I am completing the hackerrank's 10 days of javascript challenge. The question:
write a function to take an array as an argument and then return the second largest element in the array.
I have written the code but my code is returning the largest element and not the second largest as asked.
function getSecondLargest(nums) {
// Complete the function
var largest=nums[0];
for(let i=1;i<nums.length;++i)
{
if(nums[i]>largest)
largest=nums[i];
}
var large=nums[0];
for(let j=1;j<nums.length;++j)
{
if(large<nums[j]&&large<largest)
large=nums[j];
}
return large;
}
When input array nums={2,3,6,6,5} the result is coming 6 while expected output is 5. Please help and point out the errors in the function code below.
should not initialize large with first value var large=nums[0]; because it may appear the biggest value and won't work
should use nums[j]<largest instead of large<largest as mentioned above
I think don't need second loop as all checks can be done in first loop, and you can assign prev largest to large whenever you change it:
function getSecondLargest(nums) {
var largest = nums[0];
var large;
for (let i = 1; i < nums.length; ++i) {
if (nums[i] > largest) {
large = largest;
largest = nums[i];
} else if (nums[i] > large || typeof large === 'undefined') {
large = nums[i]
}
}
return large;
}
console.log(getSecondLargest([5,1-2,3]))
console.log(getSecondLargest([-5,1,-2,3]))
GET SECOND LARGEST
first, I create new array with unique values.
let arr = [...new Set(nums)];
second, sort value using built-in function .sort().
note : by default .sort() always sorts asciibetically, but for some testcase, it doesn't work. So, I put (a, b) => { return a - b } to make sure it will work properly.
arr = arr.sort((a, b) => { return a -b });
third, get the value from arr
let result = arr[arr.length - 2] || arr[0];
finally, return the result
return result
function getSecondLargest(nums) {
let arr = [...new Set(nums)];
//Javascript's array member method .sort( always sorts asciibetically.
arr = arr.sort((a, b) => { return a - b });
let result = arr[arr.length - 2] || arr[0];
return result
}
Just one minor change:
Use nums[j]<largest instead of large<largest in the second for loop
function getSecondLargest(nums) {
// Complete the function
var largest=nums[0];
for(let i=1;i<nums.length;++i)
{
if(nums[i]>largest)
largest=nums[i];
}
var large;
//To ensure that the selected number is not the largest
for(let j=0;j<nums.length;++j)
{
if (nums[j] !== largest){
large = nums[j];
break;
}
}
for(let j=1;j<nums.length;++j)
{
if(large<nums[j]&&nums[j]!=largest)
large=nums[j];
else
console.log(large)
}
return large;
}
var secondLargest = getSecondLargest([6,3,6,6,5]);
console.log("Second largest number", secondLargest);
If you want to avoid using library functions like #ifaruki suggests, this line
if(large<nums[j]&&large<largest)
should read
if (large<nums[j] && nums[j] < largest)
Sorting and picking the second or second-to-last value fails when there are duplicates of the highest value in the input array.
Another easiest logic is to remove duplicates from the array and sort.
let givenArray = [2, 3, 6, 6, 5];
let uniqueArray = [...new Set(givenArray)];
console.log("The second largets element is", uniqueArray.sort()[uniqueArray.length - 2]);
I know you had your question answered, just thought I would provide my solution for any future users looking into this.
You can use reduce to go through the array while remembering the two largest numbers so far.
You just make a simple reduction function:
function twoMax(two_max, candidate)
{
if (candidate > two_max[0]) return [candidate,two_max[0]];
else if (candidate > two_max[1]) return [two_max[0],candidate];
else return two_max;
}
And then you use it for example like this:
let my_array = [0,1,5,7,0,8,12];
let two_largest = my_array.reduce(twoMax,[-Infinity,-Infinity]);
let second_largest = two_largest[1];
This solution doesn't require sorting and goes through the array only once.
If you want to avoid using **sort method. I think here's the easiest logic to do that, which will also work in arrays where there's duplicates of largest integer exists.
function getSecondLargest(arr) {
const largest = Math.max.apply(null, arr);
for (let i = 0; i < arr.length; i++) {
if (largest === arr[i]) {
arr[i] = -Infinity;
}
}
return Math.max.apply(null, arr);
}
console.log(getSecondLargest([5, 7, 11, 11, 11])); //7
I am pulling my hair out with this algorithm. The goal is to take number array as an input and output a string that has all sequential numbers displayed as a range. For example if my input were [1,2,4,6,7,8,] my output should be "1-2,4,6-8". My issue lies with the variable assignment in my nested for loop's if statement. IndexEn is overridden each time j increments and should ultimately exit the inside loop as the highest value. The problem is that when I try to call the variable outside of the for loop it passes each value of IndexEn instead of the maximum. I don't understand how IndexEn is able to be outside of the scope of the inner loop while the inner loop is still running? Can someone please help me fix and understand whats happening here?
function bkInd(arr){
var bookSt="";
var indexSt;
var indexEn;
for(var i =0;i<arr.length-1;i++){
if(arr[i+1] !== (arr[i]+1)) {
if(i===0 || (i>0 && arr[i]) !== (arr[i-1]+1) ){
bookSt+= arr[i]+",";
}
// check to see if number are sequential and SHOULD output starting index value - last value
}else{
for(var j=i+1;j<arr.length;j++){
var count=0;
if(arr[j]==(arr[i +count]+1)){
indexSt = arr[i];
indexEn = arr[j];
count+=1;
}
}
//concatenate string
//console.log(indexEn); for value of index
bookSt+= indexSt+"-"+indexEn+",";
}
}
return bookSt;
}
var bookList = [1,3,4,5,7,9,10,11];
document.write(bkInd(bookList));
I don't like nested looping (can be slow on large datasets) so I took a different approach, hope you don't mind:
var bkInd = function(arr) {
var result = [];
var seq = []; // 'Temporary' array
for(var i = 0; i < arr.length; i++) {
seq.push(arr[i])
if(arr[i] + 1 !== arr[i + 1]) {
result.push(seq.length > 1 ? seq[0] + '-' + seq.pop() : arr[i]);
seq = [];
}
}
return result.join(', ')
}
// => '1, 3-5, 7, 9-11'
It instead 'builds up' a temporary array of numbers (seq) as it loops through arr. If the next number is sequential, seq keeps the last number and the loop continues. If the next number is more than one above the current one, seq is pushed to result, but if seq is more than one index long it will concatenate these to a hyphenated string. After seq get pushed, it gets reset to an empty array. Finally, it joins the result array with commas.
Code is updated according to your requirement :
function bkInd(arr){
var bookSt="";
var indexSt;
var indexEn;
for(var i =0;i<arr.length-1;i++){
if(arr[i+1] !== (arr[i]+1)) {
if(i===0 || (i>0 && arr[i]) !== (arr[i-1]+1) ){
bookSt+= arr[i]+",";
}
// check to see if number are sequential and SHOULD output starting index value - last value
}else{
var count=1;
indexSt = arr[i];
for(var j=i;j<arr.length;j++){
if(arr[j]==(indexSt+count)){
indexEn = arr[j];
count+=1;
i++;
}
}
//concatenate string
//console.log(indexEn); for value of index
bookSt+= indexSt+"-"+indexEn+",";
}
}
return bookSt;
}
var bookList = [1,3,4,5,7,9,10,11];
console.log(bkInd(bookList));
Your inner for loop logic was incorrect. I have updated the code to meet the requirement.
Scope is the region of the code within which a variable is directly accessible.
In JS scope is defined by functions, blocks (i.e. { and }) for let, const and by catch blocks.
The following seems to work:
function runnify(arr) {
var runFrom;
return arr.reduce((p,c,i)=>{
if (c + 1 === arr[++i]) { // true if next item is sequential
if (runFrom == null ) { // We are at start of run
runFrom = c;
}
return p;
}
if (runFrom) { // We are at the end of a run
p.push(`${runFrom}-${c}`);
runFrom = null;
return p;
}
p.push(c); // Not sequential
return p;
}, []).join(',');
}
var seq = [1, 3, 4, 5, 7, 9, 10, 11];
document.write(runnify(seq));
Basically, you could collect all items with a check, if consecutive in an array and join it at the end, for consecutive elements with a dash and the rest with comma.
function consecutive(array) {
return array.reduce(function (r, a,i) {
var last = r[r.length - 1];
if (!i || last[last.length - 1] + 1 !== a) {
r.push([a]);
} else {
last[1] = a;
}
return r;
}, []).map(function (a) { return a.join('-'); }).join();
}
var array = [1, 3, 4, 5, 7, 9, 10, 11],
grouped = consecutive(array);
console.log(grouped);
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
return total + current;
},'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;
}
return total;
}
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;
}
return total;
};
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
return total + current;
},'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