The problem is to find the unique number in a array such as [2,2,2,5].
The output should be 5 as it is the 1 unique element in the array.
I have attempted this:
function findUniq(arr) {
var b= arr[0];
var c;
for(var i=0; i<arr.length; i++)
{
if(arr[i]===b )
{
b=arr[i]
}
else
{
c=arr[i];
}
}
return c
console.log(findUniq([3, 5, 3, 3, 3]))
This works fine unless the unique number is the first element in the array. How do I fix this?
You can use indexOf and lastIndexOf to see if a value occurs more than once in the array (if it does, they will be different), and if so, it is not the unique value. Use filter to process the array:
let array = [2,2,2,5];
console.log(array.filter(v => array.indexOf(v) === array.lastIndexOf(v)));
array = [5,3,3,3,3];
console.log(array.filter(v => array.indexOf(v) === array.lastIndexOf(v)));
array = [4,4,5,4];
console.log(array.filter(v => array.indexOf(v) === array.lastIndexOf(v)));
You can create a recursive function that will take the first element of the array and see if it exists in the rest of it, if it does, it will take the next element and do the same, return the element if it doesn't exist in the rest of the array :
const arr = [3, 3, 3, 5, 3];
const find = arr => {
const [f, ...rest] = arr;
if(rest.includes(f))
return find(rest);
else
return f;
}
const result = find(arr);
console.log(result);
Note that this will return the last element if all of them are the same [3,3,3] will return 3
Try something like this using a set, which only stores unique elements:
var set = new Set(arr);
// count instances of each element in set
result = {};
for(var i = 0; i < a.length; ++i) {
if(!result[arr[i]])
result[arr[i]] = 0;
++result[arr[i]];
}
for (var value in result) {
if (value == 1) {
return value;
}
}
// if there isn't any
return false;
This should work, please tell me if it doesn't.
This is another implementation that is surely less efficient than that of #Nick's, but it is a valid algorithm anyway:
function findUniq(arr) {
var elemCount = new Map();
var uniq = [];
// Initialize elements conts
for (var k of arr.values()) {
elemCount.set(k, 0);
}
// Count elements
for (var k of arr.values()) {
elemCount.set(k, elemCount.get(k) + 1);
}
// Add uniq elements to array
for (var [k, v] of elemCount.entries()) {
if (v === 1) uniq.push(k);
}
return uniq;
}
console.log(findUniq([3, 5, 3, 3, 3]))
if you prefer .reduce over .map for your use case (for performance/etc. reasons):
function existance(data) {
return data.reduce((a, c) => (data.indexOf(c) === data.lastIndexOf(c)) ? a.concat(c) : a, []);
}
console.log(existance([1,1,1,2]));
console.log(existance([1,1,2,3,4,5,5,6,6,6]));
Related
I have this code that pairs same elements in an array, with the expectation that the array will have an odd length and it should return the only element that couldn't get a pair. So I wrote the code just well, and it works fine for smaller arrays, but with very large big integers of over 1 billion, the time complexity became O(N**2) and then the need to refactor my code to get a much better performance for large arrays and large array elements. Here is my code below;
function solution(A) {
if(!Array.isArray(A)) return 0;
var temp = new Array(A.length);
var position = 0;
for(let i=0; i<A.length; i++){
if(temp.includes(A[i])){
position = temp.indexOf(A[i]);
index = A.indexOf(A[i]);
delete temp[position];
delete A[index];
delete A[i];
}else{
temp[i] = A[i];
}
}
for(let j=0; j<A.length; j++){
if(A[j] !== undefined) return A[j];
else continue;
}
}
To test it, source data can look like [2,3,6,7,3,5,5,6,2] and it will give an output of 7. But when the array is so large up to [1,2,....] with length n = n=999,999, or n = 5000,000,000, the time complexity increases exponentially.
You might use Object to store non-paired elements only.
Please note that you don't need to store all the array elements and their counts in the Object and then filter by count (like #StepUp does).
Everything's been done in a single loop.
The function returns Array of all non-paired elements:
const solution = A => Array.isArray(A) ?
Object.keys(
A.reduce((r, k) => {
r[k] = r[k] || 0;
if (++r[k] > 1) delete r[k];
return r;
}, {})
) : [];
console.log(solution([2, 3, 6, 7, 3, 5, 5, 6, 2]))
We can try to find odd occurrences for one iteration by using great features of object. Object is key - value pair. So access to object key is O(1). So when we meet the same element, then we just increment value:
const hashMap = arr.reduce((a, c)=> {
a[c] = a[c] || 0;
a[c] += 1;
return a;
},{})
const result = Object.keys(hashMap).filter(key => hashMap[key] === 1);
An example:
let arr = [2, 3, 6, 7, 3, 5, 5, 6, 2];
const hashMap = arr.reduce((a, c)=> {
a[c] = a[c] || 0;
a[c] += 1;
return a;
},{})
const result = Object.keys(hashMap).filter(key => hashMap[key] === 1);
console.log(result);
My two 100% JavaScript solutions with optimized time complexity. The first one is using Set:
function solution(A) {
const pairs = new Set();
for (const num of A) {
if (pairs.has(num)) {
pairs.delete(num);
} else {
pairs.add(num);
}
}
const [unpaired] = pairs;
return unpaired;
}
The second one is using bitwise XOR:
function solution(A) {
let unpaired;
for (const num of A) {
unpaired ^= num;
}
return unpaired;
}
How can I get one unique value in an array or string? Only the first value. Pure JS only.
My example:
function searchValue () {
let inputText = [1,1,4,2,2,2,3,1];
let foundedValue;
for (let i = 0; i < inputText.length; i++) {
if (i === inputText.indexOf(inputText[i]) && i === inputText.lastIndexOf(inputText[i])) {
foundedValue = inputText[i];
break;
} else {
foundedValue = "not founded.";
}
}
return foundedValue;
}
console.log("Search value: "+ searchValue())
Answer is 4.
But, I need a short solution. Using the find() and filter() functions.
You can find the first unique item in your array using find() and comparing indexOf() to lastIndexOf() to determine whether or not there is more than one instance of an item in the array. If you need to find unique characters in a string, then you can first split it into an array and then use the same approach.
const arr = [1, 1, 4, 2, 2, 2, 3, 1];
const result = arr.find((x) => arr.indexOf(x) === arr.lastIndexOf(x));
console.log(result);
// 4
const text = 'aadbbbca';
const textarr = text.split('');
const textresult = textarr.find((x) => textarr.indexOf(x) === textarr.lastIndexOf(x));
console.log(textresult);
// d
You can try this.
const arr = [1, 1, 4, 2, 2, 2, 3, 1];
let r = {};
arr.map(a => r[a] = (r[a] || 0) +1)
var res = arr.find(a => r[a] === 1 )
console.log(res)
You can use js Set() object.
At first you could create a Set of duplicated elements.
const inputText = [1,1,4,2,2,2,3,1];
const duplicatesSet= inputText.reduce((dupSet, el) =>
inputText.filter(arrEl => arrEl === el).length > 1 ?
dupSet.add(el) : dupSet
, new Set());
Second you could use array.find. It returns first duplicated element.
const firstDupElement = inputText.find(el => duplicatesSet.has(el));
const searchValue = (_param) => {
for (let i= 0; i < _param.length; i+= 1) {
if (_param.indexOf(_param[i]) === _param.lastIndexOf(_param[i])) return _param[i];
}
return "not founded.";
}
let arr = [1,1,2,2,2,1,3,1,4,4,5]
const dupelearray = (array) => {
let arr2 =[...arr]
let ele = []
let state = false
arr2.map((i,index)=>{
arr2.splice(index,1)
arr.map((j)=>{
return arr2.includes(j) ? null : state=true
})
state && ele.push(i)
state=false
arr2.splice(index,0,i)
})
return console.log(arr.indexOf(ele[0]))
}
dupelearray(arr)
wow i didnt knew lastindexof method and was making this algo so difficult
btw this solution also works but definitely i am new in algo so this will take much more time but the solution still works!!!!!! damn should remember more methods or you have to think so much -_-
I'm trying to create a function that puts each array element in its own array, recursively.
I think my base case is correct, but my recursive call doesn't appear to be working. any insight?
function ownList(arr){
if (arr.length === 1) {
arr[0] = [arr[0]];
return;
} else {
return arr[0].concat(ownList(arr.slice(1)));
}
}
var arr = [1,2,3]
console.log(ownList(arr))// returns []
//should return [[1],[2],[3]]
Here I'm trying to put each pair in it's own list (recursive only). This code below is correct (update)
function ownListPair(arr){
if (arr.length === 0)
return arr;
else if(arr.length === 1)
return [[arr[0], 0]];
else
return [[arr[0], arr[1]]].concat(ownListPair(arr.slice(2)));
}
// var arr = [3,6,8,1,5]
var arr = [2,7,8,3,1,4]
//returns [ [ 2, 7 ], [ 8, 3 ], [ 1, 4 ]]
console.log(ownListPair(arr))
I prefer this solution for several reasons:
function ownList(a) {
return a.length == 0
? []
: [[a[0]]].concat(ownList(a.slice(1)))
}
It's shorter and more concise
It works for empty arrays as well
The actual wrapping happens only once in the last line. Treating length == 1 separately -- as suggested by others -- is not necessary.
It would more appropriate to make a length of 0 be the null case. Then you just have to get the brackets right. The thing on the left side of the concat should be an array consisting of the array containing the first element.
function ownList(arr) {
return arr.length ? [[arr[0]]].concat(ownList(arr.slice(1))) : [];
}
Here's an alternative, take your pick:
function ownList(arr) {
return arr.length ? [[arr.shift()]] . concat(ownList(arr)) : [];
}
Using a bit of ES6 magic for readability:
function ownList([head, ...tail]) {
return head === undefined ? [] : [[head]] . concat(ownList(tail));
}
Here the [head, ...tail] is using parameter destructuring which pulls the argument apart into its first element (head) and an array of remaining ones (tail).
Instead of concat you could also use the array constructor:
function ownList([head, ...tail]) {
return head === undefined ? [] : Array([head], ...ownList(tail));
}
I think your basic assumption is wrong. What you need to do is check if each item in the array is an array, if not just add the item to the new array, if so have the function run itself on the array item.
That is recursion.
This code does that kind of recursion...
function ownList(arr)
{
var newArr = [];
var length = arr.length;
for (var i = 0; i < length; i++) {
if (typeof(arr[i]) === 'object') {
newArr.push(ownList(arr[i]));
continue;
}
newArr.push([arr[i]]);
}
return newArr;
}
var arr = [1, 2, 3];
console.log(ownList(arr));
Would something like this work:
var arr = [1, 2, 3, ["a", "b", "c", ["str"]]],
result = [];
function flatten(input){
input.forEach(function(el){
if(Array.isArray(el)){
flatten(el)
}else{
result.push([el]);
}
});
}
flatten(arr);
console.log(JSON.stringify(result));
//[[1],[2],[3],["a"],["b"],["c"],["str"]]
JSBIN
Edit:
var result = [];
function flatten(input){
if (input.length === 0){
console.log( "result", result ); //[[1],[2],[3],["a"],["b"],["c"],["str"]]
return;
}
//if zeroth el of input !array, push to result
if (!Array.isArray(input[0])){
result.push(input.splice(0, 1));
flatten(input);
}else{
flatten(input[0]); //else, give input[0] back to flatten
}
}
window.onload = function(){
var arr = [1, 2, 3, ["a", "b", "c", ["str"]]];
flatten(arr);
}
JSBIN
After struggling through this today, turns out that this works :)
function ownList(arr){
//base case:
if (arr.length === 1) {
return [arr];
}
//recurse
//have to do two brackets here --> (arr.slice(0,1)) since length > 1
return [arr.slice(0,1)].concat(ownList(arr.slice(1)));
}
var arr = [1,2,3]
console.log(ownList(arr))// returns [[1],[2],[3]]
Using pure javascript, starting with an array, I would like to return and array by removing values that match any value in a second array.
I have solved this problem, but I believe with more code than is really necessary.
I am hoping for a more concise or elegant solution using only javascript.
function removeValues(arr){
array = arguments[0];
args = Array.prototype.slice.call(arguments);
len = arguments.length;
filtered = array.filter(function(n){
x = true;
for (var i = 1; i < len; i++) {
if (n == args[i]) { x = false; }
}
return x;
});
return filtered;
}
removeValues([1,2,3,1,2,3],2,3);
Should use a function that removes values from the first argument (an array) using values in one or more additional arguments.
When you're working with the filter function is not necessary to use loops because you're already in a loop. After converting the arguments into an array with [].slice.call(arguments), you could use indexOf that is responsible for returning the position of a value in an array, if a value is not exists, this returns -1, so we will take all the results that are -1
Your code could be reduced as well:
function removeValues(arr){
return arr.filter(function(val){
return [].slice.call(removeValues.arguments).slice(1).indexOf(val) === -1
})
}
console.log(removeValues([1,2,3,1,2,3],2,3))
ES6 Method: Using Rest parameters and Arrow Functions
var removeValues = (arr, ...values) => arr.filter(val => values.indexOf(val) === -1)
Try this instead...
function removeValues(){
var args = Array.prototype.slice.call(arguments).slice(1);
return arguments[0].filter(function(value) {
return args.indexOf(value) === -1;
});
}
removeValues([1, 2, 3, 1, 2, 3], 2, 3);
It does the exact same thing, but tidies it slightly.
Try like this:
var array1 = [ 1, 2, 3, 4, 5 ];
var array2 = [ 2, 3 ];
var result = array1.filter( function ( elem ) {
return array2.indexOf( elem ) === -1;
});
See example: Running code
I have 2 array:
var array1 = [[5,10],[6,10],[7,10],[8,10],[9,10]];
var array2 = [[1,10],[2,10],[3,10],[4,10],[5,40],[6,40]];
Want to get 1 merged array with the sum of corresponding keys;
var array1 = [[1,10],[2,10],[3,10],[4,10],[5,50],[6,50],[7,10],[8,10],[9,10]];
Both arrays have unique keys, but the corresponding keys needs to be summed.
I tried loops, concat, etc but can't get the result i need.
anybody done this before?
You can use .reduce() to pass along an object that tracks the found sets, and does the addition.
DEMO: http://jsfiddle.net/aUXLV/
var array1 = [[5,10],[6,10],[7,10],[8,10],[9,10]];
var array2 = [[1,10],[2,10],[3,10],[4,10],[5,40],[6,40]];
var result =
array1.concat(array2)
.reduce(function(ob, ar) {
if (!(ar[0] in ob.nums)) {
ob.nums[ar[0]] = ar
ob.result.push(ar)
} else
ob.nums[ar[0]][1] += ar[1]
return ob
}, {nums:{}, result:[]}).result
If you need the result to be sorted, then add this to the end:
.sort(function(a,b) {
return a[0] - b[0];
})
This is one way to do it:
var sums = {}; // will keep a map of number => sum
// for each input array (insert as many as you like)
[array1, array2].forEach(function(array) {
//for each pair in that array
array.forEach(function(pair) {
// increase the appropriate sum
sums[pair[0]] = pair[1] + (sums[pair[0]] || 0);
});
});
// now transform the object sums back into an array of pairs
var results = [];
for(var key in sums) {
results.push([key, sums[key]]);
}
See it in action.
a short routine can be coded using [].map()
var array1 = [[5,10],[6,10],[7,10],[8,10],[9,10]];
var array2 = [[1,10],[2,10],[3,10],[4,10],[5,40],[6,40]];
array1=array2.concat(array1).map(function(a){
var v=this[a[0]]=this[a[0]]||[a[0]];
v[1]=(v[1]||0)+a[1];
return this;
},[])[0].slice(1);
alert(JSON.stringify(array1));
//shows: [[1,10],[2,10],[3,10],[4,10],[5,50],[6,50],[7,10],[8,10],[9,10]]
i like how it's just 3 line of code, doesn't need any internal function calls like push() or sort() or even an if() statement.
Try this:
var array1 = [[5,10],[6,10],[7,10],[8,10],[9,10]];
var array2 = [[1,10],[2,10],[3,10],[4,10],[5,40],[6,40]];
var res = [];
someReasonableName(array1, res);
someReasonableName(array2, res);
function someReasonableName(arr, res) {
var arrLen = arr.length
, i = 0
;
for(i; i < arrLen; i++) {
var ar = arr[i]
, index = ar[0]
, value = ar[1]
;
if(!res[index]) {
res[index] = [index, 0];
}
res[index][1] += value;
}
}
console.log(JSON.stringify(res, null, 2));
So, the result may have holes. Just like 0th index. Use the below function if you want to ensure there are no holes.
function compact(arr) {
var i = 0
, arrLen = arr.length
, res = []
;
for(i; i < arrLen; i++) {
var v = arr[i]
;
if(v) {
res[res.length] = v;
}
}
return res;
}
So, you can do:
var holesRemoved = compact(res);
And finally if you don't want the 0th elem of res. Do res.shift();
Disclaimer: I am not good with giving reasonable names.
The simple solution is like this.
function sumArrays(...arrays) {
const n = arrays.reduce((max, xs) => Math.max(max, xs.length), 0);
const result = Array.from({ length: n });
return result.map((_, i) => arrays.map(xs => xs[i] || 0).reduce((sum, x) => sum + x, 0));
}
console.log(...sumArrays([0, 1, 2], [1, 2, 3, 4], [1, 2])); // 2 5 5 4