I have two examples of recursion, one calculate the sum of all elements of the array and the second return the count, both examples behave differently and i can't relate!
First Example:
const sum = (list) => {
if (list.length === 0) {
return 0;
}
return list[0] + sum(list.slice(1));
};
console.log(sum([1, 2, 3, 4])); // 10
The second:
const count = (list) => {
if (list.length === 0) {
return 0;
}
return 1 + count(list.slice(1));
};
console.log(count([0, 1, 2, 3, 4, 5])); // 6
Why the first recursion go through all array elements adding every element while the other added 1 to each element then returned just the final value?? i assumed it would do the same and the difference would be just adding 1 to the sum!!
The recursion is the same.
It visits all elements of the array and returns zero if no element is available.
For each element, it returns either the item for summing or one for counting.
Look at return 1 + count(list.slice(1)); in the count function. It would just ignore the 1st element in the list you pass recursively to the function and always use 1. Versus the sum function does consider that element. That effectively returns the count of times the function has been called which is 6
1st - sum, 2nd - count,
If you wish them to do the same thing you should use only one of them.
Related
I'm trying to solve a coding challenge
it gives me an array and waits for the answer ,then gives me the next array and so on.
https://www.codewars.com/kata/5648b12ce68d9daa6b000099/train/javascript
I am trying to take this input: var busStops = [[10,0],[3,5],[5,8]] and return 5.
the code is supposed to add the first in each pair to a total while subtracting the second in each pair from the total eg: 10 - 0 + 3 - 5 + 5 - 8 = 5
First my code loops through the inner arrays and outer array ,pushing it into myarr as a regular array eg: [10,0,3,5,5,8].
It then adds the value if it is index is 0 or even and subtracts it if the index is odd.
This actually works!
Until it is given a second array eg: [[3,0],[9,1],[4,10],[12,2],[6,1],[7,10]]
It is still calculating the total correctly but is still remembering the total from the first array meaning it is returning 22 instead of 17
Why?
There is a var answer = 0 that is being executed ahead of the second loop
It should forget the value of the previous answer.
Right?
Edit: I figured out my problem. I just needed to empty myarr after the total was calculated!
let myarr = [];
var number = function (busStops) {
for (var i = 0; i < busStops.length; i++) {
for (var j = 0; j < busStops[i].length; j++) {
/*console.log(busStops[i][j]);*/
myarr.push(busStops[i][j]);
}
}
console.log("test");
var answer = 0;
console.log("test again");
for (let t = 0; t < myarr.length; t++) {
if (t == 0 || t % 2 == 0) {
answer = answer + myarr[t];
} else {
answer = answer - myarr[t];
}
}
console.log(answer);
return answer;
};
The task at your hand tries to somehow find a value (an integer) from an array of arrays (multidimensional array). That task seems to be reducing that multidimensional array into a single integer.
Luckily, JavaScript has some powerful array methods and one of them is the reduce method:
The reduce() method executes a user-supplied "reducer" callback function on each element of the array, in order, passing in the return value from the calculation on the preceding element. The final result of running the reducer across all elements of the array is a single value. Source: MDN
The reduce method tries to reduce an array's elements into a single value and in your case we want to reduce that multidimensional array into a single value that is the number persons who are still in the bus.
Before typing some code, let's dig a bit deeper into the reduce method:
It accepts 2 parameters, a callback function that acts as the reducer and the initial value to be used in the first iteration of the reduce method.
The reducer callback function, on its own, accepts 4 parameters that are supplied by the reduce method. You may learn more about those parameters here as am only going to focus on the first 2 parameters the reducer accepts:
previousValue: hold the value from the previous reducer call. On first call, it contains the value you set to the initial value parameter of the reduce method or, if you didn't supply an initial value, previousValue shall hold the value of your array's first element (arr[0]).
currentValue: hold the current reduce iteration's item.
Now, let's get back to the task we have, we need to calculate the number of persons who are still in the bus based on a supplied multidimensional array. Each item in that multidimensional array is an array of two values where the result we need at the end is: the sum of the differences between each array, in the multidimensional array, first and second values (sum = multiDim[i][0] - multiDim[i][1] + multiDim[i + 1][0] + multiDim[i + 1][1] etc...).
To solve this task, we'll reduce that multidimensional array into a single number (let's call it result) by using a simple reducer function that will start by an initial value of 0 (as we're calculating a sum in our case) and will add, to the result, the difference between the first and the second values of the array supplied by the reduce at each iteration.
To illustrate, here's a live demo:
/**
* a function that calculates and returns the number of person who are still in the bus or precisely, the sum of the differences between "c[0]" and "c[1]"
* busArray: the supplied multidimensional array to reduce
* the reducer accepts two parameters:
* r: the result from the last call to the reducer function (initially set to 0, the initial value (see second parameter passed to the "reduce" method))
c: hold the current iteration's array.
*/
const calculateWhoAreInTheBus = busArray => busArray.reduce((r, c) => r += c[0] - c[1], 0);
/** test the created "calculateWhoAreInTheBus" function */
console.log(calculateWhoAreInTheBus([
[10, 0],
[3, 5],
[5, 8]
])); // should print: 5
console.log(calculateWhoAreInTheBus([
[3, 0],
[9, 1],
[4, 10],
[12, 2],
[6, 1],
[7, 10]
])); // should print: 17
console.log(calculateWhoAreInTheBus([
[3, 0],
[9, 1],
[4, 8],
[12, 2],
[6, 1],
[7, 8]
])); // should print: 21
console.log(calculateWhoAreInTheBus([
[0, 0],
[0, 0]
])); // should print: 0
I would advice you to use Array.prototype.reduce instead. For example like this:
const reducer = (previous, current) => previous + current[0] - current[1];
const answer = busStops.reduce(reducer, 0);
It is very brief (although this is not a goal in and of itself) and the reducer function does almost trivial work, so it does not complicate unneccesarily. Best of all it encapsulates the functionality with a minimal need of extra variables.
Othwerwise you could simplify your function a bit but use the let keyword to keep variables locked to scope like:
function number(busStops) {
let answer = 0;
for (let bs of busStops) {
answer += bs[0] - bs[1];
}
return answer;
}
Don't understand why I have to make the assignment on line 28 to make the code work! Lines 17 to 21 is where I get the results.This is a very basic solution - more interested in why the code works.
//Given a list of non-negative integers and a target sum, find a pair of numbers that sums to the target sum.
function findPairForSum(integers, target)
{
// your solution here
var integerPairArray = [];
// first have the first element as one of the pairs
var firstInteger = integers[0];
// will store the sum of integers
var sum = 0;
// iterate through the remaining elements
for(var index = 1; index < integers.length; index++)
{
// add each to the first element and compare the result with the target
sum = firstInteger + integers[index];
// if true then return the pair in an array
if(sum === target)
{
integerPairArray.push(firstInteger, integers[index]);
break;
}
// if the end of the array is reached and there is no result
if(index === integers.length - 1 && sum !== target)
{
// remove the first element from the array
integers.shift();
// repeat the loop
integerPairArray = findPairForSum(integers, target);
}
}
return integerPairArray;
}
// example
var pair = findPairForSum([3, 34, 4, 12, 5, 2], 17);
console.log(pair); // --> [12, 5]
Since the question has comments, and no answers, I will endeavor to provide an answer based on the comments made .#Solomon P Byer really helped me understand my code! Thank you!
The final return statement from the findPairForSum(integers, target) declaration will go to this function call/expression - var pair = findPairForSum([3, 34, 4, 12, 5, 2], 17); - made outside the function.
Because I made a recursive call within the findPairForSum(integers, target) function, the final return statement will initially go to the call within the function. And because it is a return statement, it has to be ‘caught’ as #Solomon P Byer mentioned.
It is caught by the integerPairArray variable and then the final return statement in the function returns the answer to the function call made outside the function and we get the answer! Whew!
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])];
}
}
}
}
The MDN defines reduce() as below:
The reduce() method executes a reducer function (that you provide) on each element of the array, resulting in a single output value.
Let's not take anything for granted and just analyse the syntax of below statements:
const array1 = [1, 2, 3, 4];
const reducer = (y, x) => y + x;
// 1 + 2 + 3 + 4
console.log(array1.reduce(reducer));
// expected output: 10
// 5 + 1 + 2 + 3 + 4
console.log(array1.reduce(reducer, 5));
reducer is a function that takes two parameters and return their sum
reducer is executed on each element of the array, but 'each element of the array' is only ONE parameter, why can i assume that the SUM is the other parameter and cached somewhere waiting for the next add operation? and why can i assume reduce is returning the sum at the end?
Could someone answer these questions? From someone coming from other language background e.g. C which also has function concept. I am often confused by Javascript's syntax.
and that's how i get more confused when i see:
const pipeline = [
array => { array.pop(); return array; },
array => array.reverse()
];
pipeline.reduce((xs, f) => f(xs), [1, 2, 3]);
because again, according to MDN, The reduce() method executes a reducer function (that you provide) on each element of the array, resulting in a single output value.
only this time,
reducer: (xs, f) => f(xs)
accumulator: [1, 2, 3]
array1: pipeline
then how do we explain its behaviour similar to our first example in English?
reducer is executed on each element of the array, but 'each element of the array' is only ONE parameter, why can i assume that the SUM is the other parameter and cached somewhere waiting for the next add operation?
The callback provided is called multiple times, once for each item in the array (or, for length - 1 times, in case no initial value is provided). You could easily implement this yourself:
const array1 = [1, 2, 3, 4];
const reducer = (y, x) => y + x;
Array.prototype.myReduce = function(callback, initialValue) {
let accum = initialValue === undefined
? this.shift()
: initialValue;
for (let i = 0; i < this.length; i++) {
accum = callback(accum, this[i], i, this);
}
return accum;
}
// 1 + 2 + 3 + 4
console.log(array1.myReduce(reducer));
// expected output: 10
// 5 + 1 + 2 + 3 + 4
console.log(array1.myReduce(reducer, 5));
Just because you pass one function to .reduce (or to any other function) doesn't put a limit on how many times that function can be called.
and why can i assume reduce is returning the sum at the end?
That's just how the method is defined - like with the implementation above, the accumulator (or accum) gets reassigned for every iteration, and passed to the next invocation of the callback.
The browser's native implementation of the method isn't actually written in Javascript like above, but its functionality is the same (for the most part).
The pipeline works the same way. For every element of the array provided, the accumulator is reassigned, and the next element is called with the new accumulator. Here, there's an array of functions which are being called, and each function's return value is being used as the next accumulator, and the value returned by the last function call is what the whole .reduce call resolves to.
const pipeline = [
array => { array.pop(); return array; },
array => array.reverse()
];
pipeline.reduce((xs, f) => f(xs), [1, 2, 3]);
Item 1: initial value (accumulator) is [1, 2, 3]. Plugging into array => { array.pop(); return array; } and you .pop() its last value (the 3, resulting in [1, 2], then you return the array.
Item 2: Accumulator (return value of last iteration) is [1, 2]. Plug it into array => array.reverse(), and you get the same array, reversed: [2, 1].
There are no more items in the array, so this [2, 1] is the value that the whole reduce call evaluates to.
In studying the reduce method I'm not quite sure why the callback passed in needs the third and fourth parameters, index and array. In the example from the MDN:
[0, 1, 2, 3, 4].reduce(function(previousValue, currentValue, index, array) {
return previousValue + currentValue;
});
A lot of other uses for the array reduce method or underscore reduce function I've studied only make use of the first two parameters for the callback: previousValue (sometimes seen as accumulator) and currentValue (aka elem, the value of the current index).
Why is reduce sometimes written with callback taking four parameters and sometimes written with only previousValue and currentValue?
What would be a case where the index and array parameters are needed?
Should all four parameters always be given in a function definition for reduce in case an application of reduce requires the third or fourth parameters?
here is a (slightly) less-contrived example to sum up unique values in an array, skipping duplicates, using the index and array arguments to find unique values:
[0, 1, 2, 3, 2, 1, 0].reduce(function(previousValue, currentValue, index, array) {
return array.indexOf(currentValue) === index ? // value used already?
previousValue + currentValue : // not used yet, add to sum
previousValue; // used already, skip currentValue
}); // == 6 ( 0+1+2+3 )
live demo: http://pagedemos.com/fn764n659ama/1/
side note: [].reduce() runs quite a bit faster in V8 by specifying all four arguments as formal parameters on the callback, regardless of if they are used by the code inside the function.
If you only need to use the first two parameters, it is perfectly fine to leave the last two out in the function arguments list. In that case the last two arguments will just be ignored. To sum the array, this is a perfectly acceptable way to do it:
[0, 1, 2, 3, 4].reduce(function(previousValue, currentValue) {
return previousValue + currentValue;
});
The last two parameters, index and array just give you extra information in case you need it. For example, say you wanted to sum up the all the elements in the array multiplied by their mirror element, that is, given the array [0, 1, 2, 3, 4] you wanted to output
(0 * 4) + (1 * 3) + (2 * 2) + (3 * 1) + (4 * 0)
then you would have to use the last two parameters:
[0, 1, 2, 3, 4].reduce(function(previousValue, currentValue, index, array) {
return previousValue + currentValue * array[array.length - index - 1];
});
Admittedly, this is a somewhat contrived example, but I'm having a hard time coming up with an example that doesn't seem contrived. In any event, the last two parameters do occasionally come in handy.