I am working on a problem on LeetCode and having some troubles
https://leetcode.com/problems/relative-sort-array/
Instructions:
Given two arrays arr1 and arr2, the elements of arr2 are distinct, and all elements in arr2 are also in arr1.
Sort the elements of arr1 such that the relative ordering of items in arr1 are the same as in arr2. Elements that don't appear in arr2 should be placed at the end of arr1 in ascending order.
Example 1:
Input: arr1 = [2,3,1,3,2,4,6,7,9,2,19], arr2 = [2,1,4,3,9,6]
Output: [2,2,2,1,4,3,3,9,6,7,19]
my attempt:
var relativeSortArray = function(arr1, arr2) {
let arr =[]
let end =[]
for (i=0; i<arr2.length; i++){
for (j=0; j<arr1.length; j++){
if(arr2[i] == arr1[j]){
arr.push(arr1[j])
}else{
end.push(arr1[j])
}
}
}
end.sort((a,b) => a-b)
console.log(end)
return arr
};
The If conditional works but the else condition isn't and I can't figure out why.
I think console.log(end) should give me the two numbers not in arr2 but it instead gives me:
[
1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3,
3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 6,
6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 9, 9,
9, 9, 9, 19, 19, 19, 19, 19, 19
]
Why is this happening?
Thanks!!!
You could take an object for the position of a value and take a large value like Number.MAX_VALUE as default value. If the delta is zero sort by value.
Taking a delta is a standard by using Array#sort. This returns a value smaller than zero, zero or greater than zero, depending on the values. The sort method receives this values and keeps or swaps the values.
const
relativeSort = (array, given) => {
const order = Object.fromEntries(given.map((v, i) => [v, i + 1]));
return array.sort((a, b) =>
(order[a] || Number.MAX_VALUE) - (order[b] || Number.MAX_VALUE) ||
a - b
);
};
console.log(...relativeSort([2, 3, 1, 3, 2, 4, 6, 7, 9, 2, 19], [2, 1, 4, 3, 9, 6]));
In each iteration of arr2:
all the numbers that are different from the current number are pushed to the end array
For example,
First iteration - compare number (2) and you will end up with:
arr: [2,2,2]
end: [3,1,3,4,6,7,9,19]
Second iteration - compare number (1) and you will end up with:
arr: [2,2,2,1]
end: [3,1,3,4,6,7,9,19] + [2,3,3,2,4,6,7,9,2,19]
try to debug your code to follow the flow
class Solution:
def relativeSortArray(self, arr1: list[int], arr2: list[int]) -> list[int]:
arr = []
for i in arr2:
value = arr1.count(i)
for j in range(value):
arr.append(i)
arr1.remove(i)
arr1.sort()
return arr+arr1
obj = Solution()
arr1 = [28,6,22,8,44,17]
arr2 = [22,28,8,6]
result = obj.relativeSortArray(arr1,arr2)
print(result)
Related
I tried to make something that works as Set() using a couple tools that I learned. It worked, but I noticed a bug: it sorted my array! Can explain me someone why, please?
Here's my code:
function uniteUnique(...arr) {
let array = arr.flat()
let newArr = [];
console.log(array) // [ 1, 3, 2, 5, 2, 1, 4, 2, 1 ]
let myMap = array
.map(elem => {
if (!newArr.includes(elem))
return newArr.push(elem)
})
.filter(Boolean)
console.log(myMap) // [ 1, 2, 3, 4, 5 ]
}
uniteUnique([1, 3, 2], [5, 2, 1, 4], [2, 1]);
I know that for you might be too simple, but I ask so I can understand what is happening here.
console.log(myMap) // [ 1, 2, 3, 4, 5 ]
The result of your log is the number of pushed elements but accidentally, you thought they are sorted.
Also, if you return the mapped list you will end up with an array that contains an integer and boolean values. Instead of this, you need to return newArr.
Your code will be like this :
function uniteUnique(...arr) {
let flattedArray = arr.flat()
let set = [];
flattedArray.forEach(elem => !set.includes(elem) && set.push(elem))
return set
}
const set = uniteUnique([1, 3, 2], [5, 2, 1, 4], [2, 1]);
console.log(set)
in your code MyMap holds your newArr length as array.push returns the length of your array
so every time it returns the count:
for example if you tried to run this code
let newArr = []
console.log(newArr.push(20)) // the output is 1
and that's what your myMap holds => the length of your newArr
so if you want the filtered array you should use newArr
let array = [ 1, 3, 2, 5, 2, 1, 4, 2, 1 ]
let newArr = [];
let myMap = array.map(elem => {
if (!newArr.includes(elem))
return newArr.push(elem)
}).filter(Boolean)
console.log(newArr) //[1, 3, 2, 5, 4]
I couldn't find an answer to this specific question on S.O.
Let's say I have an array of strings, or in this case, numbers:
var x = [1, 1, 1, 2, 3, 3, 5, 3, 3, 5, 4, 5];
I'd like the output to be:
var output = [[1,1,1], [2], [3,3,3,3,3], [4], [5, 5, 5]];
I was hoping to use Lodash but most of that stuff tends to remove duplicates rather chunk them together into their own array. Maybe some kind of .map iterator?
The order of the output doesn't really matter so much. It just needs to chunk the duplicates into separate arrays that I'd like to keep.
You can use reduce to group the array elements into an object. Use Object.values to convert the object into an array.
var x = [1, 1, 1, 2, 3, 3, 5, 3, 3, 5, 4, 5];
var result = Object.values(x.reduce((c, v) => {
(c[v] = c[v] || []).push(v);
return c;
}, {}));
console.log(result);
Shorter version:
var x = [1, 1, 1, 2, 3, 3, 5, 3, 3, 5, 4, 5];
var result = Object.values(x.reduce((c, v) => ((c[v] = c[v] || []).push(v), c), {}));
console.log(result);
You can do this with Array.reduce in a concise way like this:
var x = [1, 1, 1, 2, 3, 3, 5, 3, 3, 5, 4, 5]
let result = x.reduce((r,c) => (r[c] = [...(r[c] || []), c],r), {})
console.log(Object.values(result))
The exact same with lodash would be:
var x = [1, 1, 1, 2, 3, 3, 5, 3, 3, 5, 4, 5]
let result = _.values(_.groupBy(x))
console.log(result)
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>
Using _.values to extract the values of the grouping object and _.groupBy to get the actual groupings
Use Array#prototype#reduce to group them:
const x = [1, 1, 1, 2, 3, 3, 5, 3, 3, 5, 4, 5];
let helperObj = {};
const res = x.reduce((acc, curr) => {
// If key is not present then add it
if (!helperObj[curr]) {
helperObj[curr] = curr;
acc.push([curr]);
}
// Else find the index and push it
else {
let index = acc.findIndex(x => x[0] === curr);
if (index !== -1) {
acc[index].push(curr);
}
}
return acc;
}, []);
console.log(res);
Since you're hoping to use Lodash, you might be interested in groupBy. It returns on object, but the _.values will give you the nested array:
var x = [1, 1, 1, 2, 3, 3, 5, 3, 3, 5, 4, 5];
let groups = _.values(_.groupBy(x))
console.log(groups)
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>
Here's an imperative solution:
var x = [1, 1, 1, 2, 3, 3, 5, 3, 3, 5, 4, 5];
x.sort();
var res = [];
for (const [i, n] of x.entries()) {
if (n !== x[i-1]) res.push([n]);
else res[res.length-1].push(n);
}
console.log(res);
I would like to interleave two arrays, BUT only return pairs when a certain condition is met. As an example:
first_array = [1, 2, 3, 4, 5, 6, 7, 8];
second_array = [, , , , 1, , 0, 1];
I need to return ONLY pairs where array-2 is non-null, in other words, the output I need is:
interleaved = [5, 1, 7, 0, 8, 1];
I have an interleave function that works:
function send_values() {
let interleaved = [];
for (let i = 0; i < first_array.length; i++) {
interleaved.push(first_array[i], second_array[i]);
}
}
...but the output is, obviously:
interleaved = [1, , 2, , 3, , 4, , 5, 1, 6, , 7, 0, 8, 1];
...which is not what I need. Suggestions?
You could iterate the sparse array and take only the values with the values at the same index from array one.
var array1 = [1, 2, 3, 4, 5, 6, 7, 8],
array2 = [, , , , 1, , 0, 1],
result = array2.reduce((r, v, i) => r.concat(array1[i], v), []);
console.log(result);
Here's a generic functional solution:
pairs = (a, b) => a.map((_, i) => [a[i], b[i]])
flat = a => [].concat(...a)
valid = x => x === 0 || Boolean(x)
array_1 = [1, 2, 3, 4, 5, 6, 7, 8];
array_2 = [ , , , , 1, , 0, 1];
result = flat(
pairs(array_1, array_2)
.filter(x => x.every(valid))
)
console.log(result)
Works both ways, that is, it doesn't matter which array contains the null values. Also, you can redefine valid if there are other things to exclude.
As a bonus, if you replace pairs with zip, you can make it work for any number of arrays.
I have an array, which contains duplicate values. How can I push duplicates in to another array?
let arr1 = [1, 5, 3, 6, 9, 5, 1, 4, 2, 7, 9], and duplicates array should be dupArr = [1, 5, 9]
You could filter the array by storing the previous checked values in a Set, which is here a closure.
var array = [1, 5, 3, 6, 9, 5, 1, 4, 2, 7, 9],
duplicates = array.filter((s => v => s.has(v) || !s.add(v))(new Set));
console.log(duplicates);
I was playing around with JavaScript's array prototype map and I don't understand why I'm getting the following values:
console.log(
[1,2,2,5,6].map(function(a,b){
return a+b;
})
); /*This returns [ 1, 3, 4, 8, 10 ]*/
Shouldn't the above code return [1,3,4,7,11] ?
Here's another snippet which shows that elements in the array are added sequentially and are correct at least I believe so.
console.log(
[1,2,3,4,5,6].map(function(a,b){
return a+b;
})
); /*[ 1, 3, 5, 7, 9, 11 ]*/
This is just a curious question more along the lines of why the first snippet of code seems.
It's because map passes you the value as the first parameter and the index as the second. So, what you're adding is: 1 + 0, 2+1, 2+2, 5+3, etc.
Your a value is the value out of the array: 1, 2, 2, 5, 6
Your b value is actually an index, not a value out of the array: 0, 1, 2, 3, 4
Let's sum:
[1, 2, 2, 5, 6] // Your values
+ [0, 1, 2, 3, 4] // Indices
-----------------
[1, 3, 4, 8,10] // Result
[1, 2, 3, 4, 5, 6] // Your values
+ [0, 1, 2, 3, 4, 5] // Indices
--------------------
[1, 3, 5, 7, 9,11] // Result
The results are correct.
I think you are confusing map with reduce:
var arr = [];
[1,2,2,5,6].reduce(function(a,b){
arr.push(a+b);
return b;
}, 0);
arr; // [1, 3, 4, 7, 11]
[0, 1, 2, 2, 5] // Values returned in previous iteration
+ [1, 2, 2, 5, 6] // Your values
-----------------
[1, 3, 4, 7,11] // Result
so the first parameter 'a' is value and the 'b' is index. so adding together it shows the corect value only. so 'a' contains [1, 2, 2, 5, 6] and 'b' contains [0, 1, 2, 3, 4]