What is an efficient way of looping through two arrays to produce an alternating output? In JavaScript.
If I have two arrays like this:
var oddNumbers = [1, 3, 5, 7, 9]
var evenNumbers = [2, 4, 6, 8, 10, 12, 14]
NB: The arrays may not be the same length
How can I get the following output?
Output: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14
I would have thought this would work:
if (oddNumber.length > evenNumbers.length) {
var large = oddNumbers;
} else {
var large = evenNumbers;
}
for(var i = 0; i < large.length; i++){
if (evenNumbers.length >= i && oddNumbers.length >= i) {
console.log(oddNumbers[i] + ", " + evenNumbers[0]);
} elseif (evenNumbers.length >= i) {
console.log(evenNumbers[0]);
} else {
console.log(oddNumbers[0]);
}
}
But it's pretty messy, any better way of approaching this?
NOTE: These may not necessarily be in a numerical order, or in fact numbers
I would rather do it as follows if you just want to output them:
var oddNumbers = [1, 3, 5, 7, 9];
var evenNumbers = [2, 4, 6, 8, 10, 12, 14];
for (var i=0, j=0; i < oddNumbers.length || j < evenNumbers.length;) {
if (i < oddNumbers.length) {
console.log(oddNumbers[i++]);
}
if (j < evenNumbers.length) {
console.log(evenNumbers[j++]);
}
}
If you want to get the merge result as another array you can replace console.log with result.push to push result values on an array named result as follows:
var oddNumbers = [1, 3, 5, 7, 9];
var evenNumbers = [2, 4, 6, 8, 10, 12, 14];
var result = [];
for (var i=0, j=0; i < oddNumbers.length || j < evenNumbers.length;) {
if (i < oddNumbers.length) {
result.push(oddNumbers[i++]);
}
if (j < evenNumbers.length) {
result.push(evenNumbers[j++]);
}
}
console.log(result);
This way you iterate both arrays as long as one of them has an element that we haven't visited yet and also prevents iterating over the same index of same array twice. Please note that I used increment in if blocks to save 2 lines of code. You can also move them to the for loop since they won't break if statements.
var oddNumbers = [1, 3, 5, 7, 9];
var evenNumbers = [2, 4, 6, 8, 10, 12, 14];
var oLength = oddNumbers.length;
var eLength = evenNumbers.length;
var n = oLength > eLength ? oLength : eLength;
var rez=[];
for(i=0;i<n;i++){
if (i< oLength) rez.push(oddNumbers[i])
if (i<eLength) rez.push(evenNumbers[i])
}
console.log(rez);
var odd = ["A", "C","E","G"];
var even = ["B","D","F"];
var rez=[];
for(i=0;i<(odd.length > even.length ? odd.length : even.length);i++){
if (i< odd.length) rez.push(odd[i])
if (i<even.length) rez.push(even[i])
}
console.log(rez);
The following function accepts two arrays and returns their interleaved values as a new array:
function interleaveArrays(a, b) {
var array = [],
limit = a.length >= b.length ? a.length : b.length;
index = 0;
while (index < limit) {
a[index] && array.push(a[index]);
b[index] && array.push(b[index]);
index += 1;
}
return array;
}
Calling the function like so:
var oddNumbers = [1, 3, 5, 7, 9],
evenNumbers = [2, 4, 6, 8, 10, 12, 14];
console.log(interleaveArrays(oddNumbers, evenNumbers));
Yields:
[ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14 ]
You can then output this in your preferred manner; e.g:
var interleaved = interleaveArrays(oddNumbers, evenNumbers);
// as a loop
interleaved.forEach(function (n) {
console.log(n);
})
// or as a string
console.log(interleaved.join(', '));
// etc.
Hope this helps :)
I'd do something like this.
large = (oddNumber.length >= evenNumbers.length) ? oddNumbers : evenNumbers;
small = (oddNumber.length < evenNumbers.length) ? oddNumbers : evenNumbers;
for(var i = 0; i < large.length; i++){
if(small.length <= i + 1){
console.log(small[i] + ", "+ large[i]);
}
else {
console.log(large[i]);
}
}
A long-hand example of how it can be done. The code can be shrunk for a final solution. The basic principle I'm using is to even out the lengths to take care of the alternating then tag on the tail
var oddNumbers = [1, 3, 5, 7, 9];
var evenNumbers = [2, 4, 6, 8, 10, 12, 14];
var oLength = oddNumbers.length;
var eLength = evenNumbers.length;
var oTemp, eTemp, remainder;
if(oLength > eLength) {
eTemp = evenNumbers;
oTemp = oddNumbers.slice(0, eLength);
remainder = oddNumbers.slice(eLength);
} else if (eLength > oLength) {
eTemp = evenNumbers.slice(0, oLength);
oTemp = oddNumbers;
remainder = evenNumbers.slice(oLength);
} else {
eTemp = evenNumbers;
oTemp = oddNumbers;
remainder = [];
}
var final = [];
for(var i=0; i < eTemp.length; i++) {
final.push(oTemp[i]);
final.push(eTemp[i]);
}
final = final.concat(remainder);
alert(final);
I would simply merge the two array and sort it
var oddNumbers = [1, 3, 5, 7, 9];
var evenNumbers = [2, 4, 6, 8, 10, 12, 14];
var mergedArr=oddNumbers.concat(evenNumbers );
console.log(mergedArr.sort(function(a,b){return a-b;}));
See No loop.. No hassle. Very Simple
There will be an extra , on the screen. Add an if statement if you don't want that
for(var i = 0; i < large.length; i++){
if(i<evenNumbers.length)
console.log(evenNumbers[i]+",");
if(i<oddNumber.length)
console.log(evenNumbers[i]+",");
}
try this it will work always either number Array or String Array:
var oddNumber = [1, 3, 5, 7, 9]
var evenNumber = [2, 4, 6, 8, 10, 12, 14]
var margedNumbers = oddNumber.concat(evenNumber);
console.log("before: "+margedNumbers);
margedNumbers.sort(function(a, b){return a-b})
console.log("after: "+margedNumbers)
My solution
var oddNumbers = [1, 3, 5, 7, 9]
var evenNumbers = [2, 4, 6, 8, 10, 12, 14]
var extraElements = (oddNumbers.length > evenNumbers.length) ? oddNumbers.slice(evenNumbers.length) : evenNumbers.slice(oddNumbers.length);
var finalArr = [];
var small = (oddNumbers.length < evenNumbers.length) ? oddNumbers : evenNumbers;
small.forEach((each, index) => {
// merge elements in desired order
finalArr.push(oddNumbers[index]);
finalArr.push(evenNumbers[index]);
})
finalArr = finalArr.concat(extraElements);
alert(finalArr);
Extract the extra elements which makes both array of same length. Then, in a simple iteration, push elements from both array with same index.
Related
const diagonalSum = function (arr) {
var length=arr.length -1;
var sum=0;
for(let i=0;i<arr.length;i++){//1<3
sum+= arr[i][i]+arr[i][length-i]//[1][0]+
}
return sum;
};
tried this , but 2nd and 3rd test cases are not getting passed. Any other logic?
const diagonalSum = function (arr) {
var length=arr.length -1;
var sum=0;
for(let i=0;i<arr.length;i++){//1<3
sum+= arr[i][i]+arr[i][length-i]//[1][0]+
}
return sum;
};
searching any other logic
This will work for you.
// An efficient Javascript program to find
// sum of diagonals
function printDiagonalSums(mat,n)
{
let principal = 0, secondary = 0;
for (let i = 0; i < n; i++) {
principal += mat[i][i];
}
document.write("Principal Diagonal:"
+ principal+"<br>");
}
// Driver code
let a = [[ 1, 2, 3, 4, 5],
[5, 6, 7, 8, 5 ],
[ 1, 2, 3, 4, 5 ],
[ 5, 6, 7, 8, 5],
[ 5, 6, 7, 8, 5]];
printDiagonalSums(a, 5);
I have an array: const arr = [1, 2, 5, 10];
How I can transform it to const arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]?;
A simple way to do it without hard-coding the number of iterations is to get the minimum value from the array and the maximum and then fill the numbers in-between them.
Here's one way to do it
const arr = [1, 2, 5, 10];
var highest = Math.max(...arr);
var minimum = Math.min(...arr);
var output = [];
for(var i = minimum; i <= highest; i++){
output.push(i);
}
console.log(output);
Here's a one liner solution based on Adriani6 answer:
const arr = [1, 2, 5, 10];
var highest = Math.max(...arr);
var minimum = Math.min(...arr);
const newArr = ([...Array(highest+1).keys()]).slice(minimum);
console.log(newArr);
const arr = [1, 2, 5, 10];
for(let i = 1; i <= 10; i++) { // Loop from 1 till 10, including 10
if (!arr.includes(i)) { // If arr does not include 'i'
arr.splice(i - 1, 0, i); // We insert it into the array
// -1, because arrays start at 0
}
}
You could take a nested while statement and splice missing items.
var array = [1, 2, 5, 10],
i = array.length;
while (i--) while (array[i - 1] + 1 < array[i]) array.splice(i, 0, array[i] - 1);
console.log(...array);
I'm trying to create a sum function. When I run it through two different arrays (with same values), it's giving me different results. I can't really tell where I did wrong. It seems when I'm generating the array using the 'range' function, it's looping twice.
var myArr = [];
var tempArr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
function range(start, end) {
for (i = start; i <= end; i++) {
myArr.push(start);
start = start + 1;
}
return myArr;
}
function sum(arr) {
var sumArr = 0;
for (i = 0; i < arr.length; i++) {
sumArr = sumArr + arr[i];
//console.log(sumArr);
}
return sumArr;
}
console.log(range(1, 10)); //[ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ]
console.log(tempArr); //[ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ]
console.log(sum(range(1, 10))); //110
console.log(sum(tempArr)); //55
Any help would be appreciated. Thanks!
The reason is that var myArr = []; was a global variable. So pushed elements in the first console attempt will be there until they are cleared. You can use local variable in the function instead.
var tempArr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
function range(start, end) {
var myArr = [];
for (i = start; i <= end; i++) {
myArr.push(start);
start = start + 1;
}
return myArr;
}
function sum(arr) {
var sumArr = 0;
for (i = 0; i < arr.length; i++) {
sumArr = sumArr + arr[i];
//console.log(sumArr);
}
return sumArr;
}
console.log(range(1, 10)); //[ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ]
console.log(tempArr); //[ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ]
console.log(sum(range(1, 10))); //55
console.log(sum(tempArr)); //55
Using lodash :
You can use ._sum function.
var tempArr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
sum = _.sum(tempArr)
Don't forget to add the library if you want to use it.
<script src="https://cdn.jsdelivr.net/lodash/4.5.1/lodash.min.js"></script>
Demo
here is my problem: I have 5 arrays of integer like these in javascript:
array1 = [0, 1, 2, 3, 4];
array2 = [9, 1, 2, 3, 4];
array3 = [10, 1, 2, 11, 4];
array4 = [12, 1, 2, 13, 4];
array5 = [14, 1, 2, 15, 4];
I have to find the longest common subarray. In this case I have to retrieve the following subarray: [1, 2, 4].
For the records, I won't find repetitions inside arrays and my main goal is not execution speed.
thanks
here is the solution using Set in Javascript
var myArray = [array1 , array2 ,array3 , array4 ,array5];
let keys = new Set();
myArray.forEach(arr => arr.forEach(el => keys.add(el) ))
var common = [...keys].filter(key => myArray.every(arr => arr.includes(key)))
console.log(common);
#define MAX(a,b) a>b?a:b
int main(int argc, char* argv[])
{
if(argc < 2)
return -1;
int x = strlen(argv[1])+1;
int y = strlen(argv[2])+1;
int i,j,k,l;
int longest =0;
char* LCS = (char*)malloc(sizeof(char)*MAX(x,y));
int** arr = (int**)malloc(sizeof(int*)*x);
for(i=0;i<=y;i++)
arr[i] =(int*) malloc(sizeof(int)*y);
for(i=0;i<=x;i++)
for(j=0;j<=y;j++)
{
arr[i][j] = 0;
}
for(i=0;i<x;i++)
for(j=0;j<y;j++)
{
if(argv[1][i] == argv[2][j])
arr[i+1][j+1] = arr[i][j]+1;
if(arr[i+1][j+1] > longest)
{
longest =arr[i+1][j+1];
memset(LCS,0,MAX(x,y));
for( k=0,l=i;k<=longest;k++,l--)
LCS[k] = argv[1][l];
}
}
printf(" %s",argv[2]);
for(i=0;i<x;i++)
{
printf("\n%c",argv[1][i]);
for(j=0;j<y;j++)
{
printf("%d",arr[i][j]);
}
}
printf("\nLongest Common Subarray : %s\n",LCS);
return 0;
}
Try this:
var array1 = [0, 1, 2, 3, 4];
var array2 = [9, 1, 2, 3, 4];
var array3 = [10, 1, 2, 11, 4];
var array4 = [12, 1, 2, 13, 4];
var array5 = [14, 1, 2, 15, 4];
// join everything into one array
var all = array1.join(',')+','+array2.join(',')+','+array3.join(',')+','+array4.join(',')+','+array5.join(',');
all = all.split(',');
// get an object with all unique numbers as keys
var keys = {};
for(var i=0; i<all.length; i++) keys[all[i]] = 1;
console.log(keys);
// generate an array with values present in all arrays
var common = [];
for(var x in keys) {
if(array1.indexOf(parseInt(x)) != -1 && array2.indexOf(parseInt(x)) != -1 && array3.indexOf(parseInt(x)) != -1 && array4.indexOf(parseInt(x)) != -1 && array5.indexOf(parseInt(x)) != -1) {
common.push(x);
}
}
console.log(common);
I guess this can give you a good start:
My script will return you an object with the count of each elements. But for now, it takes the first array as base.
var array1 = [0, 1, 2, 3, 4];
var array2 = [9, 1, 2, 3, 4];
var array3 = [10, 1, 2, 11, 4];
var array4 = [12, 1, 2, 13, 4];
var array5 = [14, 1, 2, 15, 4];
var array6 = [13, 1, 2, 18, 4];
var mainArr = [array1, array2, array3, array4, array5, array6]
function getCommonElement(arr){
var subLength = arr[0].length;
var resultArr = new Array();
var ret = new Object();
for(var k=0;k<subLength;k++){
var temp = new Array();
for(var i=0;i<arr.length;i++){
temp.push(arr[i][k]);
}
resultArr.push(temp);
}
for(var i=0;i<arr[0].length;i++){
ret[arr[0][i]+''] = resultArr[i].join('').split(arr[0][i]+'').length - 1;
}
return ret;
}
Cheers.
/**
longest common subarray b/w 2 arrays
a = [2,3,4,5,6,7,8], b = [6,7,8,4,5,2,3]
ans = 6,7,8
basically create a 2d arr and if elements match dp[i][j] = 1 + dp[i-1][j-1];
if dp[i][j] > maxLen, update maxLen and store the index
Now that we have the maxLen, subarray will be from (index - maxLen) till index.
*/
int[] finMaxCommon(int[] a, int[] b){
int m = a.length, n = b.length, maxLen = 0;
int[][] dp = new int[m+1][n+1];
// i want a 0th row why? m->out of bounds; comparing i-1; i->1 then i-1 will be 0
for (int i = 1; i<=m; i++){
for(int j = 1; j<=n; j++){
if(a[i-1] == b[j-1]) {
dp[i][j] = 1 + dp[i-1][j-1];
maxLen = Math.max(maxLen, dp[i][j]);
}
}
}
// endIndex = 6, 3, a[6-3+1], a[6]
return new int[]{a[endIndex-maxLen+1], [endIndex]};
}
dry run
0,6,7,8,4,5,2,3
0, 0 //
2, 1 // (2,2) i = 1, j = 6 1 + dp[0][5]
3, 2 // (3,3) i = 2, j = 7 1 + dp[1][6]
4,
5,
6, 1
7, 2
8, 3
I have an array here:
var myArr = [1, 1, 2, 5, 5, 7, 8, 9, 9];
Now I want to remove both appearances of a duplicate. So the desired result is not:
var myArr = [1, 2, 5, 7, 8 ,9];
but
var myArr = [2, 7, 8];
Basically I know how to remove duplicates, but not in that that special way. Thats why any help would be really appreciated!
Please note: My array is filled with strings. The numbers here were only used as an example.
jsfiddle for this code:
var myArr = [1, 1, 2, 5, 5, 7, 8, 9, 9];
var newArr = myArr;
var h,i,j;
for(h = 0; h < myArr.length; h++) {
var curItem = myArr[h];
var foundCount = 0;
// search array for item
for(i = 0; i < myArr.length; i++) {
if (myArr[i] == myArr[h])
foundCount++;
}
if(foundCount > 1) {
// remove repeated item from new array
for(j = 0; j < newArr.length; j++) {
if(newArr[j] == curItem) {
newArr.splice(j, 1);
j--;
}
}
}
}
Here's my version
var a = [1, 1, 2, 5, 5, 7, 8, 9, 9];
function removeIfduplicate( arr ) {
var discarded = [];
var good = [];
var test;
while( test = arr.pop() ) {
if( arr.indexOf( test ) > -1 ) {
discarded.push( test );
continue;
} else if( discarded.indexOf( test ) == -1 ) {
good.push( test );
}
}
return good.reverse();
}
x = removeIfduplicate( a );
console.log( x ); //[2, 7, 8]
EDITED with better answer:
var myArr = [1, 1, 2, 5, 5, 7, 8, 9, 9];
function removeDuplicates(arr) {
var i, tmp;
for(i=0; i<arr.length; i++) {
tmp = arr.lastIndexOf(arr[i]);
if(tmp === i) {
//Only one of this number
} else {
//More than one
arr.splice(tmp, 1);
arr.splice(i, 1);
}
}
}
Using Hashmap
create hashmap and count occurencies
filter where hashmap.get(value) === 1 (only unique values)
const myArray = [1, 1, 2, 5, 5, 7, 8, 9, 9];
const map = new Map();
myArray.forEach(v => map.set(v, map.has(v) ? map.get(v)+1 : 1));
myArray.filter(v => map.get(v) === 1);
Old version (slower but valid too)
Heres a short version using Array.filter(). The trick is to first find all values that are NOT uniqe, and then use this array to reject all unique items in the original array.
let myArr = [1, 1, 2, 5, 5, 7, 8, 9, 9];
let duplicateValues = myArr.filter((item, indx, s) => s.indexOf(item) !== indx);
myArr.filter(item => !duplicateValues.includes(item));
// => [2, 7, 8]
Wherever removing duplicates is involved, it's not a bad idea to use a set data structure.
JavaScript doesn't have a native set implementation, but the keys of an object work just as well - and in this case help because then the values can be used to keep track of how often an item appeared in the array:
function removeDuplicates(arr) {
var counts = arr.reduce(function(counts, item) {
counts[item] = (counts[item] || 0) + 1;
return counts;
}, {});
return Object.keys(counts).reduce(function(arr, item) {
if (counts[item] === 1) {
arr.push(item);
}
return arr;
}, []);
}
var myArr = [1, 1, 2, 5, 5, 7, 8, 9, 9];
console.log(removeDuplicates(myArr), myArr);
Check out the example on jsfiddle.
Alternately, you could not use calls to reduce(), and instead use for and for(item in counts) loops:
function removeDuplicates(arr) {
var counts = {};
for(var i=0; i<arr.length; i++) {
var item = arr[i];
counts[item] = (counts[item]||0)+1;
}
var arr = [];
for(item in counts) {
if(counts[item] === 1) {
arr.push(item);
}
}
return arr;
}
Check out the example on jsfiddle.
If it's just alphanumeric, duplicates are case-sensitive, and there can be no more than two of any element, then something like this can work:
var a = [2, 1, "a", 3, 2, "A", "b", 5, 6, 6, "B", "a"],
clean_array = $.map(a.sort(), function (v,i) {
a[i] === a[i+1] && (a[i] = a[i+1] = null);
return a[i];
});
// clean_array = [1,3,5,"A","B","b"]
In this example,we are taking two arrays as function arguments, from this we are going to print only unique values of both arrays hence deleting the values that are present in both arrays.
first i am concatenating both the arrays into one. Then I taking each array value at a time and looping over the array itself searching for its no of occurrence. if no of occurrence(i.e.,count) equal to 1 then we are pushing that element into the result array. Then we can return the result array.
function diffArray(arr1, arr2) {
var newArr = [];
var myArr=arr1.concat(arr2);
var count=0;
for(i=0;i<myArr.length;i++){
for(j=0;j<myArr.length;j++){
if(myArr[j]==myArr[i]){
count++;
}
}
if(count==1){
newArr.push(myArr[i]);
}
count=0;
}
return newArr;
}
EDIT: Here is the jspref http://jsperf.com/deleting-both-values-from-array
http://jsfiddle.net/3u7FK/1/
This is the fastest way to do it in two passes without using any fancy tricks and keeping it flexible. You first spin through and find the count of every occurance and put it into and keyvalue pair. Then spin through it again and filter out the ones where the count was greater than 1. This also has the advanatage of being able to apply other filters than just "greater than 1"; as well as the having the count of occurances if you needed that as well for something else.
This should work with strings as well instead of numbers.
http://jsfiddle.net/mvBY4/1/
var myArr = [1, 1, 2, 5, 5, 7, 8, 9, 9];
var map = new Object();
for(var i = 0; i < myArr.length; i++)
{
if(map[myArr[i]] === undefined)
{
map[myArr[i]] = 1;
}
else
{
map[myArr[i]]++;
}
}
var result = new Array();
for(var i = 0; i < myArr.length; i++)
{
if(map[myArr[i]] > 1)
{
//do nothing
}
else
{
result.push(myArr[i]);
}
}
alert(result);
You can use Set (available in IE 11+) as below
const sourceArray = [1, 2, 3, 4, 5, 5, 6, 6, 7, 7, 8];
const duplicatesRemoved = new Set();
sourceArray.forEach(element => {
if (duplicatesRemoved.has(element)) {
duplicatesRemoved.delete(element)
} else {
duplicatesRemoved.add(element)
}
})
console.log(Array.from(duplicatesRemoved))
N.B. Arrow functions are not supported in older browsers. Use normal function syntax for that instead. However, Array.from can easily be polyfilled for older browsers.
Try it here.