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Move zeroes to end of array in javascript - how to return nothing?
(21 answers)
Closed last year.
Trying to remove all 0's from an array and return them in the same array
So for example if I have this as an array
let arrThree = [9,0,9,1,2,1,1,3,1,9,0,0,9,0,0,0,0,0,0,0]
I would get this:
let arrThree = [9,9,9,1,2,1,1,3,1,9,9,0,0,0,0,0,0,0,0,0,0,0]
This is what I wrote:
var remove = function (arr) {
let test = [];
for(let i = 0; i < arr.length; i++){
arr[i] === 0 ? test.push(arr.splice(i,1)) : false //if i is 0 then we want to push that into another array
}
arr.push(...test)
return [].concat(...arr)
}
When I run this function I get this
[
9, 9, 1, 2, 1, 1, 3,
1, 9, 0, 9, 0, 0, 0,
0, 0, 0, 0, 0, 0
]
Not sure where I am going wrong?
You need only two iterations, one for finding non zero values and another to put zeroes to the right side until end of the array. You need no array pushing or splicing.
const
move0 = array => {
let i = 0,
j = 0;
while (i < array.length) {
if (array[i]) array[j++] = array[i];
i++;
}
while (j < array.length) {
array[j] = '#'; // in reality it is zero
j++;
}
},
array = [9, 0, 9, 1, 2, 1, 1, 3, 1, 9, 0, 0, 9, 0, 0, 0, 0, 0, 0, 0];
move0(array)
console.log(...array);
If you don't need to have any sorting on the non-zero elements, a rather efficient solution would be to swap the left-most 0 with the right-most non-zero element. Once your indices for tracking where you are on the left and right cross you'll know you're done.
function moveZeros(arr) {
let i = 0;
let j = arr.length - 1;
while(i < j) {
// Found a 0 to the left of a non-zero, swap.
if(arr[i] == 0 && arr[j] != 0) {
let tmp = arr[j];
arr[j] = arr[i];
arr[i] = tmp;
}
// Find a zero
if(arr[i] != 0) {
i++;
}
// Find a non-zero
if(arr[j] == 0) {
j--;
}
}
return arr;
}
console.log(moveZeros([1,2,0,3,0,0,0]));
console.log(moveZeros([9,0,9,1,2,1,1,3,1,9,0,0,9,0,0,0,0,0,0,0]));
The issue with your code is you change the length of arr with arr.splice(i,1). This messes up the loop condition i < arr.length.
To fix your code, you can loop backwards through the array. So as the length of the array shortens, i is still valid.
let arrThree = [9,0,9,1,2,1,1,3,1,9,0,0,9,0,0,0,0,0,0,0];
var remove = function(arr) {
let test = [];
for(let i = arr.length - 1; i >= 0; --i) {
if (arr[i] === 0) test.push(arr.splice(i, 1)[0]);
}
return arr.concat(test);
}
console.log(remove(arrThree));
There are a few other ways to do this. One is to use filter to create 2 arrays - one with no 0 and one with all 0 and join them:
let arrThree = [9,0,9,1,2,1,1,3,1,9,0,0,9,0,0,0,0,0,0,0];
let remove = (arr) => arr.filter(i => i != 0).concat(arr.filter(i => i == 0));
console.log(remove(arrThree));
I'm tackling this problem and I can't seem to arrive at the correct solution. The question is:
"Given a list of n integers arr[0..(n-1)], determine the number of different pairs of elements within it which sum to k. If an integer appears in the list multiple times, each copy is considered to be different; that is, two pairs are considered different if one pair includes at least one array index which the other doesn't, even if they include the same values.
My approach is that I'm building a map that contains each number in the array and the number of times it occurs. Then I iterate over the map to find my answer.
function numberOfWays(arr, k) {
let output = 0;
let map = {};
// put values and # of occurences into map
for(let i = 0; i < arr.length; i++) {
let key = arr[i];
if(!(key in map)) {
map[key] = 1;
} else {
map[key]++;
}
}
for(let key in map) {
let difference = k-key
if((difference) in map) {
if(k/2 === key) {
output += map[key]*(map[key]-1)/2;
} else {
output += map[key] * map[key] / 2; // divide by 2 so that pairs aren't counted twice
}
}
}
return output;
}
The two test cases are:
var arr_1 = [1, 2, 3, 4, 3]; expected result: [2] -- I'm getting [3]
var arr_2 = [1, 5, 3, 3, 3]; expected result: [4] -- I'm getting [5.5]
I'm definitely doing something wrong in my calculations, but I can't seem to wrap my ahead around it.
This is one way to nest the loops to find the pairs in array "arr" with the sum "k".
function numberOfWays(arr, k) {
let output = 0;
for (i = 0; i < arr.length; i++) {
for (n = i+1; n < arr.length; n++) {
if (arr[i] + arr[n] == k)
output++;
}
}
return output;
}
You could count the smaller and greater values for building k and then taker either the product or if only two of the same value is building the sum take factorial of the cound divided by two.
function numberOfWays(array, k) {
const
f = n => +!n || n * f(n - 1),
pairs = {};
for (const value of array) {
const smaller = Math.min(value, k - value);
pairs[smaller] ??= { one: 2 * smaller === k, min: 0, max: 0 };
pairs[smaller][value === smaller ? 'min' : 'max']++;
}
let count = 0;
for (const k in pairs) {
const { one, min, max } = pairs[k];
if (one) {
if (min > 1) count += f(min) / 2;
} else if (min && max) {
count += min * max;
}
}
return count;
}
console.log(numberOfWays([1, 2, 3, 4, 3], 6)); // 2
console.log(numberOfWays([1, 5, 3, 3, 3], 6)); // 4
function numberOfWays(items, k) {
// Clone as to not mutate original array
const arr = [...items]
let count = 0
// Stop comparing when no items left to compare
while (arr.length) {
for (let i = 0; i < arr.length; i++) {
// Compare each item to the first item
const sum = arr[0] + arr[i + 1]
if (sum === k) {
count++
}
}
// Remove the first item after comparing to the others
arr.shift()
}
return count
}
console.log(numberOfWays([1, 2, 3, 4, 3], 6))
console.log(numberOfWays([1, 5, 3, 3, 3], 6))
console.log(numberOfWays([1, 1, 1, 1, 1], 2))
import math
from math import factorial as f
def get_number_of_combination(n,r):
return f(n)//(f(n-r)*f(r))
def numberOfWays(arr, k):
num_count = {}
num_ways = 0
for i in arr:
old_count = num_count.get(i,0)
num_count.update({i: old_count+1})
for i in list(num_count.keys()):
if i == k - i and num_count.get(i,0) > 1:
num_ways += (get_number_of_combination(num_count.get(i,0),2))
num_count.update({i:0})
else:
i_n = num_count.get(i, 0)
ki_n = num_count.get(k-i, 0)
num_ways += i_n * ki_n
num_count.update({i:0,k-i:0})
return num_ways
Problem: find an index N where the sum of the integers to the left of N is equal to the sum of the integers to the right of N. If there is no index that would make this happen, return -1.
My solution
function findEvenIndex(arr) {
var sum = i => i.reduce((a, b) => a + b),
l = arr.length;
for (let j = 0; j <= l; j++) {
if (sum(arr.slice(0, j - 1)) === sum(arr.slice(j, l))) {
return j
} else {
continue;
}
}
return -1
}
console.log(
findEvenIndex([1, 2, 3, 4, 3, 2, 1])
)
When I run this on say findEvenIndex([1,2,3,4,3,2,1]), It doesn't return anything? Where is the error that prevents 3 from being returned in the case of this example?
I've set the for loop procedure as follows to see what's going on
for(let j = 0; j <= arr.length; j++){
var left = arr.slice(0, j-1), right = arr.slice(j)
console.log(left, right)
}
/* returns
[1] [3,4,3,2,1]
[1,2] [4,3,2,1]
[1,2,3] [3,2,1]
as expected
*/
However, when try to console.log the sum of these arrays:
function sum(i){ return i.reduce((a, b) => a+b)}
var l = arr.length;
for(let j = 0; j <= l; j++){
var left = arr.slice(0, j-1), right = arr.slice(j)
console.log(sum(left), sum(right))
}
Using the snippet above, findEvenIndex([1,2,3,4,3,2,1]) returns "15 16"?
The main issue with your code is that calling sum([]) throws an error (which you will find in the console during debugging):
Reduce of empty array with no initial value
The reduce method does not know what to return if your array doesn't have any values. You solve it by passing the initial value as a second argument to .reduce:
const add = (a, b) => a + b;
[1, 2, 3].reduce(add); // add(add(1, 2), 3)
[1, 2].reduce(add); // add(1, 2)
[1].reduce(add); // 1
[].reduce(add); // ERROR: Reduce of empty array
// with no initial value
[1, 2].reduce(add, 0); // add(add(0, 1), 2)
[1].reduce(add, 0); // add(0, 1)
[].reduce(add, 0); // 0
Once you fix that, it's easier to debug the rest of the code.
Fixing it
Here's an example that I think does what it should do:
function findEvenIndex(arr) {
// Add a seed value --v
var sum = i => i.reduce((a, b) => a + b, 0),
l = arr.length;
for (let j = 0; j <= l; j++) {
const left = arr.slice(0, j);
const right = arr.slice(j + 1);
const leftSum = sum(left);
const rightSum = sum(right);
console.log(
{ left, right, leftSum, rightSum }
);
if (leftSum === rightSum) {
return j
}
}
return -1
}
console.log(
findEvenIndex([1]), // 0
findEvenIndex([1, 2, 3, 4, 3, 2, 1]), // 3
findEvenIndex([10, 0, 5, 5]), // 1
findEvenIndex([3, 2, 1]) // -1
)
Another approach
Note that looping over all elements of the array for every index is quite expensive! A more efficient approach would be:
Take the sum of the source array, store it as rightSum
Define leftSum as 0
Look at the integer value at index 0 and subtract it from rightSum
If leftSum === rightSum, return 0
Else, add value to leftSum and increment index
Once you've reached the final index, return -1
const findEvenIndex = (arr) => {
let leftSum = 0;
let rightSum = arr
.reduce((a, b) => a + b, 0);
for (let i = 0; i < arr.length; i += 1) {
const n = arr[i];
rightSum -= n;
if (leftSum === rightSum) return i;
leftSum += n;
}
return -1;
}
console.log(
findEvenIndex([1]), // 0
findEvenIndex([1, 2, 3, 4, 3, 2, 1]), // 3
findEvenIndex([10, 0, 5, 5]), // 1
findEvenIndex([3, 2, 1]) // -1
)
You can get the index like following using reduce(). Your implementation regarding reduce() is not correct.
function findEvenIndex(arr)
{
for(let i = 0; i < arr.length; i++) {
let leftSum = arr.slice(0, i).reduce((accumulator, current) => accumulator + current, 0);
let rightSum = arr.slice(i + 1).reduce((accumulator, current) => accumulator + current, 0);
if (leftSum === rightSum) {
return i;
}
}
return -1;
}
console.log(
findEvenIndex([1, 2, 3, 4, 3, 2, 1])
)
Please check following blog to find out how Array reduce() work
https://www.javascripttutorial.net/javascript-array-reduce/
Upon finishing my solution, I noticed that it's effectively the same as #Abu's answer above. The idea is to brute force the way through the array, comparing the two halves as you go.
/*
Find an index N where the sum of the integers to the left of N is equal to the sum of the integers to the right of N. If there is no index that would make this happen, return -1
*/
const array = [10, 90, 10, 1, 10, 90, 10];
// incrementTotal :: (Number t, Number n) -> t
incrementTotal = (total, number) => total + number;
// indexIsEqual :: (Array a, Number c) -> Boolean
function indexIsEqual(array, count) {
let chunkL = array.slice(0, count-1);
let chunkR = array.slice(count , );
return chunkL.reduce(incrementTotal) === chunkR.reduce(incrementTotal);
}
// findEvenIndex :: (Array a) -> (a[x] || -1)
function findEvenIndex(array) {
for (let count = 2; count < array.length; count++) {
if (indexIsEqual(array, count)) {
return array[count-1];
}
}
return -1;
}
console.log(findEvenIndex(array));
Using JavaScript, I'm trying to find a way to find the longest occurrence of the same number (in this case, 1) in an array.
For instance, here's a sample array:
[2,5,3,1,1,1,3,7,9,6,4,1,1,1,1,1,4,7,2,3,1,1,4,3]
I'd like to write a function that would return "5", since the number 1 occurs 5 times in a row. (It also occurs 3 and 2 times in a row, but I'm after the longest occurrence).
So far, I have written:
function streak(arr) {
var i,
temp,
streak,
length = arr.length;
for(i=0; i<length; i++) {
if (arr[i] === 1) {
streak += 1;
} else {
temp = streak;
break;
}
}
}
I know I need some way of knowing where I left off if I find an occurrence, but I'm feeling kind of stuck.
Any pointers?
I've modified your function slightly. You need to store the highest streak as a separate variable from the current streak, and overwrite that where necessary in your loop - finally returning that variable at the end of your function.
function streak(arr) {
var i,
temp,
streak,
length = arr.length,
highestStreak = 0;
for(i = 0; i < length; i++) {
// check the value of the current entry against the last
if(temp != '' && temp == arr[i]) {
// it's a match
streak++;
} else {
// it's not a match, start streak from 1
streak = 1;
}
// set current letter for next time
temp = arr[i];
// set the master streak var
if(streak > highestStreak) {
highestStreak = streak;
}
}
return highestStreak;
}
var array = [2,5,3,1,1,1,3,7,9,6,4,1,1,1,1,1,4,7,2,3,1,1,4,3];
console.log(streak(array)); // 5
And if you want to also track what the value of the highest streak was, define another variable at the start of your function, save the value of it when you save the highest streak, and return it as an array:
// set the master streak var
if(streak > highestStreak) {
highestStreakValue = temp;
highestStreak = streak;
}
}
return [highestStreak, highestStreakValue];
var array = [2,5,3,1,1,1,3,7,9,6,4,'a','a','a','a','a',4,7,2,3,1,1,4,3];
console.log(streak(array)); // [5, "a"]
Demo returning both
An alternative approach. I'm converting the array to a string. The regular expression has a backrefence, which ensures that only sequences of the same character are matched. Also when exec is used with the g flag, repeated executions will continue from the end of last match, and not from the beginning.
var arr = [2,5,3,1,1,1,3,7,9,6,4,1,1,1,1,1,4,7,2,3,1,1,4,3];
var str = arr.join('');
var regex = /(.)\1*/g;
var match;
var largest = '';
while (match = regex.exec(str)) {
largest = match[0].length > largest.length ? match[0] : largest;
}
console.log(largest.length);
Your problems:
You don't store current streak
You don't specify when streak is more then older streak
Use this:
function streak(arr) {
var i,
temp,
streak = 1,
maxStreak = 0,
prevNumber,
length = arr.length;
for(i=1; i<length; i++) {
prevNumber = arr[i-1];
if (arr[i] == prevNumber) {
streak += 1;
} else {
if(streak > maxStreak) {
maxStreak = streak;
streak = 1;
}
}
}
return maxStreak;
}
Demo
You will need another two arrays here.
Store the distinct numbers from your source array using a loop
Make a second set of array which is equal to the length of the first set of array which has the distinct numbers.
Make a loop equal to the length of the first set of array and then push the values to the second set of array according to its index.
Make a loop again using the second set of array and there you will find the most occurence using the index of the second array
Finally, get from the first set of array the number using the index you got from step 4.
I did not make the code for you to try it yourself first since you are asking only for some pointers
Alternative: use regexp and converting the array to a string.
var arr = [2,5,3,1,1,1,3,7,9,6,4,1,1,1,1,1,4,7,2,3,1,1,4,3];
var str = arr.join('').match(/1+/g);
console.log(process ? process.sort().pop() : "No ocurrences");
You could take Array#reduce and return the start index of the actual same item sequence. Then check and update the counter if the item is not equal.
var array = [2, 5, 3, 1, 1, 1, 3, 7, 9, 6, 4, 1, 1, 1, 1, 1, 4, 7, 2, 3, 1, 1, 4, 3],
maxCount = 0,
maxValues;
array.reduce(function (j, a, i, aa) {
if (aa[j] === a) {
return j;
}
if (i - j === maxCount){
maxValues.push(aa[j]);
}
if (i - j > maxCount) {
maxCount = i - j;
maxValues = [aa[j]];
}
return i;
}, -1);
console.log(maxCount);
console.log(maxValues);
My proposal:
function getLongestRow(inputArray) {
// Initialize dummy variables
var start = inputArray[0], curRowLen = 0, maxRowLen = 0, maxRowEle = 0;
// Run through the array
for(var i = 0;i < inputArray.length;i++) {
// If current Element does not belong to current row
if(inputArray[i] != start) {
// If current row is longer than previous rows, save as new longest row
if(curRowLen > maxRowLen) {
maxRowLen = curRowLen;
maxRowEle = start;
curRowLen = 1;
}
// Start new row
start = inputArray[i];
} else {
// Current element does belongt to current row, increase length
curRowLen++;
}
}
// Check whether last row was longer than previous rows
if(curRowLen > maxRowLen) {
maxRowLen = curRowLen;
maxRowEle = start;
}
// Return longest row & element longest row consits of
console.log('The longest row in your array consists of '+maxRowLen+' elements of '+maxRowEle+'.');
}
JsFiddle: http://jsfiddle.net/hdwp5/
Here's a way to do it:
var values = function(obj) {
var res = [];
for (var i in obj) {
if (obj.hasOwnProperty(i)) {
res.push(obj[i]);
}
}
return res;
};
var countStreak = function(xs) {
var res = xs.reduce(function(acc, x, i) {
if (x === xs[i+1]) {
acc[x] = acc[x]+1 || 2;
} else {
acc[x] = acc[x]-1 || 0;
}
return acc;
},{})
return Math.max.apply(0, values(res));
};
var ns = [2,5,3,1,1,1,3,7,9,6,4,1,1,1,1,1,4,7,2,3,1,1,4,3]
countStreak(ns) //=> 5
You can use fewer iterations by looking ahead at all matches from a given index,
and jumping ahead to the next non-matching item's index.
You can also quit when there are less items left than the maximum you have found.
function maxRepeats(arr){
var L= arr.length, i= 0,
max= 1, count= 0;
while(L-i > max){
while(arr[i+count]=== arr[i])++count;
if(count > max) max= count;
i+= count;
count= 0;
}
return max;
}
var A= [2, 5, 3, 1, 1, 1, 3, 7, 9, 6, 4, 1,
1, 1, 1, 1, 4, 7, 2, 3, 1, 1, 4, 3];
maxRepeats(A); returns 5
Finding multiple items that repeat the max number of times is not so easy,
since you have to find the max number before you can list them.
If you really only need the max number, ignore this:
function mostRepeats(arr, maximum){
var i= 0, max= maximum || 1,
L= arr.length-max,
count= 0, index= [];
while(i<L){
while(arr[i+count]=== arr[i])++count;
if(count=== maximum) index.push(arr[i]+' starting at #'+i);
else if(count > max) max= count;
i+= count;
count= 0;
}
if(max===1) return 'No repeats';
return maximum? max+' repeats of: '+index.join(', '): mostRepeats(arr, max);
}
var A= [2, 5, 3, 1, 1, 1, 3, 7, 9, 6, 4, 1, 1, 1,
1, 1, 4, 7, 2, 3, 3, 3, 3, 3, 1, 1, 4, 3];
mostRepeats(A);returns:
5 repeats of: 1 starting at #11, 3 starting at #19
Unfortunately I can't comment yet due to lack of reputation so I will post this as an answer. For my task Robbie Averill's solution was perfect, but it contains a little bug. I had array that consisted of 2 values - 0 & 1.5, but above-mentioned code was counting only "1.5" values although I had "0" repeating in a higher streak. Problem was that value wasn't doing strict comparison here:
if(temp != '' && temp == arr[i]) {
and the fix was simple: if(temp !== '' && temp == arr[i]) {
I've updated Robbie's jsfiddler with this fix: http://jsfiddle.net/d5X2k/5/
Unfortunatly, a question has been marked as duplicate, but it was not the same as this one. So I must put my answer here, sorry…
let tab = [0,0,0,1,1,1,0,0,0,0,1,0,1,1,1,1,1]
, arr = []
, n = 0
, res = null ;
for(let i of tab)
{
if ( i ) { ++ n }
else if ( n ) { arr.push(n) ; n = 0 }
}
arr.push(n) ;
res = Math.max(...arr);
console.log("Streak with 1 is ", Math.max(...arr));
It's a better solution than with reduce, slower, as you can see:
let tab = [0,0,0,1,1,1,0,0,0,0,1,0,1,1,1,1,1];
let arr = [];
let n = 0;
let res = null;
let loop = 0;
let start = new Date().getTime();
while (loop < 1000000){
++ loop;
arr = [];
for(let i of tab)
{
if ( i ) { ++ n }
else if ( n ) { arr.push(n) ; n = 0 }
}
arr.push(n);
res = Math.max(...arr);
}
let end = new Date().getTime();
console.log("laps old fashion = ", end - start);
loop = 0;
let streaks = null;
start = new Date().getTime();
while (loop < 1000000){
++ loop;
streaks = tab.reduce((res, n) =>
(n ? res[res.length-1]++ : res.push(0), res)
, [0]);
res = Math.max(...streaks);
}
end = new Date().getTime();
console.log("laps reduce = ", end - start);
console.log("Streak with 1 is ", Math.max(...arr));
Input array:
const seq = [
0, 0, 0,
1, 1, 1,
1, 1, 1, 1, 1,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0,
1, 1, 1, 1, 1,
0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0,
1, 1, 1, 1, 1,
];
Shortest solutions:
console.log(Math.max(...Array.from(seq.join("").matchAll(/(.)\1+/g), m=>m[0].length)))
Alternative with regexp (spoiler: it's ~25%, slower than solution with reduce(). See "Modern approach with reduce()" below):
const longestSeq = (seq) => {
let max = 0;
seq.join("").replace(/(.)\1+/g, m=> max = Math.max(max, m.length));
return max;
};
Straightforward, old-school style, human readable and fastest solution:
let longestSeq = () => {
let maxCount = 0,
curCount = 0,
curItem, prevItem,
l = seq.length+2, // +1+1 to finish last sequence and compare 'undefined' with previous
i = 0;
for (; i < l; ++i) {
curItem = seq[i];
if (curItem === prevItem) ++curCount;
else {
if (curCount > maxCount) maxCount = curCount;
curCount = 1;
prevItem = curItem;
}
}
return maxCount;
}
Modern approach with reduce() (just very little slower than old-school code above):
const longestSeq = (seq) => seq
.reduce(
({count, max}, item) => item === 0
? { count: ++count, max: Math.max(count, max) }
: { count: 0, max: max },
{ count: 0, max: 0} )
.max;
Performance test, Reduce() vs old-school for(): https://jsbench.me/ifkgsin56z/1