(array.length / 2); i++) {
[array[i], array[array.length - i - 1]] = [
array[array.length - i - 1],
array[i]
];
}
return array;
};
let array = [1,2,3,4,5,6];
const reverse = (array) => {
let reversed = [];
for(let i = array.length - 1; i >= 0; i--){
reversed[array.length - i] = array[i];
}
return reversed;
}
console.log(reverse(array))
you can use the two pointers approach
example
function reverseArrayTwoPointers(arr = [1, 2, 3, 4, 5]) {
let p1 = 0;
let p2 = arr.length - 1;
while (p2 > p1) {
const temp = arr[p1];
arr[p1] = arr[p2];
arr[p2] = temp;
p1++;
p2--;
}
return arr;
}
to return [5,4,3,2,1]
example on vscode
let checkValue = ["h","a","p","p","y"]
let reverseValue = [];
checkValue.map((data, i) => {
x = checkValue.length - (i + 1);
reverseValue[x] = data;
})
function reverse(str1) {
let newstr = [];
let count = 0;
for (let i = str1.length - 1; i >= 0; i--) {
newstr[count] = str1[i];
count++;
}
return newstr;
}
reverse(['x','y','z']);
Array=[2,3,4,5]
for(var i=0;i<Array.length/2;i++){
var temp =Array[i];
Array[i]=Array[Array.length-i-1]
Array[Array.length-i-1]=temp
}
console.log(Array) //[5,4,3,2]