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I am trying to solve a leetcode type problem that is a practice problem that came with an upcoming code test I need to do for a job and I am having trouble with it. Can anyone help me understand whats going wrong?
I am essentially looking for the brute force option as I dont know algos/DS.
PROBLEM:
Write a function:
function solution(A);
that, given an array A of N integers, returns the smallest positive integer (greater than 0) that does not occur in A.
For example, given A = [1, 3, 6, 4, 1, 2], the function should return 5.
Given A = [1, 2, 3], the function should return 4.
Given A = [−1, −3], the function should return 1.
Write an efficient algorithm for the following assumptions:
N is an integer within the range [1..100,000];
each element of array A is an integer within the range [−1,000,000..1,000,000].
HERE IS MY SOLUTION:
function solution(A) {
let newArray = A.sort(function(a, b){return a-b})
let lowestNumber = 1
for(i=0; i < newArray.length; i++) {
if(lowestNumber > newArray[0]) {
return lowestNumber
}
if(lowestNumber == newArray[i]) {
lowestNumber = lowestNumber + 1
}
if(i = newArray.length - 1) {
return lowestNumber
}
}
}
The below snippet isnt working like I expect it to. lowestNumber isnt being increased and also the loop is exiting here I believe.
if(lowestNumber == newArray[i]) {
lowestNumber = lowestNumber + 1
Thanks for your help!
You can do this in O(N) using a Map():
First set every number in the array.
Then starting from 1 look for and return the missing number in the sequence.
function solution(arr) {
const seen = new Map();
for (let i = 0; i < arr.length; i++) {
seen.set(arr[i]);
}
for (let i = 1; i <= arr.length + 1; i++) {
if (!seen.has(i)) return i;
}
return 1;
}
console.log(solution([1, 3, 6, 4, 1, 2])); //-> 5
console.log(solution([1, 2, 3])); //-> 4
console.log(solution([-1, -3])); //-> 1
I think your > should be <, and the = in if(i = newArray.length - 1) should be ===.
And lowestNumber > newArray[0] will always be true if the array contains a negative number, so 1 will be returned.
Your effort seems careless, so you are going to have to up your game for the interview.
const integers = [5, -345, 562456, 95345, 4, 232, 1, 2, 3, 7, -457];
function solution(A) {
let newArray = A.sort((a, b) => a - b);
let lowestNumber = 1;
for (let i = 0; i < newArray.length; i++) {
const n = newArray[i];
if (n > 0) {
if (lowestNumber < n) {
return lowestNumber;
} else {
lowestNumber = n + 1;
}
}
}
return lowestNumber;
}
console.log(solution(integers));
The fastest solution
function solution(A) {
// write your code in JavaScript (Node.js 8.9.4)
if (!A) return 1;
A.sort();
if (A[A.length - 1] < 1) return 1;
const setA = new Set(A);
let length = setA.size;
for (let i = 1; i <= length; i++) {
if (!setA.has(i)) {
return i;
}
}
return length + 1;
}
I have worked same problem for nowadays, and regardless the original answer, here is my version of finding least positive number which is missing the in the array.
function findLeastPositive(array) {
const numbersSet = new Set(array);
let leastPositiveNumber = 1;
while(numbersSet.has(leastPositiveNumber)) {
leastPositiveNumber++;
}
return leastPositiveNumber;
}
let result = findLeastPositive([1,2,3,4,5,6,7,8,9,0]);
console.log(result);
result = findLeastPositive([10,11,12,13,14,15,16,17,18,19]);
console.log(result);
There are sure similar answers floating on the internet but using given array length disturbing me of which I can't explain properly why we have to create second loop starts with 1 (known the least positive number) and to N.
Using hash table (I am using Set here) for lookup table is fine idea, as in it effect to overall performance O(N) (probably initialize the Set with array) and O(1) for checking if the value in the Set or not.
Then we need to set second loop for obvious reason that checking the the smallest positive number existence, starting from 1..N range. This is the part bugged me, so I decided to go for while loop. It's obvious rather why there's a for..loop starts from 1..N on which N is the length of the array.
Here is 100% code
function solution(A) {
/**
* create new array of positive numbers in given array,
* if there sis no specific number in given array, in result array
* that index will be undefine
*/
const c = A.reduce((arr, cur) => {
if(cur > 0) arr[cur] = 1;
return arr;
} , [1] )
/**
* return first undefined index
*/
for(let i = 0; i < c.length; i++)
if(!c[i]) return i;
// otherwise return the maximum index in array
return c.length;
}
function solution(arr) {
for (let i = 1; i <= arr.length + 1; i++) {
if (!arr.includes(i)) return i;
}
return 1;
}
console.log(solution([1, 3, 6, 4, 1, 2])); //-> 5
console.log(solution([1, 2, 3])); //-> 4
console.log(solution([-1, -3])); //-> 1
I need to create a function that take as parameter an array and a target. It should return an array of arrays where the sum of these numbers equals to the target
sumPairs(array, target) {
}
For example:
sumPairs([1, 2, 3, 4, 5], 7) // output : [[2, 5], [3, 4]]
I know I have to use map(), and probably reduce(), set(), or filter() maybe (I read their documentation in MDN but still cant find out). I tried some ways but I can't get it.
If you guys could help me to find out how to dynamically create arrays and push them into a new array..
I read there some solutions (Split array into arrays of matching values) but I hate to just use created functions without knowing what they really do or how they work.
Some very basic code for achieving it, Just run all over combinations and conditionally add the items you want.
function sumPairs(array, target) {
var res = [];
for(var i = 0; i < array.length; i++){
for(var j = 0; j < array.length; j++){
if(i!=j && array[i]+array[j]==target &&
res.filter((x)=> x[0] == array[j] && x[1] == array[i]).length == 0 )
res.push([array[i], array[j]]);
}
}
return res;
}
var result = sumPairs([1, 2, 3, 4, 5], 7);
console.log(result);
Option 2 - see this answer for more options (like using reduce)
function sumPairs(array, target) {
return array.flatMap(
(v, i) => array.slice(i+1).filter(w => (v!=w && v+w==target)).map(w=> [w,v])
);
}
var result = sumPairs([1, 2, 3, 4, 5], 7);
console.log(result);
"The exercise says that it sould be arrays of pairs that sum the
target value so I think only 2 items"
If you need a pair that matches a sum and you pick any number from the list, you are left with
the following equation to solve num + x = sum where we want to find x. E.g. if you picked 7 and the target sum is 10 then you know you are looking for a 3.
Therefore, we can first construct a counting map of the numbers available in our list linear (O(n)) time and then search for matches in linear time as well rather than brute forcing with a quadratic algorithm.
const nums = [1, 2, 3, 4, 5];
console.log(findSumPairs(nums, 7));
function findSumPairs(nums, sum) {
const countByNum = countGroupByNum(nums);
return nums.reduce((pairs, num) => {
countByNum[num]--;
const target = sum - num;
if (countByNum[target] > 0) {
countByNum[target]--;
pairs.push([num, target]);
} else {
countByNum[num]++;
}
return pairs;
}, []);
}
function countGroupByNum(nums) {
return nums.reduce((acc, n) => (acc[n] = (acc[n] || 0) + 1, acc), {});
}
Here's another implementation with more standard paradigms (e.g. no reduce):
const nums = [1, 2, 3, 4, 5];
console.log(findSumPairs(nums, 7));
function findSumPairs(nums, sum) {
const countByNum = countGroupByNum(nums);
const pairs = [];
for (const num of nums) {
const target = sum - num; //Calculate the target to make the sum
countByNum[num]--; //Make sure we dont pick the same num instance
if (countByNum[target] > 0) { //If we found the target
countByNum[target]--;
pairs.push([num, target]);
} else {
countByNum[target]++; //Didin't find a match, return the deducted num
}
}
return pairs;
}
function countGroupByNum(nums) {
const countByNum = {};
for (const num of nums) {
countByNum[num] = (countByNum[num] || 0) + 1;
}
return countByNum;
}
You can also sort your array and find all the pairs with given sum by using two pointer method. Place the first pointer to the start of the array and the second pointer to the end.
if the sum of the values at the two places is :
More than target: Decrement your second pointer by 1
Less than target: Increment your first pointer by 1
Equal to target: This is one possible answer, push them to your answer array and increment your first pointer by 1 and decrement your second pointer by 1.
This is more performant solution with complexity O(n*log(n))
I'm looking for an efficient JavaScript utility method that in O(n) will remove a set of items from an array in place. You can assume equality with the === operator will work correctly.
Here is an example signature (written in TypeScript for type clarity)
function deleteItemsFromArray<T>(array: T[], itemsToDelete: T[]) { ... }
My thought is to do this in two passes. The first pass gets the indexes that need to be removed. The second pass then compacts the array by copying backwards from the current index being removed through the next index being removed.
Does anyone have code like this already handy or a more efficient way to do this?
P.S. Please don't point out the filter function as that creates a copy of the array, it does not work in place.
Iterate over the array, copying elements that aren't in itemsToDelete to the next destination index in the array. When you delete an element, you don't increment this index.
Finally, reset the length of the array at the end to get rid of the remaining elements at the end.
function deleteItemsFromArray(array, itemsToDelete) {
let destIndex = 0;
array.forEach(e => {
if (!itemsToDelete.includes(e)) {
array[destIndex++] = e;
}
});
array.length = destIndex;
}
const array = [1, 2, 3, 4, 5, 6, 7];
deleteItemsFromArray(array, [3, 5]);
console.log(array);
function deleteItems(array,items){
let itemsToDelete=0;
let indexToSwap = array.length-1;
for(let i = array.length-1,currItem ; i>=0 ; i--){
if(items.includes(array[i]){
[array[i] , array[indexToSwap]] = [array[indexToSwap] , array[i]];
--indexToSwap;
++itemsToDelete;
}
}
array.splice(array.length-itemsToDelete,itemsToDelete);
}
This should work, I haven't tested it.
The idea is to swap the elements to delete to the end of the array. You can remove them at the end how I do in my code or you could use too the pop() function every time.
It's very, very simple - transform itemsToDelete to a Set. This ensures O(1) lookups. Then walk through the array backwards and remove items with Array#splice. In total, that gives you a linear O(n+m) time complexity:
function deleteItemsFromArray<T>(array: T[], itemsToDelete: T[]) {
const setToDelete = new Set(itemsToDelete);
for (let i = array.length - 1; i >= 0; i--) {
const item = array[i];
const shouldBeDeleted = setToDelete.has(item);
if (shouldBeDeleted) {
array.splice(i, 1);
}
}
}
You can save the conversion step if you just make the function accept a set to begin with and change the signature to:
function deleteItemsFromArray<T>(array: T[], itemsToDelete: Set<T>)
function deleteItemsFromArray(array, itemsToDelete) {
for (let i = array.length - 1; i >= 0; i--) {
const item = array[i];
const shouldBeDeleted = itemsToDelete.has(item);
if (shouldBeDeleted) {
array.splice(i, 1);
}
}
}
const arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14];
deleteItemsFromArray(
arr, new Set([1, 3, 5, 7, 8, 9])
)
console.log(arr);
Using copyWithin method will be helpful here.
method 1
Traverse from last index to first index, when ever we find the item to remove just copy/move elements to left.
Though we will be doing times the copy/move, But still have some unnecessary moving elements.
method 2
Traverse from first index to last index, when ever we find the item to remove, identify the group of elements to copy/move. This will avoid the unnecessary moving elements as in method 1.
function deleteItemsFromArray(array, delete_list) {
for (let i = array.length - 1; i > -1; i--) {
if (delete_list.includes(array[i])) {
array.copyWithin(i, i + 1).pop();
}
}
}
// Alternate way, with minimal copy/move group of elements.
function deleteItemsFromArrayAlternate(array, delete_list) {
let index = -1;
let count = 0;
for (let i = 0; i <= array.length; i++) {
if (delete_list.includes(array[i]) || !(i in array)) {
if (index > -1) {
array.copyWithin(index - count + 1, index + 1, i);
}
count += 1;
index = i;
}
}
array.length = array.length - delete_list.length;
}
const array = [9, 12, 3, 4, 5, 6, 7];
deleteItemsFromArray(array, [9, 6, 7]);
console.log(array);
I've noticed my algorithm solutions often include ugly nested for loops when better options are available. In the example below, how can I determine the prime numbers up to the given parameter without nested loops? One constraint of the assignment was to not utilize any external functions.
function sumPrimes(num) {
var p = [];
for (var i = 2; i <= num; i++)
{
p.push(i);
for (var j = 2; j < i; j++)
{
if (i % j === 0) //not a prime
{
p.pop(i);
break;
}
}
}
return p;
I don't know if there is any other good solution but I came up with this. There is still one for loop ;)
function primes(num) {
return Array.apply(null, {length: (num + 1)}).map(Number.call, Number)
.filter(function(n) {
for(var i = 2; i < n; i++) {
if(n % i === 0) return false;
}
return n !== 0 && n !== 1;
})
}
Explanation:
Array.apply(null, {length: (num + 1)}).map(Number.call, Number)
This line creates an Array with the range of the passed argument. If num would be 5 it would create the following Array: [0, 1, 2, 3, 4, 5]. I found this here.
Then I use the filter function to remove all non prime numbers. If the filter function returns false the number will be removed from the array. So inside the filter function we check if the current number is a prime number. If so, true is returned. As you can see I still use a normal for loop here.
No you can call the function as follows: primes(28);
This will return the following Array: [ 2, 3, 5, 7, 11, 13, 17, 19, 23 ]
In this specific case I would say to use normal for loops is absolutely ok. But always consider functions like map, reduce or filter when you operate on arrays.
So, I was working on this challenge to return the third largest number in an array. I had got it worked out until I realized that I must account for repeat numbers. I handled this by adding 3 layers of for loops with variables i, j, and k. You'll see what I mean in the code. This is not terribly efficient or scalable.
My question is, how can I optimize this code? What other methods should I be using?
function thirdGreatest (arr) {
arr.sort(function(a, b) {
if (a < b) {
return 1;
} else if (a > b) {
return -1;
} else {
return 0;
}
});
for ( var i = 0; i < arr.length; i++) {
for (var j = 1; j < arr.length; j++) {
for (var k = 2; k < arr.length; k++) {
if (arr[i] > arr[j]) {
if (arr[j] > arr[k]) {
return arr[k];
}
}
}
}
}
}
console.log(thirdGreatest([5, 3, 23, 7,3,2,5,10,24,2,31, 31, 31])); // 23
console.log(thirdGreatest([5, 3, 23, 7,3,2,5,10,24,2,31])) // 23
console.log(thirdGreatest([5, 3, 7, 4])); // 4
console.log(thirdGreatest([2, 3, 7, 4])); // 3
Since you already sorted the array, it seems like you should be fine iterating over the list and keep track of the numbers you have already seen. When you have seen three different numbers, return the current one:
var seen = [arr[0]];
for (var i = 1; i < arr.length; i++) {
if (arr[i] !== seen[0]) {
if (seen.length === 2) {
return arr[i];
}
seen.unshift(arr[i]);
}
}
function thirdGreatest (arr) {
arr.sort(function(a, b) {
return b - a;
});
var seen = [arr[0]];
for (var i = 1; i < arr.length; i++) {
if (arr[i] !== seen[0]) {
if (seen.length === 2) {
return arr[i];
}
seen.unshift(arr[i]);
}
}
}
console.log(thirdGreatest([5, 3, 23, 7,3,2,5,10,24,2,31, 31, 31])); // 23
console.log(thirdGreatest([5, 3, 23, 7,3,2,5,10,24,2,31])) // 23
console.log(thirdGreatest([5, 3, 7, 4])); // 4
console.log(thirdGreatest([2, 3, 7, 4])); // 3
Note: You can simplify the sort callback to
arr.sort(function(a, b) {
return b - a;
});
// With arrow functions:
// arr.sort((a, b) => b - a);
The callback has to return a number that is larger, smaller or equal to 0, it doesn't have to be exactly -1 or 1.
A one-"line"r using Set to remove duplicates
Array.from(new Set(arr)).sort(function(a, b) {
return b - a;
})[2];
Set now has reasonable browser support
The optimal solution is to do this in a single pass O(n) time. You do not need to sort the array - doing so makes your solution at-least (n log n).
To do this in as single pass, you simply need three temporary variables: largest, secondLargest, thirdLargest. Just go through the array and update these values as necessary (i.e. when you replace largest it becomes second largest, etc...). Lastly, when you see duplicates (i.e. currentValue == secondLargest), just ignore them. They don't affect the outcome.
Don't forget to check for edge cases. You cannot provide an answer for [2, 2, 2, 2, 2] or [3, 2].
Try to think about what data structure you can use here. I suggest a set. Every time you add a nested loop your function gets exponentially slower.
Edited:
function thirdGreatest(arr) {
var s = Array.from(new Set(arr)).sort(function(a, b) {
return a - b;
})
return s[2] || s[1] || s[0] || null;
}
Working Example
We need to be able to handle:
[1,2,1,2] // 2
[1,1,1,1] // 1
[] // null
This assumes that you get an array passed in.
If you do not have a third largest number, you get the second.
If you do not have a second largest you get the first largest.
If you have no numbers you get null
If you want the 3rd largest or nothing, return s[2] || null
Many of the other answers require looping through the initial array multiple times. The following sorts and deduplicates at the same time. It's a little less terse, but is more performant.
const inputArray = [5,3,23,24,5,7,3,2,5,10,24,2,31,31,31];
const getThirdGreatest = inputArray => {
const sorted = [inputArray[0]]; // Add the first value from the input to sorted, for our for loop can be normalized.
let migrated = false;
let j;
for(let i = 1; i<inputArray.length; i++) { // Start at 1 to skip the first value in the input array
for(j=0; j<sorted.length; j++) {
if(sorted[j] < inputArray[i]) {
// If the input value is greater than that in the sorted array, add that value to the start of the sorted array
sorted.splice(j,0,inputArray[i]);
migrated = true;
break;
} else if(sorted[j] === inputArray[i]) {
// If the input value is a duplicate, ignore it
migrated = true;
break;
}
}
if(migrated === false) {
// If the input value wasn't addressed in the loop, add it to the end of the sorted array.
sorted[sorted.length] = inputArray[i];
} else {
migrated = false;
}
}
// Return the third greatest
return sorted[2];;
};
const start = performance.now();
getThirdGreatest(inputArray); // 23
const end = performance.now();
console.log('speed: ', end - start); // 0.1 - 0.2ms
One single iteration O(n) and very fast method of doing this is making your own Set like object. The advantageous point is making no comparisons at all when constructing our "sorted" list of "unique" elements which brings enormous efficiency. The difference is very noticeable when dealt with huge lists like in the lengths exceeding 1,000,000.
var arr = [5, 3, 23, 7,3,2,5,10,24,2,31, 31, 31],
sorted = Object.keys(arr.reduce((p,c)=> (p[c] = c, p),Object.create(null))),
third = sorted[sorted.length-3];
document.write(third);
If you think Object.keys might not return a sorted array (which i haven't yet seen not) then you can just sort it like it's done in the Set method.
Here i tried it for 1,000,000 item array and returns always with the correct result in around 45msecs. A 10,000,000 item array would take like ~450msec which is 50% less than other O(n) solutions listed under this question.
var arr = [],
sorted = [],
t0 = 0,
t1 = 0,
third = 0;
for (var i = 0; i<1000000; i++) arr[i] = Math.floor(Math.random()*100);
t0 = performance.now();
sorted = Object.keys(arr.reduce((p,c)=> (p[c] = c, p),Object.create(null)));
third = sorted[sorted.length-3];
t1 = performance.now();
document.write(arr.length + " size array done in: " + (t1-t0) + "msecs and the third biggest item is " + third);