## Javascript - Counting array elements by reduce method until specific value occurs doesn't give a correct output - javascript

### JavaScript sum array using recursion

```My task is to sum elements of an array and add it to second parameter (number) using recursion.
Return only gives me last value of sum. I would appreciate any feedback :)
const getArraySum = (numbersArray, initialValue) => {
// let sum = initialValue
// for (let i = 0; i < numbersArray.length; i++) {
// sum += numbersArray[i]
// } return sum
if (numbersArray.length === 0 ) {
return initialValue
} else {
let sum = 0
sum += numbersArray[numbersArray.length-1]
console.log (numbersArray)
numbersArray.pop()
console.log (sum)
getArraySum (numbersArray)
return sum + initialValue
}
};
const result1 = getArraySum([4,7,10], 5)
console.log (result1)
```
```You're ignoring the return value of the recursive call to getArraySum. Instead, you should add it to the returned value:
const getArraySum = (numbersArray, initialValue) => {
if (numbersArray.length === 0) {
return initialValue
}
return numbersArray.pop() + getArraySum(numbersArray, initialValue);
};
const result = getArraySum([4,7,10], 5)
console.log (result)
Note that the initialValue should only be taken into consideration once, in the end condition where the array is empty.
```
```The idea is to split an array into head (=the first element or null for an empty array) and tail (everything else). Then, establish that the sum is head + sum(tail):
let sum = ([head = null, ...tail]) =>
console.log(sum([1,2,3,4]))
Having an initial value is a silly requirement, but it goes the same way:
let sum = ([head = null, ...tail], init = 0) =>
console.log(sum([1, 2, 3, 4], 100))
```
```You could take the single values a summing variable and return immediately if the array is empty.
Otherwise return the function call with a new shorter array and hte sum of sum and the first element.
const
getSum = (numbers, sum = 0) => {
if (!numbers.length) return sum;
return getSum(numbers.slice(1), sum + numbers[0]);
};
console.log (getSum([4, 7, 10], 5));
```
```As long as the array has an element, return last element (pop) plus getArraySum([rest of the array]), otherwise return initial value:
const getArraySum = (numbersArray, initialValue) => {
if (numbersArray.length === 0 ) {
return initialValue
} else {
return numbersArray.pop() + getArraySum(numbersArray, initialValue);
}
};
const result1 = getArraySum([4,7,10], 5)
console.log (result1)```

### Calculating second largest number in array (Javascript) : Seems to work in Sandbox but fails Hackerrank testing

```Tried to determine the second largest number in an array (Javascript) on CodeSandbox. It seems to work fine, but it fails the CodeWars testing. I have added a dummy array just to run my own tests in Sandbox.(Have mercy, I'm a beginner and this is my first StackOverFlow question)
const nums = [3, 100.3, 88, 1, -2.4, 9, 18];
const getSecondLargest = (nums) => {
const descending = nums.sort((a, b) => b - a);
return descending[1];
};
console.log(getSecondLargest(nums)); // console returns 88
EDIT: Okay so I with my super-tired brain I said CodeWars, when I actually meant Hackerrank (so sorry!). I realized they didn't necessarily test with NaNs, but they did have repeating numbers, so using the index of [1] isn't ideal. The exercise is from the 10 Days of Javascript - Day 3: Arrays https://hackerrank.com/domains/tutorials/10-days-of-javascript
So I now tried this code below, and it passes...but my code seems a bit janky, is there a cleaner way to write this, and can I combine it with the isNan logic then?
const nums = [3, 100, 88, 100, -2.4, 9, 18];
const getSecondLargest = (nums) => {
const ascending = nums.sort((a, b) => a - b);
if (ascending[ascending.length - 2] === ascending[ascending.length - 1]) {
return ascending[ascending.length - 3];
} else {
return ascending[ascending.length - 2];
}
};
console.log(getSecondLargest(nums)); // console returns 88
```
```It looks like there maybe strings in the array and you need to handle that. Here are a few ways:
One is to filter the non-numerical stuff out before sorting. You can use isNaN() to test if an object "is not a number".
const getSecondLargest = (nums) => {
const descending = nums
.filter(n => !isNaN(n))
.sort((a, b) => b - a);
return descending.length < 2 ? undefined : descending[1];
};
Another option is to handle the strings in sorting. Push them to the end of the array:
const getSecondLargest = (nums) => {
const descending = nums.sort((a, b) => {
if (isNaN(a) && isNaN(b)) return 0;
if (isNaN(a)) return 1;
if (isNaN(b)) return -1;
return b - a;
});
return descending.length < 2 || isNaN(descending[1]) ? undefined : descending[1];
};
A third way is a simple for loop that keeps track of the 2 highest values:
const getSecondLargest = (nums) => {
let max1 = undefined;
let max2 = undefined;
for (let n of nums) {
if (isNaN(n)) continue;
if (max2 === undefined || n > max2) {
if (max1 === undefined || n > max1 ) {
max2 = max1;
max1 = n;
}
else {
max2 = n;
}
}
}
return max2;
}```

### Javascript: Remove element from array

```I have an array and I want to delete some elements. My array is arrayAuxiliar. My code is the following one:
for( var i = 0; i < arrayAuxiliar.length; i++)
{
if (arrayAuxiliar[i] == valueFirts) {
arrayAuxiliar.splice(i,1);
}
else if(arrayAuxiliar[i] == value){
arrayAuxiliar.splice(i,1);
}
}
Example of what is happening:
Initial values: arrayAuxiliar = [3,2,1], valueFirst = 1, value = 2
Final values: arrayAuxiliar = [3,1]
I know this happens because splice() changes the original array and that's for this reason the comparasion between arrayAuxiliar[i] == valueFirts is never true. I've also tried to use remove arrayAuxiliar[i] but it returns [3,,].
How can I solve this situation in order to get the only element that does not verify the conditions which is 3? The idea was the final result be [3] and I could get it by a arrayAuxiliar[0] command.
```
```Use filter and includes method
const array = [3, 2, 1]
const remove_list = [2, 1]
const res = array.filter(num => !remove_list.includes(num))
console.log(res)
```
```I think using filter would be more appropiate. You can try:
let arrayAuxiliar = [3, 2, 1]
const valueFirst = 1
const value = 2
arrayAuxiliar = arrayAuxiliar.filter(i => i !== valueFirst && i !== value)
console.log(arrayAuxiliar)
```
```I would use Array.prototype.filter():
const arr = [3,2,1];
const valueFirst = 1;
const value = 2;
console.log(arr.filter(item => ![value, valueFirst].includes(item)));
```
```Also you can use reduce function
function test(arrayAuxiliar, valueFirts, value) {
return arrayAuxiliar.reduce((acc,rec) => {
if ((rec === value) || (rec === valueFirts)) {
return [...acc]
}
return [...acc, rec]
}, [])
}
console.log(test([3,2,1,4,1,2,2,3,1,4,5,8,2,5], 1, 2))```

### Count the repetition of an element in an array using a function with one parameter

```Good Day, I am trying to count how many times a particular element in an array appears. I tried but my code below counts only one of the array even if it appears more than once (this is not the problem). I want it to return the amount of time each element appears. For example
let arr = [1, 3, 2, 1];
this should return
{1:2} {3:1} {2:1}
My code returns 3 (as in it just doesn't count one twice)
Below is my code
function numberCount(number) {
let count = 0;
number.forEach(function (item, index) {
if (number.indexOf(item) == index) count++;
});
console.log(count);
}
```
```While iterating over number (better to call it arr, it's an array, not a number), use an object to keep track of the number of times each number has occured so far. Then, iterate over the resulting object's entries to create the objects desired:
let arr = [1, 3, 2, 1];
function numberCount(arr) {
let count = 0;
const obj = arr.reduce((a, num) => {
a[num] = (a[num] || 0) + 1;
return a;
}, {});
return Object.entries(obj).map(([key, val]) => ({ [key]: val }));
}
console.log(numberCount(arr));
Numeric keys always come in numeric order in an object. If you want the objects in the output to come in insertion order (eg, the object with key 3 before the object with key 2), then use a Map instead of an object (map keys will be iterated over in insertion order):
let arr = [1, 3, 2, 1];
function numberCount(arr) {
let count = 0;
const map = arr.reduce((a, num) => (
a.set(num, (a.get(num) || 0) + 1)
), new Map());
return [...map.entries()]
.map(([key, val]) => ({ [key]: val }));
}
console.log(numberCount(arr));
```
```You should filter out these numbers, then use the length:
let arr = [1, 3, 2, 1];
function itemCount(array) {
var sorted = array.sort()
var uniqueCount = sorted.filter((v, i, a) => a.indexOf(v) == i);
var count = [];
uniqueCount.forEach(item => {
var itemCount = sorted.filter(e => e == item).length;
count.push({[item]: itemCount});
});
return count;
}
console.log(itemCount(arr));
```
```I would suggest not reinventing the wheel, and instead use lodash which already has this function. Using countBy() you will get an object you can then convert into your desired result. For example:
const arr = [1, 3, 2, 1]
const count = _.countBy(arr)
const result = Object.keys(count).map(k => ({ [k]: count[k] }))
console.log(result)
<script src="https://cdn.jsdelivr.net/npm/lodash#4.17.11/lodash.min.js"></script>```

### How to early break reduce() method?

```How can I break the iteration of reduce() method?
for:
for (var i = Things.length - 1; i >= 0; i--) {
if(Things[i] <= 0){
break;
}
};
reduce()
Things.reduce(function(memo, current){
if(current <= 0){
//break ???
//return; <-- this will return undefined to memo, which is not what I want
}
}, 0)
```
```You CAN break on any iteration of a .reduce() invocation by mutating the 4th argument of the reduce function: "array". No need for a custom reduce function. See Docs for full list of .reduce() parameters.
Array.prototype.reduce((acc, curr, i, array))
The 4th argument is the array being iterated over.
const array = ['apple', '-pen', '-pineapple', '-pen'];
const x = array
.reduce((acc, curr, i, arr) => {
if(i === 2) arr.splice(1); // eject early
return acc += curr;
}, '');
console.log('x: ', x); // x: apple-pen-pineapple
WHY?:
The one and only reason I can think of to use this instead of the many other solutions presented is if you want to maintain a functional programming methodology to your algorithm, and you want the most declarative approach possible to accomplish that. If your entire goal is to literally REDUCE an array to an alternate non-falsey primitive (string, number, boolean, Symbol) then I would argue this IS in fact, the best approach.
WHY NOT?
There's a whole list of arguments to make for NOT mutating function parameters as it's a bad practice.
UPDATE
Some of the commentators make a good point that the original array is being mutated in order to break early inside the .reduce() logic.
Therefore, I've modified the answer slightly by adding a .slice(0) before calling a follow-on .reduce() step, yielding a copy of the original array.
NOTE: Similar ops that accomplish the same task are slice() (less explicit), and spread operator [...array] (slightly less performant). Bear in mind, all of these add an additional constant factor of linear time to the overall runtime ... + O(n).
The copy, serves to preserve the original array from the eventual mutation that causes ejection from iteration.
const array = ['apple', '-pen', '-pineapple', '-pen'];
const x = array
.slice(0) // create copy of "array" for iterating
.reduce((acc, curr, i, arr) => {
if (i === 2) arr.splice(1); // eject early by mutating iterated copy
return (acc += curr);
}, '');
console.log("x: ", x, "\noriginal Arr: ", array);
// x: apple-pen-pineapple
// original Arr: ['apple', '-pen', '-pineapple', '-pen']
```
```Don't use reduce. Just iterate on the array with normal iterators (for, etc) and break out when your condition is met.
```
```You can use functions like some and every as long as you don't care about the return value. every breaks when the callback returns false, some when it returns true:
things.every(function(v, i, o) {
// do stuff
if (timeToBreak) {
return false;
} else {
return true;
}
}, thisArg);
Edit
A couple of comments that "this doesn't do what reduce does", which is true, but it can. Here's an example of using every in a similar manner to reduce that returns as soon as the break condition is reached.
// Soruce data
let data = [0,1,2,3,4,5,6,7,8];
// Multiple values up to 5 by 6,
// create a new array and stop processing once
// 5 is reached
let result = [];
data.every(a => a < 5? result.push(a*6) : false);
console.log(result);
This works because the return value from push is the length of the result array after the new element has been pushed, which will always be 1 or greater (hence true), otherwise it returns false and the loop stops.
```
```There is no way, of course, to get the built-in version of reduce to exit prematurely.
But you can write your own version of reduce which uses a special token to identify when the loop should be broken.
var EXIT_REDUCE = {};
function reduce(a, f, result) {
for (let i = 0; i < a.length; i++) {
let val = f(result, a[i], i, a);
if (val === EXIT_REDUCE) break;
result = val;
}
return result;
}
Use it like this, to sum an array but exit when you hit 99:
reduce([1, 2, 99, 3], (a, b) => b === 99 ? EXIT_REDUCE : a + b, 0);
> 3
```
```Array.every can provide a very natural mechanism for breaking out of high order iteration.
const product = function(array) {
let accumulator = 1;
array.every( factor => {
accumulator *= factor;
return !!factor;
});
return accumulator;
}
console.log(product([2,2,2,0,2,2]));
// 0
```
```You can break every code - and thus every build in iterator - by throwing an exception:
function breakReduceException(value) {
this.value = value
}
try {
Things.reduce(function(memo, current) {
...
if (current <= 0) throw new breakReduceException(memo)
...
}, 0)
} catch (e) {
if (e instanceof breakReduceException) var memo = e.value
else throw e
}
```
```You can use try...catch to exit the loop.
try {
Things.reduce(function(memo, current){
if(current <= 0){
throw 'exit loop'
//break ???
//return; <-- this will return undefined to memo, which is not what I want
}
}, 0)
} catch {
// handle logic
}
```
```As the promises have resolve and reject callback arguments, I created the reduce workaround function with the break callback argument. It takes all the same arguments as native reduce method, except the first one is an array to work on (avoid monkey patching). The third [2] initialValue argument is optional. See the snippet below for the function reducer.
var list = ["w","o","r","l","d"," ","p","i","e","r","o","g","i"];
var result = reducer(list,(total,current,index,arr,stop)=>{
if(current === " ") stop(); //when called, the loop breaks
},'hello ');
console.log(result); //hello world
function reducer(arr, callback, initial) {
var hasInitial = arguments.length >= 3;
var total = hasInitial ? initial : arr[0];
var breakNow = false;
for (var i = hasInitial ? 0 : 1; i < arr.length; i++) {
var currentValue = arr[i];
var currentIndex = i;
var newTotal = callback(total, currentValue, currentIndex, arr, () => breakNow = true);
if (breakNow) break;
total = newTotal;
}
}
And here is the reducer as an Array method modified script:
Array.prototype.reducer = function(callback,initial){
var hasInitial = arguments.length >= 2;
var total = hasInitial ? initial : this[0];
var breakNow = false;
for (var i = hasInitial ? 0 : 1; i < this.length; i++) {
var currentValue = this[i];
var currentIndex = i;
var newTotal = callback(total, currentValue, currentIndex, this, () => breakNow = true);
if (breakNow) break;
total = newTotal;
}
};
var list = ["w","o","r","l","d"," ","p","i","e","r","o","g","i"];
var result = list.reducer((total,current,index,arr,stop)=>{
if(current === " ") stop(); //when called, the loop breaks
},'hello ');
console.log(result);
```
```Reduce functional version with break can be implemented as 'transform', ex. in underscore.
I tried to implement it with a config flag to stop it so that the implementation reduce doesn't have to change the data structure that you are currently using.
const transform = (arr, reduce, init, config = {}) => {
const result = arr.reduce((acc, item, i, arr) => {
if (acc.found) return acc
acc.value = reduce(config, acc.value, item, i, arr)
if (config.stop) {
acc.found = true
}
return acc
}, { value: init, found: false })
return result.value
}
module.exports = transform
Usage1, simple one
const a = [0, 1, 1, 3, 1]
console.log(transform(a, (config, acc, v) => {
if (v === 3) { config.stop = true }
if (v === 1) return ++acc
return acc
}, 0))
Usage2, use config as internal variable
const pixes = Array(size).fill(0)
const pixProcessed = pixes.map((_, pixId) => {
return transform(pics, (config, _, pic) => {
if (pic[pixId] !== '2') config.stop = true
return pic[pixId]
}, '0')
})
Usage3, capture config as external variable
const thrusts2 = permute([9, 8, 7, 6, 5]).map(signals => {
const datas = new Array(5).fill(_data())
const ps = new Array(5).fill(0)
let thrust = 0, config
do {
config = {}
thrust = transform(signals, (_config, acc, signal, i) => {
const res = intcode(
datas[i], signal,
{ once: true, i: ps[i], prev: acc }
)
if (res) {
[ps[i], acc] = res
} else {
_config.stop = true
}
return acc
}, thrust, config)
} while (!config.stop)
return thrust
}, 0)
```
```You cannot break from inside of a reduce method. Depending on what you are trying to accomplish you could alter the final result (which is one reason you may want to do this)
const result = [1, 1, 1].reduce((a, b) => a + b, 0); // returns 3
console.log(result);
const result = [1, 1, 1].reduce((a, b, c, d) => {
if (c === 1 && b < 3) {
return a + b + 1;
}
return a + b;
}, 0); // now returns 4
console.log(result);
Keep in mind: you cannot reassign the array parameter directly
const result = [1, 1, 1].reduce( (a, b, c, d) => {
if (c === 0) {
d = [1, 1, 2];
}
return a + b;
}, 0); // still returns 3
console.log(result);
However (as pointed out below), you CAN affect the outcome by changing the array's contents:
const result = [1, 1, 1].reduce( (a, b, c, d) => {
if (c === 0) {
d[2] = 100;
}
return a + b;
}, 0); // now returns 102
console.log(result);
```
```Providing you do not need to return an array, perhaps you could use some()?
Use some instead which auto-breaks when you want. Send it a this accumulator. Your test and accumulate function cannot be an arrow function as their this is set when the arrow function is created.
const array = ['a', 'b', 'c', 'd', 'e'];
var accum = {accum: ''};
function testerAndAccumulator(curr, i, arr){
this.tot += arr[i];
return curr==='c';
};
accum.tot = "";
array.some(testerAndAccumulator, accum);
var result = accum.tot;
In my opinion this is the better solution to the accepted answer provided you do not need to return an array (eg in a chain of array operators), as you do not alter the original array and you do not need to make a copy of it which could be bad for large arrays.
```
```So, to terminate even earlier the idiom to use would be arr.splice(0).
Which prompts the question, why can't one just use arr = [] in this case?
I tried it and the reduce ignored the assignment, continuing on unchanged.
The reduce idiom appears to respond to forms such as splice but not forms such as the assignment operator??? - completely unintuitive - and has to be rote-learnt as precepts within the functional programming credo ...
const array = ['9', '91', '95', '96', '99'];
const x = array
.reduce((acc, curr, i, arr) => {
if(i === 2) arr.splice(1); // eject early
return acc += curr;
}, '');
console.log('x: ', x); // x: 99195
```
```The problem is, that inside of the accumulator it is not possible to just stop the whole process. So by design something in the outer scope must be manipulated, which always leads to a necessary mutation.
As many others already mentioned throw with try...catch is not really an approach which can be called "solution". It is more a hack with many unwanted side effects.
The only way to do this WITHOUT ANY MUTATIONS is by using a second compare function, which decides whether to continue or stop. To still avoid a for-loop, it has to be solved with a recursion.
The code:
function reduceCompare(arr, cb, cmp, init) {
return (function _(acc, i) {
return i < arr.length && cmp(acc, arr[i], i, arr) === true ? _(cb(acc, arr[i], i, arr), i + 1) : acc;
})(typeof init !== 'undefined' ? init : arr[0], 0);
}
This can be used like:
var arr = ['a', 'b', 'c', 'd'];
function join(acc, curr) {
return acc + curr;
}
console.log(
reduceCompare(
arr,
join,
function(acc) { return acc.length < 1; },
''
)
); // logs 'a'
console.log(
reduceCompare(
arr,
join,
function(acc, curr) { return curr !== 'c'; },
''
)
); // logs 'ab'
console.log(
reduceCompare(
arr,
join,
function(acc, curr, i) { return i < 3; },
''
)
); // logs 'abc'
I made an npm library out of this, also containing a TypeScript and ES6 version. Feel free to use it:
https://www.npmjs.com/package/array-reduce-compare
or on GitHub:
https://github.com/StefanJelner/array-reduce-compare
```
```You could to write your own reduce method. Invoking it like this, so it follows same logic and you control your own escape / break solution. It retains functional style and allows breaking.
const reduce = (arr, fn, accum) => {
const len = arr.length;
let result = null;
for(let i = 0; i < len; i=i+1) {
result = fn(accum, arr[i], i)
if (accum.break === true) {
break;
}
}
return result
}
const arr = ['a', 'b', 'c', 'shouldnotgethere']
const myResult = reduce(arr, (accum, cur, ind) => {
accum.result = accum.result + cur;
if(ind === 2) {
accum.break = true
}
return accum
}, {result:'', break: false}).result
console.log({myResult})
Or create your own reduce recursion method:
const rcReduce = (arr, accum = '', ind = 0) => {
const cur = arr.shift();
accum += cur;
const isBreak = ind > 1
return arr.length && !isBreak ? rcReduce(arr, accum, ind + 1) : accum
}
const myResult = rcReduce(['a', 'b', 'c', 'shouldngethere'])
console.log({myResult})
```
```Another simple implementation that I came with solving the same issue:
function reduce(array, reducer, first) {
let result = first || array.shift()
while (array.length > 0) {
result = reducer(result, array.shift())
if (result && result.reduced) {
return result.reduced
}
}
return result
}
```
```If you want to chain promises sequentially with reduce using the pattern below:
return [1,2,3,4].reduce(function(promise,n,i,arr){
return promise.then(function(){
// this code is executed when the reduce loop is terminated,
// so truncating arr here or in the call below does not works
return somethingReturningAPromise(n);
});
}, Promise.resolve());
But need to break according to something happening inside or outside a promise
things become a little bit more complicated because the reduce loop is terminated before the first promise is executed, making truncating the array in the promise callbacks useless, I ended up with this implementation:
function reduce(array, promise, fn, i) {
i=i||0;
return promise
.then(function(){
return fn(promise,array[i]);
})
.then(function(result){
if (!promise.break && ++i<array.length) {
return reduce(array,promise,fn,i);
} else {
return result;
}
})
}
Then you can do something like this:
var promise=Promise.resolve();
reduce([1,2,3,4],promise,function(promise,val){
return iter(promise, val);
}).catch(console.error);
function iter(promise, val) {
return new Promise(function(resolve, reject){
setTimeout(function(){
if (promise.break) return reject('break');
console.log(val);
if (val==3) {promise.break=true;}
resolve(val);
}, 4000-1000*val);
});
}
```
```I solved it like follows, for example in the some method where short circuiting can save a lot:
const someShort = (list, fn) => {
let t;
try {
return list.reduce((acc, el) => {
t = fn(el);
console.log('found ?', el, t)
if (t) {
throw ''
}
return t
}, false)
} catch (e) {
return t
}
}
const someEven = someShort([1, 2, 3, 1, 5], el => el % 2 === 0)
console.log(someEven)
UPDATE
Away more generic answer could be something like the following
const escReduce = (arr, fn, init, exitFn) => {
try {
return arr.reduce((...args) => {
if (exitFn && exitFn(...args)) {
throw args[0]
}
return fn(...args)
}, init)
} catch(e){ return e }
}
escReduce(
Array.from({length: 100}, (_, i) => i+1),
(acc, e, i) => acc * e,
1,
acc => acc > 1E9
); // 6227020800
give we pass an optional exitFn which decides to break or not```