I'm currently learning Javascript and would love if someone could help me understand the for loops further. I want to see if someone can give me a bit of an in depth explanation as to how this loop works.
The idea is to return the first non consecutive number in the argument, which as you can see is 6.
Because I'm still learning I wanted to get a detailed yet easy understanding of how this works for example, what's the difference between arr[i]+1 and arr[i+1]?
function firstNonConsec(arr){
for(let i = 0; i < arr.length - 1; i++){
if(arr[i] + 1 !== arr[i+1]){
return arr[i + 1];
}
}
return null
};
console.log(firstNonConsec([1,2,3,4,6,7,8]));
what's the difference between arr[i]+1 and arr[i+1]?
This is not a question about the for loop, but about arrays.
if arr is an array, then you can get the value of one of its items by doing arr[item_number]
arr[i]+1 will therefore give you the value at the place i of the table (e.g. if i equals 0, that would be the first entry in the array), plus one*
arr[i+1] will give you the value at the place i+1 of the table (e.g. if i equals 0, that would be the second entry in the array)
note that +1 can do a lot of things in Javascript, depending on type auto conversion; in your case with only numbers it will increase the number by 1
You can get reference from this workflow of firstNonConsec([1,2,3,4,6,7,8])
i : 0 1 2 3 4 5 6
arr[i] : [ 1, 2, 3, 4, 6, 7, 8 ]
arr[i] + 1 : 2 3 4 5 7 8 9
arr[i+1] : 2 3 4 6 7 8 N
if-statement : T T T F T T F // T: true F: false
return : 2 3 4 N 7 8 N // N: null
arr[i]+1 returns the value of arr's ith array position, then adds one to that value.
arr = [2, 9, 5, 1]
i = 2
arr[2] + 1
Result: 6
So it 1) finds the item in that position 2) adds 1. An array's index starts at 0: arr[n] = [0+n]
You always get the value equal to the result within the bracket.
arr[i+1] would return the i+1th value.
arr = [2, 9, 5, 1]
i = 2
print(arr[i+1]) == print(arr[3])
Result: 1
So this one changes the position itself by 1. This is how a robocaller might complete a call then select the next phone number in a list.
In your case:
if(arr[i] + 1 !== arr[i+1]){
return arr[i + 1];
if the value of the next item in the list is not equal to the previous value + 1, return that value. The function prints you a list of every non consecutive number.
for loop is basically a special type of while loop
var i = 0;
while (i < 10) {
// < code >
i++
}
is the same as
for (let i = 0; i < 10; i++) {
//< code >
}
while loops are used for many different things, so remember them, but for loops are used most. Basically the first "segment" (I'm gonna call whatever's before a semicolon a segment) runs before whatever code you wrote. It usually declares a variable. The second segment runs every time, and stops the loop if the condition isn't met. The third segment also runs every time, but it just runs some code after all the code you wrote is written..
In a lot of languages, including JavaScript, you can also loop through arrays and HashMaps. In JavaScript, you use in and of words.
use for/in for objects
for (let x in < object >) {
//< code >
}
and for each iteration, x is the key of one of the object's properties
to loop through arrays and other iterable objects (you can use for/in, but it's bad practice to), use for/of
for (var x of < array >) {
//< code >
}
this loops through the values of an array, or other such iterable object
the difference between arr[i+1] and arr[i]+1 is that arr[i+1] will access the element after the index specified, but arr[i]+1 will take the value of the index, and return that + 1 (won't change the value, use += to change value). BTW you don't always have to use of for looping thru arrays, you can do it this way which takes i and increases it every time, then takes the value of the index of i.
the answer to your problem:
function firstNonConsec(arr) {
let temp = arr[0];
for (let i = 1; i < arr.length; i++) {
temp = arr[i - 1];
if (temp !== arr[i] - 1) return arr[i];
}
return null;
}
I'm not using for of because you didn't, but comment if you want me to write with for of
Note I used let most of the time because you don't want there to be a random variable that you don't need, and let is limited to the scope of the loop. Also, iterating variables are commonly named i, j, x, y, z and such so you might use i a lot.
Related
Im just wondering who can explain the algorithm of this solution step by step. I dont know how hashmap works. Can you also give a basic examples using a hashmap for me to understand this algorithm. Thank you!
var twoSum = function(nums, target) {
let hash = {};
for(let i = 0; i < nums.length; i++) {
const n = nums[i];
if(hash[target - n] !== undefined) {
return [hash[target - n], i];
}
hash[n] = i;
}
return [];
}
Your code takes an array of numbers and a target number/sum. It then returns the indexes in the array for two numbers which add up to the target number/sum.
Consider an array of numbers such as [1, 2, 3] and a target of 5. Your task is to find the two numbers in this array which add to 5. One way you can approach this problem is by looping over each number in your array and asking yourself "Is there a number (which I have already seen in my array) which I can add to the current number to get my target sum?".
Well, if we loop over the example array of [1, 2, 3] we first start at index 0 with the number 1. Currently, there are no numbers which we have already seen that we can add with 1 to get our target of 5 as we haven't looped over any numbers yet.
So, so far, we have met the number 1, which was at index 0. This is stored in the hashmap (ie object) as {'1': 0}. Where the key is the number and the value (0) is the index it was seen at. The purpose of the object is to store the numbers we have seen and the indexes they appear at.
Next, the loop continues to index 1, with the current number being 2. We can now ask ourselves the question: Is there a number which I have already seen in my array that I can add to my current number of 2 to get the target sum of 5. The amount needed to add to the current number to get to the target can be obtained by doing target-currentNumber. In this case, we are currently on 2, so we need to add 3 to get to our target sum of 5. Using the hashmap/object, we can check if we have already seen the number 3. To do this, we can try and access the object 3 key by doing obj[target-currentNumber]. Currently, our object only has the key of '1', so when we try and access the 3 key you'll get undefined. This means we haven't seen the number 3 yet, so, as of now, there isn't anything we can add to 2 to get our target sum.
So now our object/hashmap looks like {'1': 0, '2': 1}, as we have seen the number 1 which was at index 0, and we have seen the number 2 which was at index 1.
Finally, we reach the last number in your array which is at index 2. Index 2 of the array holds the number 3. Now again, we ask ourselves the question: Is there a number we have already seen which we can add to 3 (our current number) to get the target sum?. The number we need to add to 3 to get our target number of 5 is 2 (obtained by doing target-currentNumber). We can now check our object to see if we have already seen a number 2 in the array. To do so we can use obj[target-currentNumber] to get the value stored at the key 2, which stores the index of 1. This means that the number 2 does exist in the array, and so we can add it to 3 to reach our target. Since the value was in the object, we can now return our findings. That being the index of where the seen number occurred, and the index of the current number.
In general, the object is used to keep track of all the previously seen numbers in your array and keep a value of the index at which the number was seen at.
Here is an example of running your code. It returns [1, 2], as the numbers at indexes 1 and 2 can be added together to give the target sum of 5:
const twoSum = function(nums, target) {
const hash = {}; // Stores seen numbers: {seenNumber: indexItOccurred}
for (let i = 0; i < nums.length; i++) { // loop through all numbers
const n = nums[i]; // grab the current number `n`.
if (hash[target - n] !== undefined) { // check if the number we need to add to `n` to reach our target has been seen:
return [hash[target - n], i]; // grab the index of the seen number, and the index of the current number
}
hash[n] = i; // update our hash to include the. number we just saw along with its index.
}
return []; // If no numbers add up to equal the `target`, we can return an empty array
}
console.log(twoSum([1, 2, 3], 5)); // [1, 2]
A solution like this might seem over-engineered. You might be wondering why you can't just look at one number in the array, and then look at all the other numbers and see if you come across a number that adds up to equal the target. A solution like that would work perfectly fine, however, it's not very efficient. If you had N numbers in your array, in the worst case (where no two numbers add up to equal your target) you would need to loop through all of these N numbers - that means you would do N iterations. However, for each iteration where you look at a singular number, you would then need to look at each other number using a inner loop. This would mean that for each iteration of your outer loop you would do N iterations of your inner loop. This would result in you doing N*N or N2 work (O(N2) work). Unlike this approach, the solution described in the first half of this answer only needs to do N iterations over the entire array. Using the object, we can find whether or not a number is in the object in constant (O(1)) time, which means that the total work for the above algorithm is only O(N).
For further information about how objects work, you can read about bracket notation and other property accessor methods here.
You may want to check out this method, it worked so well for me and I have written a lot of comments on it to help even a beginner understand better.
let nums = [2, 7, 11, 15];
let target = 9;
function twoSums(arr, t){
let num1;
//create the variable for the first number
let num2;
//create the variable for the second number
let index1;
//create the variable for the index of the first number
let index2;
//create the variable for the index of the second number
for(let i = 0; i < arr.length; i++){
//make a for loop to loop through the array elements
num1 = arr[i];
//assign the array iteration, i, value to the num1 variable
//eg: num1 = arr[0] which is 2
num2 = t - num1;
//get the difference between the target and the number in num1.
//eg: t(9) - num1(2) = 7;
if(arr.includes(num2)){
//check to see if the num2 number, 7, is contained in the array;
index1 = arr.indexOf(num2);
//if yes get the index of the num2 value, 7, from the array,
// eg: the index of 7 in the array is 1;
index2 = arr.indexOf(num1)
//get the index of the num1 value, which is 2, theindex of 2 in the array is 0;
}
}
return(`[${index1}, ${index2}]`);
//return the indexes in block parenthesis. You may choose to create an array and push the values into it, but consider space complexities.
}
console.log(twoSums(nums, target));
//call the function. Remeber we already declared the values at the top already.
//In my opinion, this method is best, it considers both time complexity and space complexityat its lowest value.
//Time complexity: 0(n)
function twoSum(numbers, target) {
for (let i = 0; i < numbers.length; i++) {
for (let j = i + 1; j < numbers.length; j++) {
if (numbers[i] + numbers[j] === target) {
return [numbers.indexOf(numbers[i]), numbers.lastIndexOf(numbers[j])];
}
}
}
}
I need help understanding what and how to use hash maps in javascript. I have an example where
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
Can someone breakdown what this hashmap solution is doing and why it's better? Also if someone would be kind of enough to give me a similar problem to practice with that would extremely helpful.
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
My BruteForce Solution
for (var i = 0; i < nums.length; i++) {
for (var j = i + 1; j < nums.length; j++) {
if (nums[i] + nums[j] === target) {
result.push(i);
result.push(j);
}
}
}
return result;
}
console.log(twoSum([2, 7, 11, 15], 9));
HashMapSolution
function twoSumBest(array, target) {
const numsMap = new Map();
for (let i = 0; i < array.length; i++) {
if(numsMap.has(target - array[i])) {
return [numsMap.get(target - array[i], i)];
// get() returns a specified element associated with the specified key from the Map object.
} else {
numsMap.set(array[i], i);
// set() adds or updates an element with a specified key and value to a Map object.
}
}
}
If you want to reach 10 and you have 7, you already know that the other number that is needed is 3. So for every number in the array, you only have to check wether the complementary number is in the array, you don't necessarily have to search for it.
If we go over all array entries once (let's call them a) and add them to a hashmap with their index, we can go over the array again and for each entry (b), we can check if the hashmap contains an a (where a + b = target). If that entry is found, the index of b is known, and the index of a can be retrieved from the hashmap¹. Now using that method we only iterate the array twice (or just once, doesn't matter that much), so if you have 1000 numbers, it'll iterate 1000 times. Your solution will iterate 1000 * 1000 times (worst case). So the hashtable approach is way faster (for larger arrays).
¹ Hashmaps are very special, as the time it takes to look up a key takes a constant amount of time, so it does not matter wether the array has 10 or 10 million entries (the time complexity is constant = O(1)). Thus, looking up in an hashtable is way better than searching inside of an array (which takes more time with more entries = O(n)).
I have the following question (this is not school -- just code site practice questions) and I can't see what my solution is missing.
A non-empty array A consisting of N integers is given. The array contains an odd number of elements, and each element of the array can be paired with another element that has the same value, except for one element that is left unpaired.
Assume that:
*N is an odd integer within the range [1..1,000,000];
*each element of array A is an integer within the range [1..1,000,000,000];
*all but one of the values in A occur an even number of times.
EX: A = [9,3,9,3,9,7,9]
Result: 7
The official solution is using the bitwise XOR operator :
function solution(A) {
var agg = 0;
for(var i=0; i<A.length; i++) {
agg ^= A[i];
}
return agg;
}
My first instinct was to keep track of the occurrences of each value in a Map lookup table and returning the key whose only value appeared once.
function solution(A) {
if (A.length < 1) {return 0}
let map = new Map();
let res = A[0]
for (var x = 0; x < A.length; x++) {
if (map.has(A[x])) {
map.set(A[x], map.get(A[x]) + 1)
} else {
map.set(A[x], 1)
}
}
for ([key,value] of map.entries()) {
if (value===1) {
res = key
}
}
return res;
}
I feel like I handled any edge cases but I'm still failing some tests and it's coming back with a 66% correct by the automated scorer.
You could use a Set and check if deletion deletes an item or not. If not, then add the value to the set.
function check(array) {
var s = new Set;
array.forEach(v => s.delete(v) || s.add(v));
return s.values().next().value;
}
console.log(check([9, 3, 9, 7, 3, 9, 9])); // 7
You're not accounting for cases like this:
[ 1, 1, 2, 2, 2 ] => the last 2 is left unpaired
So your condition should be if ( value % 2 ) instead of if ( value === 1 ).
I think also there is not much benefit to using a Map rather than just a plain object.
The official solution works due to the properties of the bitwise XOR (^), namely the fact that a ^ a == 0, a ^ 0 == a, and that the operation is commutative and associative. This means that any two equal elements in the array will cancel each other out to become zero, so all numbers appearing an even amount of times will be removed and only the number with an odd frequency will remain. The solution can be simplified using Array#reduce.
function findOdd(arr) {
return arr.reduce((a,c)=>a ^ c, 0);
}
You need not to make a count of each and traverse again, if you are sure that there will be exactly one number which will occur odd number of times. you can sum the array and do + when odd entry and - when even entry (to dismiss it back) and in the hash (map or object) you can just toggle for subsequent entry of each number.
Here is an example:
let inputArray1 = [10,20,30,10,50,20,20,70,20,70,50, 30,50], //50 -> 3 times
inputArray2 = [10,20,30,20,10], //30 -> 1 time
inputArray3 = [7,7,7,7,3,2,7,2,3,5,7]; //5 -> 1 time
let getOddOccuence = arr => {
let hash = {};
return arr.reduce((sum, n) => sum + ((hash[n] = !hash[n]) ? n : -n), 0);
}
console.log('Input Array 1: ', getOddOccuence(inputArray1));
console.log('Input Array 2: ', getOddOccuence(inputArray2));
console.log('Input Array 3: ', getOddOccuence(inputArray3));
In case the input contains multiple or no numbers having odd number of occurance (if you are not sure there) then you have already the hash (and you can ignore performing sum) and return the keys of hash (where value is true (and not checking with %2 and then consider as truthy or false in case of you have count))
function solution(A) {
let result = 0;
for (let element of A) {
// Apply Bitwise XOR to the current and next element
result ^= element;
}
return result;
}
const unpaired = solution([9, 3, 9, 3, 9, 7, 9]);
console.log(unpaired);
Source: https://gist.github.com/k0ff33/3eb60cfb976dee0a0a969fc9f84ae145
Hello all I am trying to understand this solution to combination sum.
function combinationSum(candidates, target) {
var result = [];
if ((candidates == null) || (candidates.length == 0)) {
return result;
}
var cur = [];
candidates = candidates.sort((a, b) => a - b)
csh(candidates, target, cur, result, 0);
return result;
};
function csh(cand, target, cur, result, j) {
//console.log(cur);
if (target == 0) {
var temp = cur.slice();
result.push(temp);
return;
}
for (var i = j; i < cand.length; i++) {
if (target < cand[i]) {
return;
}
//console.log(cur);
cur.push(cand[i]);
console.log(cur);
csh(cand, target - cand[i], cur, result, i);
cur.pop();
}
}
https://leetcode.com/problems/combination-sum/description/
While I understand the basic principles of recursion this problem is a little bit lost on me. So for example for the input:
candidates = [2,3,6,7]
target = 7
When you first enter the function cur is empty so our first iteration is:
[],
[2],
[2,2]
And then we keep adding cand[i] which is currently 2
[2,2,2]
However at this point, target = 1 which is less than cand[i] which is 2 so we return. And since we're returning we pop off the stack which pops the last 2 off the stack. Since we've returned we increment i and then we add 3 to cur
[2,2,3]
Since our target array is equal to 0 we now return again and my question is, at this point do we keep returning until cur is empty and continue the function like the following?
[2,2]
[2]
[]
[6]
[]
[7]
I'm just trying to understand what is being done in this function.
target is local to each invocation of the function. target is 0 only for some invocations of the function. Notice that the recursive invocation is:
csh(cand, target - cand[i], cur, result, i);
The target in that scope has not change, but the invocation of csh currently being entered will have a lower value for its target. When that function returns and the program flow reenters that other level, we resume using the higher value of target, insted of the reduce value target - cand[i] that was supplied to the subcall.
The algorithm will try all other possibilities on the [2,2,...] path as well, before changing the second element to the next alternative. Then it will explore the [2,3,...] space, and the [2,6,...] space, and ultimately all the [3,...], [6,...] and [7,...] possibilities as well.
The algorithm always goes as deep as a possible (i.e., as long of an array as possible) when it can do so without going over the original limit.
Note that it does not produce [2,3,2], because an earlier candidate cannot come after a later one (so a 2 can never be subsequent to a 3 in a result). It enforces this by making the for look start at i = j, where j is the array-depth of the last candidate used, so when the result-in-progress ends with the nth candidate, it only considers nth and higher candidates. this is of practical value because the algorithm only returns one permutation of each value result set: [2,2,3] and [2,3,2] contain the same set of values.
I completely understand that recursion can be very difficult to understand and to explain as well, but here is my take on it.
When csh is called for the 1st time, this is what is being passed
csh(cand, 7, [], [], 0)
Now from the for loop, i = 0, function called is
csh(cand, 5, [2], [], 0)
from the loop, i = 0, function called is
csh(cand, 3, [2,2], [], 0)
from the loop, i = 0, function called is
csh(cand, 1, [2,2,2],[],0)
from the for loop, target(1) < cand[0](2), so return to step Step 4 and pop the last 2 from [2,2,2] resulting in [2,2]
from the loop i = 1, function called is
csh(cand, 0, [2,2,3], [], 1)
here, target == 0 condition is met. so, [2,2,3] is pushed in the result. and then return to step 4. again, 3 is popped from [2,2,3].
from the loop i = 2, target(3) < cand[2](6), so return to step 3. and pop 2 from [2,2] resulting in [2].
from the loop i = 1, function called is
csh(cand, 2, [2,3], [[2,2,3]], 1)
from the loop i = 1, target(2) < cand[1](1) so return to step 9.
and so no...
Basically, each and every combination will be checked.
[2,2,2]
[2,2,3] --> added to result
[2,3,3]
[2,6]
[2,7]
[3,3,3]
[3,6]
[3,7]
[6,6]
[6,7]
[7] --> added to res
In javascript, how do I remove an element from an array of objects?
Here is the code:
$.fn.mapImage.deletePinpoint = function(image, pinpoint){
var deleted = false;
for (var i = 0; i < image.pinpoints.length; i++) {
if(image.pinpoints[i].position == pinpoint.position){
image.pinpoints.remove(i);
deleted = true;
}
if(deleted){
image.pinpoints[i].position -= 1;
}
}
$('.edit-mode').find('div.dynamic-pinpoint-area').remove();
$('.edit-mode').find('div.pinpoint-text').remove();
$('.create-mode').find('div.static-pinpoint-area').remove();
$('.create-mode').find('div.pinpoint-text').remove();
$.fn.mapImage.load(image);
}
image.pinpoints is the array of objects. Thanks again guys!
See https://developer.mozilla.org/en/Core_JavaScript_1.5_Reference/Operators/Special_Operators/delete_Operator
e.g. (from source)
var trees = ["redwood","bay","cedar","oak","maple"];
delete trees[3];
if (3 in trees) {
// this does not get executed
}
.splice is the method given at w3schools.com http://www.w3schools.com/jsref/jsref_splice.asp
To remove one element from an array with index x you would have trees.splice(x,x+1); This removes x and returns it if you need it.
I think you should rephrase the question to be more clear. From your example, it looks like multiple elements can get deleted from the image.pinpoints array if the position property matches that of pinpoint. So it will delete each image.pinpoints[i].position == pinpoint.position where i goes from 0 to (image.pinpoints.length - 1).
Since you are also iterating through the array at the same time, I wouldn't recommend using splice by itself. Instead delete each index first, and then cleanup the array in a second pass.
splice and delete will work differently as delete creates a hole in the array and sets the deleted property's value to undefined. On the other hand, splice will remove the element as if it never existed, and fix the indexes of all elements after it to be contiguous. Consider this example:
> var a = [2,3,5,7,11]; // create an array of 5 elements
> undefined
> a[2] // see the value of the third element
> 5
> delete a[2] // delete the third element using "delete"
> true
> a // log contents of a
> [2, 3, undefined, 7, 11]
> a[2] // index 2 still exists with value "undefined" now
> undefined
splice here by itself is also problematic as if you delete an element, all indexes after that element will shift one up, and you will skip checking the next element. Consider this second example:
> var a = [2,3,5,7,11]; // create array of 5 elements
> for(var i = 0; i < a.length; i++) {
if(a[i] == 3 || a[i] == 5) { // if it's 3 or 5, take it out
a.splice(i, 1);
}
}
> a
[2, 5, 7, 11]; // yikes, 5 still exists
In the above example, 5 is still present as we never checked that value. When we saw the 3, the current index was 1. After splicing the array, the next element - 5 moved up to take it's spot and became index 1. Since we're already done with index 1 at this point, we will simply move onto the next index - 2, which now has value 7, and skip 5. In general it's not a good practice to iterate using indexes, and do in-place removals.
As a solution, I would create a new array and only insert the properties which are not to be deleted in it.
$.fn.mapImage.deletePinpoint = function(image, pinpoint) {
// will hold all objects that are not to be deleted
var remainingPinpoints = [];
for (var i = 0; i < image.pinpoints.length; i++) {
// reverse condition
if(image.pinpoints[i].position != pinpoint.position) {
// add to new array
remainingPinpoints.push(image.pinpoints[i]);
}
}
// assign new array to pinpoints property
image.pinpoints = remainingPinpoints;
...
}