Develop Reference

How to insert an array to another array at each iteration of a for loop in javascript

I have a function to bubble sort and I want to save the array after each swap into another array. The bubble sort is working properly and I can log the array after each swap to the console. But I cant seem to push to the other array properly.
Here's my code :
var arr = [5, 4, 3, 2, 1];
var steps = [];
function bubbleSort() {
for (let i = 0; i < arr.length - 1; i++) {
for (let j = 0; j < arr.length - i - 1; j++) {
if (arr[j] > arr[j + 1]) {
swap(arr, j, j + 1);
}
var temp = arr;
console.log(temp);
steps.push(temp);
}
}
console.log(steps);
}
const swap = (a, x, y) => {
var temp = a[x];
a[x] = a[y];
a[y] = temp;
};
bubbleSort();
Here's a screenshot of the console :
screenshot of console
It's only when I try to use get a hold of the array at each step that it isn't showing properly? what do I do?

I think clonning the array could work ? var temp = [...arr];
var arr = [5, 4, 3, 2, 1];
var steps = [];
function bubbleSort() {
for (let i = 0; i < arr.length - 1; i++) {
for (let j = 0; j < arr.length - i - 1; j++) {
if (arr[j] > arr[j + 1]) {
swap(arr, j, j + 1);
}
var temp = [...arr];
console.log(temp);
steps.push(temp);
}
}
console.log(steps);
}
const swap = (a, x, y) => {
var temp = a[x];
a[x] = a[y];
a[y] = temp;
};
bubbleSort();

You are inserting the Reference of the Temp-Array and not the actual content. This way, you store multiple times the reference and at the End you are presented with all those reference pointing to the last version of your temp Array.
You can use the destructuring assignment of an array, to create an easy shallow copy to be stored.
var arr = [5, 4, 3, 2, 1];
var steps = [];
function bubbleSort() {
for (let i = 0; i < arr.length - 1; i++) {
for (let j = 0; j < arr.length - i - 1; j++) {
if (arr[j] > arr[j + 1]) {
swap(arr, j, j + 1);
}
var temp = arr;
console.log(temp);
// Changed to destructuring assignment
steps.push([...temp])
}
}
console.log(steps);
}
const swap = (a, x, y) => {
var temp = a[x];
a[x] = a[y];
a[y] = temp;
};
bubbleSort();

How to turn this solution from O(n^2) to O(n)?

I am having some problems turning this solution from O(n^2) to O(n). Can anyone kindly help? I am not able to think of any ways to make the time complexity O(n).
//MERGE SORTED ARRAY
/*arr1 = [0,3,4,31]
arr2 = [4,6,30]*/
const mergeSortedArrays = (arr1, arr2) => {
let arr = [];
let flag = true;
// MERGING ARRAYS
for (let i = 0; i < arr1.length; i++) {
arr.push(arr1[i]);//PUSHING ARRAY1 in arr
}
for (let i = 0; i < arr2.length; i++) {
arr.push(arr2[i]);//PUSING ARRAY2 in arr
}
//SORTING ARRAYS
while (flag) {
for (let i = 0; i < arr.length; i++) {
if (arr[i] > arr[i + 1]) {
let temp = arr[i + 1];
arr[i + 1] = arr[i];
arr[i] = temp;
flag = true;
} else {
flag = false;
}
}
}
console.log(arr1);
console.log(arr2);
console.log(arr);//FINAL MERGED & SORTED ARRAY
// return arr;
}
mergeSortedArrays([0, 3, 4, 31], [4, 6, 30]);

Try visualising it. It is as if you had two sorted stacks of cards you wanted to sort. You can compare cards at the top of each stack, and put the smaller value on a third stack. And repeat until all cards are on the third sorted stack.
You can keep up two pointers, i and j, one for each array. This will emulate a stack.
The algorithm:
Repeat until the end of both arrays is reached:
if arr1[i] <= arr2[j]
push arr1[i] to the merged array and increment i
else
push arr2[j] to the merged array and increment j
And some javascript code:
let merged = [];
let i = 0;
let j = 0;
while(i < arr1.length || j < arr2.length){
if(j == arr2.length || (i < arr1.length && arr1[i] <= arr2[j])){
merged.push(arr1[i]);
i++;
} else{
merged.push(arr2[j]);
j++;
}
}

You can use two pointer method
(this solution is based on the assumption that the two lists will always be sorted):
let p1 = 0, p2 = 0;
let arr = [];
while (p1 < arr1.length && p2 < arr2.length) {
if (arr1[p1] < arr2[p2])
arr.push(arr1[p1++]);
else
arr.push(arr2[p2++]);
}
while (p1 < arr1.length)
arr.push(arr1[p1++]);
while (p2 < arr2.length)
arr.push(arr2[p2++]);
This code will run at the time complexity of O(N).

Updated answer based on comments from using Array#sort() to:
const mergeSortedArrays = (arr1, arr2) => {
const array = { arr1, arr2 }
const index = { a1: 0, a2: 0 }
const length = { a1: array.arr1.length, a2: array.arr2.length }
const merged = []
let current = 0
while (current < length.a1 + length.a2) {
const [a, i] =
!(index.a1 >= length.a1) &&
(index.a2 >= length.a2 || array.arr1[index.a1] < array.arr2[index.a2])
? ['arr1', 'a1']
: ['arr2', 'a2']
merged[current] = array[a][index[i]]
index[i]++
current++
}
return merged
}
const result = mergeSortedArrays([0, 3, 4, 31], [4, 6, 30])
console.log(result)

hasPairsWithSum Google Interview Question

I solved this problem by iterating through the array then find the item when the sum equals to array[i] + item returning true otherwise returning false.
My Question is => How I can return the indices of those numbers that add up to sum not just true? Using the same code below:
function hasPairsWithSum(array,sum) {
for (let i = 0; i < array.length; i++) {
if (array.find((item) => {return sum === array[i] + item}
));
return true;
};
return false;
};
console.log(hasPairsWithSum([1,2,4,4],8))
Note: Time complexity must be less than O(n ^ 2).

JavaScript O(n) Solution.
function hasPairsWithSum(array, sum) {
const map = new Map ();
for(let i = 0; i < array.length; i++) {
let currVal = array[i];
if (map.has(currVal)) {
return [map.get(currVal),i]
}
// difference value = sum - current value
let diff = sum - currVal
map.set(diff,i)
}
};
console.log(hasPairsWithSum([2,2,4,4], 8))

Please refer this code.
function hasPairsWithSum(array,sum) {
let result = [];
for (let i = 0; i < array.length; i++) {
if (array.some((item, index) => {return i === index ? false : sum === array[i] + item}))
result.push(i);
};
return result;
};
console.log(hasPairsWithSum([1,2,4,4],8))
console.log(hasPairsWithSum([3,2,4],6))
console.log(hasPairsWithSum([0,4,3,0],0))

O(n) Soln ... using math concept a+b = n then if a is present in our array then need to find b = n - a is present or not ..
def hasPairsWithSum(array,sum):
d = {}
for i in range(len(array)):
if(array[i] in d):
d[array[i]].append(i)
else:
d[array[i]] = [i]
ans = []
for i in range(len(array)):
val = sum - array[i]
if(val in d):
if(d[val][0] == i):
if(len(d[val]) > 1):
ans.append((i,d[val][1]))
break
else:
continue
else:
ans.append((i,d[val][0]))
break
return ans
print(hasPairsWithSum([4, 4, 4, 4], 8))
O(nlogn) soln ....just store the index with elements .. then sort it by their values .. next step run a loop with complexity of O(n) [concept : Two pointers]
def hasPairsWithSum(array,sum):
arr = []
for i in range(len(array)):
arr.append((array[i],i))
arr.sort()
i = 0
j = len(array)-1
ans = []
while(i<j):
tmp_sum = arr[i][0] + arr[j][0]
if(tmp_sum == sum):
ans.append((arr[i][1] , arr[j][1]))
#add your logic if you want to find all possible indexes instead of break
break
elif(tmp_sum < sum):
i = i + 1
elif(tmp_sum > sum):
j = j - 1
return ans
print(hasPairsWithSum([1,2,4,4],8))
note : if you want to find all possible soln then these approaches will not work either add you own logic in while loop or another approach is use binary search with traversal on every element and store the indexes in set (worst case this will be O(n^2) as we have to find all possible values) Eg: [4,4,4,4,4,4] , sum = 8 and you want to print all possible indexes then we end up running it upto n^2 (why? reason: total possible solns. are 5+4+3+2+1 = n*(n-1)/2 ≈ n^2)

You have to iterate over the array elements checking at every iteration for every element of the array (except the last one) all the elements at the right of it like below:
function findIndexes(array, sum) {
const result = [];
for (let i = 0; i < array.length -1; ++i) {
for (let j = i + 1; j < array.length; ++j) {
if ((array[i] + array[j]) === sum) {
result.push([i, j]);
}
}
}
return result;
}
console.log(findIndexes([1, 2, 4, 4], 8));
console.log(findIndexes([3, 2, 4], 6));
Update:
It is possible to obtain a linear O(n) complexity using an auxiliary Map structure associating an integer value as key with as a value the list containing all the indexes of the elements in the array equal to the integer key like below:
function findIndexes(array, sum) {
const map = new Map();
const result = [];
for (let i = 0; i < array.length; ++i) {
const a = array[i];
const b = sum - a;
if (map.has(b)) {
for (const index of map.get(b)) {
result.push([index, i]);
}
}
const l = map.has(a) ? map.get(a) : [];
l.push(i);
map.set(a, l);
}
return result;
}
console.log(findIndexes([1, 2, 4, 4], 8));
console.log(findIndexes([3, 2, 4], 6));
console.log(findIndexes([1, 1, 1], 2));

Javascript twoSum algorithm: Given an array of integers, return indices of the two numbers such that they add up to a specific target

Given an array of integers, return indices of the two numbers such that they add up to a specific target.
Example:
Given nums = [3, 2, 4], target = 6,
Because nums[1] + nums[2] = 2 + 4 = 6
return [1, 2].
Solution
var twoSum = function(nums, target) {
for(let i = 0; i <= nums.length; i++){
for(let j = 0; j <= nums.length; j++){
if(nums[i] + nums[j] == target){
return [i, j]
}
}
}
};
The code above works in other cases but not this one.
Expected result [1,2]
Output [0,0]
For instance, I've tried to use a different array of numbers and a different target and it works even if you change the order of the numbers
Example:
New array: [15, 7, 11, 2], target = 9,
Output: [1, 3].
I don't understand what is wrong with the solution and I hope that someone can explain. Thanks

You can use a very simple technique.
Basically, you can check if the difference of target & the element in the current iteration, exists in the array.
Assuming same index cannot be used twice
nums = [3, 2, 4], target = 6
nums[0] = 3
target = 6
diff = 6 - 3 = 3
nums.indexOf[3] = 0 // FAILURE case because it's the same index
// move to next iteration
nums[1] = 2
target = 6
diff = 6 - 2 = 4
nums.indexOf(4) = 2 // SUCCESS '4' exists in the array nums
// break the loop
Here's the accepted answer by the leetcode.
/**
* #param {number[]} nums
* #param {number} target
* #return {number[]}
*/
var twoSum = function(nums, target) {
for (let index = 0; index < nums.length; index++) {
const diff = target - nums[index];
const diffIndex = nums.indexOf(diff);
// "diffIndex !== index" takes care of same index not being reused
if (diffIndex !== -1 && diffIndex !== index) {
return [index, diffIndex];
}
}
};
Runtime: 72 ms, faster than 93.74% of JavaScript online submissions for Two Sum.
Memory Usage: 38.5 MB, less than 90.55% of JavaScript online submissions for Two Sum.
Can anybody help me in reducing the memory usage also?

I don't understand what is wrong with the solution and I hope that
someone can explain ?
Here you're both inner and outer loop start from 0th so in the case [3,2,4] and target 6 it will return [0,0] as 3 + 3 is equal to target, so to take care of same index element not being used twice created a difference of 1 between outer and inner loop
Make outer loop to start from 0th index and inner loop with value i+1
var twoSum = function(nums, target) {
for(let i = 0; i < nums.length; i++){
for(let j = i+1; j < nums.length; j++){
if(nums[i] + nums[j] == target){
return [i, j]
}
}
}
};
console.log(twoSum([15, 7, 11, 2],9))
console.log(twoSum([3, 2, 4],6))

we can solve this problem in O(n) by using the map/object.
We can maintain a map or object which will save all values with index and then we can iterate the array and find target-nums[i] for each value and will find that value in map/object.
let's see this by example:-
nums=[2,7,11,15]
target = 9;
then map/object will be
mm={
2 : 0,
7: 1,
11: 2,
15: 3
}
then for each value, we will find diff and find that diff in map/object.
for i=0 diff= 9-2=7 and mm.has(7) is true so our answer is 2 and 7.
for their index we can use mm.get(7) and i.
return [mm.get(7), i]
var twoSum = function(nums, target) {
let mm=new Map();
for(let i=0;i<nums.length;i++){
mm.set(nums[i],i);
}
let diff=0;
let j;
for(let i=0;i<nums.length;i++){
diff=target-nums[i];
if(mm.has(diff) && i!=mm.get(diff)){
j=mm.get(diff);
if(j>i){
return [i,j];
}else{
return [j,i];
}
}
}
};
Runtime: 76 ms, faster than 88.18% of JavaScript online submissions for Two Sum.
Memory Usage: 41.4 MB, less than 13.32% of JavaScript online submissions for Two Sum.

Your solution works as expected. For nums = [3, 2 ,4] and target = 6, [0, 0] is a valid solution for the outlined problem as nums[0] + nums[0] = 3 + 3 = 6.
If you need two different indices (In my understanding this is not required by the task) you can add an additional check for inequality (nums[i] + nums[j] == target && i != j).

var twoSum = function(nums, target) {
for(let i=0; i<nums.length; i++){
for(let j=i+1; j<nums.length; j++){
if(nums[j] === target - nums[i]){
return [i, j];
}
}
}
};

Here's a simple method to solve this problem and its efficient using different type of inputs using JavaScript.
Like with input of ([3,3], 6) and its expected output will be [0,1] and input like ([3,2,4], 6) with expected output will be [2,4]
var twoSum = function (nums, target) {
for (let i = 0; i <= nums.length; i++) {
for (let j = 0; j <= nums.length; j++) {
if (i !== j) {
if (nums[i] + nums[j] === target) {
return [i, j];
}
}
}
}
};
console.log(twoSum([3, 2, 4], 6));

var twoSum = function (nums, target) {
var len = nums.length;
for (var i = 0; i < len; i++) {
for (var j = i + 1; j < len; j++) {
if (nums[i] + nums[j] == target) {
return [i,j];
}
}
}
};

var twoSum = function(nums, target) {
for(var i=0;i<nums.length;i++){
for(var j=i+1;j<nums.length;j++){
temp = nums[i]+nums[j];
if(temp == target){
return [i,j]
}
}
}
};
console.log(twoSum([15, 7, 11, 2],9))
console.log(twoSum([3, 2, 4],6))
console.log(twoSum([3,3],6))
This works perfectly and the Runtime: 72 ms lesser than 84ms

Another efficient solution way to solve this problem with an O(n) time complexity is not using nested loops. I commented the steps, so JS developers can easy understand. Here is my solution using golang:
func twoSum(intArray []int, target int) []int {
response := []int{-1, -1} // create an array as default response
if len(intArray) == 0 { // return default response if the input array is empty
return response
}
listMap := map[int]int{} // create a Map, JS => listMap = new Map()
for index, value := range intArray { // for loop to fill the map
listMap[value] = index
}
for i, value := range intArray { // for loop to verify if the subtraction is in the map
result := target - value
if j, ok := listMap[result]; ok && i != j { // this verify if a property exists on a map in golang. In the same line we verify it i == j.
response[0] = i
response[1] = j
return response
}
}
return response
}

var twoSum = function(nums, target) {
var numlen = nums.length;
for(let i=0; i<=numlen; i++){
for(let j=i+1;j<numlen;j++){
var num1 = parseInt(nums[i]);
var num2 = parseInt(nums[j]);
var num3 = num1 + num2;
if(num3 == target){
return[i,j]
}
}
}
};

I suppose this could be a better solution. Instead of nesting loops, this provides a linear solution.
(PS: indexOf is kinda a loop too with O(n) complexity)
var twoSum = function (nums, target) {
const hm = {}
nums.forEach((num, i) => {
hm[target - num] = i
})
for (let i = 0; i < nums.length; i++) {
if(hm[nums[i]] !== undefined && hm[nums[i]] !== i) {
return ([hm[nums[i]], i])
}
}
};

var twoSum = function(nums, target) {
for (var i = 0; i < nums.length; i++)
{
if( nums.indexOf(target - nums[i]) !== -1)
{
return [i , nums.indexOf(target - nums[i])]
}
}
};

For clearity console a & b in inner for loop . This will give you clear insight
var twoSum = function(nums, target) {
var arr=[];
for(var a=0;a<nums.length;a++){
for(var b=1;b<nums.length;b++){
let c=nums[a]+nums[b];
if(c== target && a != b){
arr[0]=a;
arr[1]=b;
}
}
}
return arr;
};

My solution:
Create a Set
Then loop the array for retrieve all the differences between the target number minus all the elements in array. Insert only which are positive (as is a Set, duplicates will not be included).
Then iterates array again, now only compare if array elements is in the set, if yes take the index from i store in the index array and return it.
public static Object solution(int[] a, int target){
Set<Integer> s = new HashSet<>();
ArrayList<Integer> indexes = new ArrayList<Integer>();
for (int e : a){
Integer diff = new Integer(target - e);
if(diff>0){
s.add(diff);
}
}
int i = 0;
for (int e : a){
if(s.contains(e)){
indexes.add(i);
}
i++;
}
return indexes;
}

def twoSum(self, nums: List[int], target: int) -> List[int]:
for i in range(len(nums)):
comp = target - nums[i]
if comp in nums:
j = nums.index(comp)
if(i!=j):
return [i,j]

var twoSum = function(nums, target) {
for(let i = 0; i < nums.length; i++){
for(let j = i+1; j < nums.length; j++){
if(nums[i] + nums[j] == target){
return [i, j]
}
}
}
};
console.log(twoSum([15, 7, 11, 2],9))
console.log(twoSum([3, 2, 4],6))

class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
indx =[]
for i in range(len(nums)):
for j in range(i+1,len(nums)):
if nums[j]==(target-nums[i]):
indx.append(i)
indx.append(j)
return indx

twoSummation = (numsArr, estTarget) => {
for(let index = 0; index < numsArr.length; index++){
for(let n = index+1; n < numsArr.length; n++){
if(numsArr[index] + numsArr[n] == estTarget){
return [index, n]
}
}
}
}
console.log(twoSummation([2,7,11,15], 9))
console.log(twoSummation([3,2,4], 6))

<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<title>Document</title>
</head>
<body>
<script>
var twoSum = function(nums, target) {
for(var i=0;i<nums.length;i++){
for(var j=i+1;j<nums.length;j++){
temp = nums[i]+nums[j];
if(temp == target){
return [nums[i],nums[j]]
}
}
}
};
console.log(twoSum([15, 7, 11, 2],9))
console.log(twoSum([3, 2, 4],6))
console.log(twoSum([3,3],6))
</script>
</body>
</html>

Solution from my leetcode submission in Java. Top 75% in runtime, top 25% in memory. It is 0(n) however uses an external HashMap.
import java.util.HashMap;
class Solution {
public int[] twoSum(int[] nums, int target) {
HashMap<Integer, Integer> numbers= new HashMap<Integer, Integer>();
for(int i=0; i<nums.length; i++){
int cur = nums[i];
numbers.put(cur, i);
}
for(int i=0; i<nums.length; i++){
//check map for diff of target - cur
int diff = target - nums[i];
if(numbers.containsKey(diff) && numbers.get(diff)!=i){
return new int[]{i, numbers.get(diff)};
}
}
return null;
}
}

How to programmatically create 3D array that increments to a defined number, resets to zero, and increments again?

Starting with this initial 2D array:
var initialArray = [[2,3],[6,7],[4,5],[1,2],[5,6],[2,3]];
I need to create this 3D array programmatically:
var fullArray = [
[[2,3],[6,7],[4,5],[1,2],[5,6],[2,3]],
[[3,4],[0,1],[5,6],[2,3],[6,7],[3,4]],
[[4,5],[1,2],[6,7],[3,4],[0,1],[4,5]],
[[5,6],[2,3],[0,1],[4,5],[1,2],[5,6]],
[[6,7],[3,4],[1,2],[5,6],[2,3],[6,7]],
[[0,1],[4,5],[2,3],[6,7],[3,4],[0,1]],
[[1,2],[5,6],[3,4],[0,1],[4,5],[1,2]],
[[2,3],[6,7],[4,5],[1,2],[5,6],[2,3]],
[[3,4],[0,1],[5,6],[2,3],[6,7],[3,4]],
[[4,5],[1,2],[6,7],[3,4],[0,1],[4,5]],
[[5,6],[2,3],[0,1],[4,5],[1,2],[5,6]]
];
See the pattern?
On each pair, the [0] position should increment to 6 (from any starting number <= 6) and then reset to 0 and then continue incrementing. Similarly, the [1] position should increment to 7 (from any starting number <= 7) and then reset to 1 and then continue incrementing.
In this example, there are 10 2D arrays contained in the fullArray. However, I need this number to be a variable. Something like this:
var numberOf2DArraysInFullArray = 12;
Furthermore, the initial array should be flexible so that initialArray values can be rearranged like this (but with the same iteration follow-through rules stated above):
var initialArray = [[6,7],[2,3],[5,6],[4,5],[1,2],[6,7]];
Any thoughts on how to programmatically create this structure?
Stumped on how to gracefully pull this off.
Feedback greatly appreciated!

Here's a solution, I've separated the methods, and I made it so if instead of pairs it's an N size array and you want the [2] to increase up to 8 and reset to 2, if that's not needed you can simplify the of the loop for(var j = 0; j < innerArray.length; j++)
var initialArray = [[2,3],[6,7],[4,5],[1,2],[5,6],[2,3]];
var create3DArray = function(array, size){
var newArray = [initialArray];
for(var i = 0; i < size; i++)
{
newArray.push(getNextArrayRow(newArray[i]));
}
return newArray;
}
var getNextArrayRow = function(array){
var nextRow = [];
for(var i = 0; i < array.length; i++)
{
var innerArray = array[i];
var nextElement = [];
for(var j = 0; j < innerArray.length; j++)
{
var value = (innerArray[j] + 1) % (7 + j);
value = value === 0 ? j : value;
nextElement.push(value);
}
nextRow.push(nextElement);
}
return nextRow;
}
console.log(create3DArray(initialArray,3));
Note, the results from running the snippet are a bit difficult to read...

var initialArray = [[2,3],[6,7],[4,5],[1,2],[5,6],[2,3]];
var numOfArrays = 10;
// get a range array [0, 1, 2, ...]
var range = [];
for (var i = 0; i < numOfArrays; i++) {
range.push(i);
}
var result = range.reduce(function(prev, index) {
if (index == 0) {
return prev;
}
prev.push(transformArray(prev[index - 1]));
return prev;
}, [initialArray])
console.log(result);
function transformArray(arr) {
return arr.map(transformSubArray)
}
function transformSubArray(arr) {
return arr.map(function(val) {
return val == 7 ? 0 : val + 1;
})
}
Here's a pretty simple functional-ish implementation