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shuffle array without consecetive item - Javascript

I currently acheive shuffle one of array without duplicated with previous value of next value.
However, I don't think my method is good. I think there are better way.
Firstly, I will describe what the array looks like.
it's an array of object with property {answer,url,category} and there are duplicated item in the array. think as [A,A,B,B,C,C] , A,B,C as object here.
As EzioMerce pointed out, in this array, object will always has equal amount of numbers. such if there are 3 A, will definitely have 3 B and C. It will not have array as such [A,A,A,B]
I need the array to shuffle until there is no consecutive object next to each other such as [A,B,C,B,A,C]
Here is my solution (which I have tested 40 times without consecutive object):
getShuffleWords(array: Card[]) {
//First shuffle..
for (let i = array.length - 1; i > 0; i--) {
const j = Math.floor( Math.random() * (i - 1));
if (j > 0) {
//make sure there are no simliar value around position i and j
if (array[i]?.answer !== array[j - 1]?.answer
&& array[i]?.answer !== array[j + 1]?.answer
&& array[i]?.answer !== array[j - 1]?.answer
&& array[i]?.answer !== array[j + 1]?.answer) {
[array[i], array[j]] = [array[j], array[i]];
}
} else {
if (array[i]?.answer !== array[j + 1]?.answer) {
[array[i], array[j]] = [array[j], array[i]];
}
}
}
const before = [...array];
console.log(before);
//Checking duplicate and swap with other value
for (let x = 0; x < array.length; x++){
if (array[x].answer == array[x + 1]?.answer) {
console.log(x,array[x])
for (let y = 0; y < array.length; y++){
//make sure there are no simliar value around position x and y
if (array[y]?.answer !== array[x]?.answer
&& array[y + 1]?.answer !== array[x].answer
&& array[y - 1]?.answer !== array[x].answer
&& array[y]?.answer !== array[x+1]?.answer
&& array[y]?.answer !== array[x-1]?.answer
) {
console.log(y, array[y]);
if (x < y) {
[array[x], array[y]] = [array[y], array[x]];
} else {
[array[y], array[x]] = [array[x], array[y]];
}
break;
}
}
}
}
//just another caculate if there is duplication left (for testing purpose)
let dup = 0;
for (let x = 0; x < array.length; x++){
if (array[x].answer == array[x + 1]?.answer) {
dup++;
}
}
console.log(array, dup);
return array;
}
However, in the //First shuffle.. there are always has some consecutive object exisited in the array. thats why I repeat another checking and replace the value.
How would I improve my method. and why my first shuffle does not work perfectly?

You can try this way. It also works if arr will be an array of objects
const arr = [1, 2, 3, 1, 2, 3, 1, 2, 3];
const shuffle = (arr) => {
const uniqueArr = [...(new Set(arr))];
const shuffledArr = [];
let countOfOneObj = 0;
for (const obj of arr) {
if (obj === arr[0]) ++countOfOneObj;
};
const getRandomObj = () => {
const i = Math.floor(Math.random() * uniqueArr.length);
return uniqueArr[i];
}
for (let i = 0; i < countOfOneObj; ++i) {
let usedObjs = []
for (let j = 0; j < uniqueArr.length; ++j) {
const obj = getRandomObj();
if (
usedObjs.includes(obj) ||
shuffledArr[shuffledArr.length - 1] === obj
) {
--j;
continue;
}
usedObjs.push(obj);
shuffledArr.push(obj);
}
usedObjs = [];
}
return shuffledArr;
}
console.log(shuffle(arr));

You could shuffle an array of indices of same grouped values until you got a result which has no same neighbour indices.
const
fy = array => { // fisher-yates, adapted from https://stackoverflow.com/a/59837259/1447675
let i = array.length;
while (--i) {
const j = Math.floor(Math.random() * (i + 1));
[array[j], array[i]] = [array[i], array[j]];
}
},
shuffle = array => {
do fy(array);
while (array.some((v, i, a) => v === a[i + 1]))
};
indices = [0, 0, 0, 1, 1, 1, 2, 2];
shuffle(indices);
console.log(...indices);

finding a triplet in an array that sums up to a given value in javascript

I am trying to solve this problem in JavaScript:
Given an array and a value in JavaScript, find if there is a triplet in array whose sum is equal to the given value. If there is such a triplet present in array, then print the triplet and return true. Else return false.
Now, I wrote some code, and for some reason, it's not working properly.
Here is the code:
A = [1, 4, 45, 6, 10, 8];
sum = 15;
x = A.length;
function find3Numbers(A, x, sum) {
for (i=0; i<(x-2); i++) {
for (j=i+1; j<(x-1); j++) {
for (k=j+1; x; k++) {
if (A[i] + A[j] + A[k] == sum) {
console.log(A[i]);
console.log(A[j]);
console.log(A[k]);
return true
}
return false
}
}
}
}
console.log(find3Numbers(A, x, sum));
Now when I run the code, I get a "false" message.
Any idea, why is that?

You immediately return false if the first triplet you try does not match, when you should only do so after all loops have finished.
A = [1, 4, 45, 6, 10, 8];
sum = 15;
x = A.length;
function find3Numbers(A, x, sum) {
for (i = 0; i < (x - 2); i++) {
for (j = i + 1; j < (x - 1); j++) {
for (k = j + 1; x; k++) {
if (A[i] + A[j] + A[k] == sum) {
console.log(A[i]);
console.log(A[j]);
console.log(A[k]);
return true
}
}
}
}
return false;
}
console.log(find3Numbers(A, x, sum));

Consider sorting the array beforehand for O(n^2) time complexity rather than O(n^3).
Note: O(nlog(n) + n^2) is asymptotically the same as O(n^2).
const find3Numbers = (nums, nums_length, target) => {
nums.sort((a, b) => a - b);
let i = 0;
while (i < nums_length - 2) {
let j = i + 1;
let k = nums_length - 1;
while (j < k) {
curr_sum = nums[i] + nums[j] + nums[k];
if (curr_sum < target) {
j++;
while (j < k && nums[j] == nums[j - 1]) {
j++;
}
} else if (curr_sum > target) {
k--;
while (j < k && nums[k] == nums[k + 1]) {
k--;
}
} else {
return [nums[i], nums[j], nums[k]];
}
}
i++;
while (i < nums_length - 2 && nums[i] == nums[i - 1]) {
i++;
}
}
return [-1, -1, -1];
}
const A = [1, 4, 45, 6, 10, 8];
const x = A.length;
const sum = 15;
console.log(find3Numbers(A, x, sum));

Find Highest Sum below/equals given Limit

I've got the following function:
var object = [12,23,14,35,24];
//debugger;
function f(objects, sum){
var curSum = objects[0];
var maxSum = 0;
var start = 0;
for (var i = 1; i < objects.length; i++) {
while(curSum > sum && start < i){
curSum -= objects[start];
start++
}
maxSum = Math.max(maxSum, curSum);
curSum += objects[i];
if(curSum <= sum){
maxSum = Math.max(curSum, maxSum);
}
}
return maxSum;
}
console.log(f(object,50));
I'd expect the highest possible sum to be 50, due to elements 12,14,24
However, I can only manage to get to 49. What am I missing here?

You could take a recursive approach by iterating the array and check if the temporary array has the right sum and check against the result for eiter replaceinf smaller sums or push to same sum parts.
function combine(array, sum) {
function fork(i, p) {
var s = p.reduce((a, b) => a + b, 0),
l = (result[0] || []).reduce((a, b) => a + b, 0);
if (i === array.length) {
if (s <= sum) {
if (l < s) {
result = [p];
}
if (l === s) {
result.push(p);
}
}
return;
}
fork(i + 1, p.concat(array[i]));
fork(i + 1, p);
}
var result = [];
fork(0, []);
return result;
}
console.log(combine([12, 23, 14, 35, 24], 50));
.as-console-wrapper { max-height: 100% !important; top: 0; }

What is the best way to shift a multidimensional array in Javascript?

I'm trying to create a javascript function that shifts an array right x units any up y units. It must keep the array size the same, and it must call unloadChunk for elements that are getting shifted off the multidimensional array. Here is my current implementation:
function shift(x, y) {
if (x > 0) {
for (var i = 0; i < chunks.length; i++) {
for (var j = chunks[i].length - 1; j >= 0; j--) {
if(j + x > chunks[i].length - 1 && chunks[i][j]) {
unloadChunk(i, j);
}
if (j < x) {
chunks[i][j] = null;
}
else {
chunks[i][j] = chunks[i][j - x];
}
}
}
}
else if (x < 0) {
for (var i = 0; i < chunks.length; i++) {
for (var j = 0; j < chunks[i].length; j++) {
if(j + x < 0 && chunks[i][j]) {
unloadChunk(i, j);
}
if (j - x >= chunks[i].length) {
chunks[i][j] = null;
}
else {
chunks[i][j] = chunks[i][j - x];
}
}
}
}
if (y > 0) {
for (var i = 0; i < chunks.length; i++) {
if (i + y >= chunks.length) {
for (var j = 0; j < chunks.length; j++) {
if(i - y < 0 && chunks[i][j]) {
unloadChunk(i, j);
}
chunks[i][j] = null;
}
}
else {
for (var j = 0; j < chunks.length; j++) {
if(i - y < 0 && chunks[i][j]) {
unloadChunk(i, j);
}
chunks[i][j] = chunks[i + y][j];
}
}
}
}
else if (y < 0) {
for (var i = chunks.length - 1; i >= 0; i--) {
if (i + y < 0) {
for (var j = 0; j < chunks.length; j++) {
if(i - y > chunks.length - 1 && chunks[i][j]) {
unloadChunk(i, j);
}
chunks[i][j] = null;
}
}
else {
for (var j = 0; j < chunks.length; j++) {
if(i - y > chunks.length - 1 && chunks[i][j]) {
unloadChunk(i, j);
}
chunks[i][j] = chunks[i + y][j];
}
}
}
}
}
If you're having trouble understanding exactly what I want the shift function to do, take a look at this fiddle and look at the html output.
My attempt at creating the shift function works, but it has 10 for loops. My question was, is there a more efficient, less verbose way to do this?

This proposal uses
Array#forEach: visit each item
Array#map: return value for each item
Array#pop: removes and return last element
Array#push: adds one or more elements at the end
Array#shift: removes and return first element
Array#unshift: adds one or more elements at the beginning
For better visibillity, I replaced the null value with 1000, 2000, 3000 and 4000.
function shift(x, y) {
while (x > 0) {
chunks.forEach(function (a) {
a.pop();
a.unshift(1000);
});
x--;
}
while (x < 0) {
chunks.forEach(function (a) {
a.shift();
a.push(2000);
});
x++;
}
while (y > 0) {
chunks.unshift(chunks.pop().map(function () { return 3000; }));
y--;
}
while (y < 0) {
chunks.push(chunks.shift().map(function () { return 4000; }));
y++;
}
}
function print(msg) {
document.body.innerHTML += '<p>' + msg + '</p>';
}
function printarr(arr) {
for (var i = 0; i < arr.length; i++) {
print(JSON.stringify(arr[i]))
}
}
var chunks = [[5, 3, 1], [9, 2, 5], [2, 3, 7]];
print("chunks: " + JSON.stringify(chunks));
shift(1, 0);
print("shifting right 1. chunks: "); printarr(chunks);
shift(-1, 0);
print("shifting left 1. chunks: "); printarr(chunks);
shift(0, 1);
print("shifting up 1. chunks: "); printarr(chunks);
shift(0, -1);
print("shifting down 1. chunks: "); printarr(chunks);

You can use pop(), push(), shift(), unshift() Array methods
var chunks = [
[5, 3, 1],
[9, 2, 5],
[2, 3, 7]
];
function shiftDown(){
chuck.pop();
chuck.unshift(Array(3));
}
function shiftUp(){
chuck.shift();
chuck.push(Array(3));
}
function shiftRight(){
chuck.foreach(function(v){
v.pop();
v.unshift(null);
})
}
function shiftLeft(){
chuck.foreach(function(v){
v.shift();
v.push(null);
})
}

If interpret Question correctly, you can use Array.prototype.forEach(), Array.prototype.splice()
var chunks = [[5, 3, 1], [9, 2, 5], [2, 3, 7]];
// `x`: Index at which to start changing array
// `y`: An integer indicating the number of old array elements to remove
function shift(arr, x, y, replacement) {
arr.forEach(function(curr, index) {
// The elements to add to the array, beginning at the start index.
// start index: `x`
curr.splice(x, y, replacement)
});
return arr
}
// e.g.,
shift(chunks, -1, 1, null);
console.log(chunks);

it must call unloadChunk for elements that should be set to null
it's not a good idea to mutate an Array, while you're iterating over the same Array. So change unloadChunk() to not change chunks, but return the new value.
in my case I'm filling all the null values with new values.
then why do you bother to insert the null-values in the first place? why don't you just insert the new values?
//one-dimensional shift
function shift(arr, offset, callback){
var i, j, len = arr.length;
if(len && (offset |= 0)){
typeof callback === "function" || (callback = function(v){return v});
if(offset < 0){
for(i=-offset,j=0; i<len;)arr[j++]=arr[i++];
while(j<len)arr[j]=callback(null,j++,arr);
}else if(offset){
for(i=len-offset,j=len; i>0;)arr[--j]=arr[--i];
for(i=0; i<j;++i)arr[i]=callback(null,i,arr);
}
}
return arr;
}
//two dimensional shift
function shift2d(matrix, offsetX, offsetY, callback){
var i, len = matrix.length, tmp, fn;
offsetY |= 0;
offsetX |= 0;
if(len && matrix[0].length && (offsetY || offsetX)){
typeof callback === "function" || (callback = function(v){return v});
fn = function(val,j){ return callback(null, [i,j], matrix) };
tmp = {length: matrix[0].length};
offsetY && shift(matrix, offsetY, function(){return tmp});
for(i = 0; i < len; ++i){
if(matrix[i] === tmp){
matrix[i] = Array.from(tmp,fn);
}else{
shift(matrix[i], offsetX, fn);
}
}
}
return matrix;
}
and the code:
var chunks = [[5, 3, 1], [9, 2, 5], [2, 3, 7]];
console.log(chunks);
console.log(shift2d(chunks, -1, 1));
console.log(shift2d(chunks, 1, -1, computeValue));
function computeValue(value, index, array){
console.log("value: %o index: %o, is chunks: %o", value, index, array === chunks);
//return the new value for this field
return JSON.stringify(index);
return Math.random();
//with ES6 array destructuring:
var [row,col] = index;
return row*3 + col;
}
shift() and shift2d() expect as the last argument an (optional) callback function that returns the new values.
For consistency reasons, I pass the same argument as Array.map and the others
value: always null, only there for consistency reasons
index: the "index" that is accessed. For shift2d this is an array of indices (take a look at array destructuring)
array: the current array/matrix. Don't mess around with this while it gets changed. The main reason to pass this argument, is to check wich Array you're currently processing.

Pair of elements from a specified array whose sum equals a specific target number

I am in mid of my JavaScript session. Find this code in my coding exercise. I understand the logic but I didn't get this map[nums[x]] condition.
function twoSum(nums, target_num) {
var map = [];
var indexnum = [];
for (var x = 0; x < nums.length; x++)
{
if (map[nums[x]] != null)
// what they meant by map[nums[x]]
{
index = map[nums[x]];
indexnum[0] = index+1;
indexnum[1] = x+1;
break;
}
else
{
map[target_num - nums[x]] = x;
}
}
return indexnum;
}
console.log(twoSum([10,20,10,40,50,60,70],50));
I am trying to get the Pair of elements from a specified array whose sum equals a specific target number. I have written below code.
function arraypair(array,sum){
for (i = 0;i < array.length;i++) {
var first = array[i];
for (j = i + 1;j < array.length;j++) {
var second = array[j];
if ((first + second) == sum) {
alert('First: ' + first + ' Second ' + second + ' SUM ' + sum);
console.log('First: ' + first + ' Second ' + second);
}
}
}
}
var a = [2, 4, 3, 5, 6, -2, 4, 7, 8, 9];
arraypair(a,7);
Is there any optimized way than above two solutions? Can some one explain the first solution what exactly map[nums[x]] this condition points to?

Using HashMap approach using time complexity approx O(n),below is the following code:
let twoSum = (array, sum) => {
let hashMap = {},
results = []
for (let i = 0; i < array.length; i++){
if (hashMap[array[i]]){
results.push([hashMap[array[i]], array[i]])
}else{
hashMap[sum - array[i]] = array[i];
}
}
return results;
}
console.log(twoSum([10,20,10,40,50,60,70,30],50));
result:
{[10, 40],[20, 30]}
I think the code is self explanatory ,even if you want help to understand it,let me know.I will be happy enough to for its explanation.
Hope it helps..

that map value you're seeing is a lookup table and that twoSum method has implemented what's called Dynamic Programming
In Dynamic Programming, you store values of your computations which you can re-use later on to find the solution.
Lets investigate how it works to better understand it:
twoSum([10,20,40,50,60,70], 50)
//I removed one of the duplicate 10s to make the example simpler
In iteration 0:
value is 10. Our target number is 50. When I see the number 10 in index 0, I make a note that if I ever find a 40 (50 - 10 = 40) in this list, then I can find its pair in index 0.
So in our map, 40 points to 0.
In iteration 2:
value is 40. I look at map my map to see I previously found a pair for 40.
map[nums[x]] (which is the same as map[40]) will return 0.
That means I have a pair for 40 at index 0.
0 and 2 make a pair.
Does that make any sense now?
Unlike in your solution where you have 2 nested loops, you can store previously computed values. This will save you processing time, but waste more space in the memory (because the lookup table will be needing the memory)
Also since you're writing this in javascript, your map can be an object instead of an array. It'll also make debugging a lot easier ;)

function twoSum(arr, S) {
const sum = [];
for(let i = 0; i< arr.length; i++) {
for(let j = i+1; j < arr.length; j++) {
if(S == arr[i] + arr[j]) sum.push([arr[i],arr[j]])
}
}
return sum
}
Brute Force not best way to solve but it works.

Please try the below code. It will give you all the unique pairs whose sum will be equal to the targetSum. It performs the binary search so will be better in performance. The time complexity of this solution is O(NLogN)
((arr,targetSum) => {
if ((arr && arr.length === 0) || targetSum === undefined) {
return false;
} else {
for (let x = 0; x <=arr.length -1; x++) {
let partnerInPair = targetSum - arr[x];
let start = x+1;
let end = (arr.length) - 2;
while(start <= end) {
let mid = parseInt(((start + end)/2));
if (arr[mid] === partnerInPair) {
console.log(`Pairs are ${arr[x]} and ${arr[mid]} `);
break;
} else if(partnerInPair < arr[mid]) {
end = mid - 1;
} else if(partnerInPair > arr[mid]) {
start = mid + 1;
}
}
};
};
})([0,1,2,3,4,5,6,7,8,9], 10)

function twoSum(arr, target) {
let res = [];
let indexes = [];
for (let i = 0; i < arr.length - 1; i++) {
for (let j = i + 1; j < arr.length; j++) {
if (target === arr[i] + arr[j] && !indexes.includes(i) && !indexes.includes(j)) {
res.push([arr[i], arr[j]]);
indexes.push(i);
indexes.push(j);
}
}
}
return res;
}
console.log('Result - ',
twoSum([1,2,3,4,5,6,6,6,6,6,6,6,6,6,7,8,9,10], 12)
);
Brute force.

const findTwoNum = ((arr, value) => {
let result = [];
for(let i= 0; i< arr.length-1; i++) {
if(arr[i] > value) {
continue;
}
if(arr.includes(value-arr[i])) {
result.push(arr[i]);
result.push(value-arr[i]);
break;;
}
}
return result;
});
let arr = [20,10,40,50,60,70,30];
const value = 120;
console.log(findTwoNum(arr, value));
OUTPUT : Array [50, 70]

function twoSum(arr){
let constant = 17;
for(let i=0;i<arr.length-2;i++){
for(let j=i+1;j<arr.length;j++){
if(arr[i]+arr[j] === constant){
console.log(arr[i],arr[j]);
}
}
}
}

let myArr = [2, 4, 3, 5, 7, 8, 9];
function getPair(arr, targetNum) {
for (let i = 0; i < arr.length; i++) {
let cNum = arr[i]; //my current number
for (let j = i; j < arr.length; j++) {
if (cNum !== arr[j] && cNum + arr[j] === targetNum) {
let pair = {};
pair.key1 = cNum;
pair.key2 = arr[j];
console.log(pair);
}
}
}
}
getPair(myArr, 7)

let sumArray = (arr,target) => {
let ar = []
arr.forEach((element,index) => {
console.log(index);
arr.forEach((element2, index2) => {
if( (index2 > index) && (element + element2 == target)){
ar.push({element, element2})
}
});
});
return ar
}
console.log(sumArray([8, 7, 2, 5, 3, 1],10))

Use {} hash object for storing and fast lookups.
Use simple for loop so you can return as soon as you find the right combo; array methods like .forEach() have to finish iterating no matter what.
And make sure you handle edges cases like this: twoSum([1,2,3,4], 8)---that should return undefined, but if you don't check for !== i (see below), you would erroneously return [4,4]. Think about why that is...
function twoSum(nums, target) {
const lookup = {};
for (let i = 0; i < nums.length; i++) {
const n = nums[i];
if (lookup[n] === undefined) {//lookup n; seen it before?
lookup[n] = i; //no, so add to dictionary with index as value
}
//seen target - n before? if so, is it different than n?
if (lookup[target - n] !== undefined && lookup[target - n] !== i) {
return [target - n, n];//yep, so we return our answer!
}
}
return undefined;//didn't find anything, so return undefined
}

We can fix this with simple JS object as well.
const twoSum = (arr, num) => {
let obj = {};
let res = [];
arr.map(item => {
let com = num - item;
if (obj[com]) {
res.push([obj[com], item]);
} else {
obj[item] = item;
}
});
return res;
};
console.log(twoSum([2, 3, 2, 5, 4, 9, 6, 8, 8, 7], 10));
// Output: [ [ 4, 6 ], [ 2, 8 ], [ 2, 8 ], [ 3, 7 ] ]

Solution In Java
Solution 1
public static int[] twoNumberSum(int[] array, int targetSum) {
for(int i=0;i<array.length;i++){
int first=array[i];
for(int j=i+1;j<array.length;j++){
int second=array[j];
if(first+second==targetSum){
return new int[]{first,second};
}
}
}
return new int[0];
}
Solution 2
public static int[] twoNumberSum(int[] array, int targetSum) {
Set<Integer> nums=new HashSet<Integer>();
for(int num:array){
int pmatch=targetSum-num;
if(nums.contains(pmatch)){
return new int[]{pmatch,num};
}else{
nums.add(num);
}
}
return new int[0];
}
Solution 3
public static int[] twoNumberSum(int[] array, int targetSum) {
Arrays.sort(array);
int left=0;
int right=array.length-1;
while(left<right){
int currentSum=array[left]+array[right];
if(currentSum==targetSum){
return new int[]{array[left],array[right]};
}else if(currentSum<targetSum){
left++;
}else if(currentSum>targetSum){
right--;
}
}
return new int[0];
}

function findPairOfNumbers(arr, targetSum) {
var low = 0, high = arr.length - 1, sum, result = [];
while(low < high) {
sum = arr[low] + arr[high];
if(sum < targetSum)
low++;
else if(sum > targetSum)
high--;
else if(sum === targetSum) {
result.push({val1: arr[low], val2: arr[high]});
high--;
}
}
return (result || false);
}
var pairs = findPairOfNumbers([1,2,3,4,4,5], 8);
if(pairs.length) {
console.log(pairs);
} else {
console.log("No pair of numbers found that sums to " + 8);
}

Simple Solution would be in javascript is:
var arr = [7,5,10,-5,9,14,45,77,5,3];
var arrLen = arr.length;
var sum = 15;
function findSumOfArrayInGiven (arr, arrLen, sum){
var left = 0;
var right = arrLen - 1;
// Sort Array in Ascending Order
arr = arr.sort(function(a, b) {
return a - b;
})
// Iterate Over
while(left < right){
if(arr[left] + arr[right] === sum){
return {
res : true,
matchNum: arr[left] + ' + ' + arr[right]
};
}else if(arr[left] + arr[right] < sum){
left++;
}else{
right--;
}
}
return 0;
}
var resp = findSumOfArrayInGiven (arr, arrLen, sum);
// Display Desired output
if(resp.res === true){
console.log('Matching Numbers are: ' + resp.matchNum +' = '+ sum);
}else{
console.log('There are no matching numbers of given sum');
}
Runtime test JSBin: https://jsbin.com/vuhitudebi/edit?js,console
Runtime test JSFiddle: https://jsfiddle.net/arbaazshaikh919/de0amjxt/4/

function sumOfTwo(array, sumNumber) {
for (i of array) {
for (j of array) {
if (i + j === sumNumber) {
console.log([i, j])
}
}
}
}
sumOfTwo([1, 2, 3], 4)

function twoSum(args , total) {
let obj = [];
let a = args.length;
for(let i = 0 ; i < a ; i++){
for(let j = 0; j < a ; j++){
if(args[i] + args[j] == total) {
obj.push([args[i] , args[j]])
}
}
}
console.log(obj)}
twoSum([10,20,10,40,50,60,70,30],60);
/* */