I am practising on code wars and am currently stuck on a kyu 8 question, all the tests seem to pass bar the last one. I will add my code and the tests below plus the output I get below.
function sumArray(array) {
if (array == null || array.length <= 2) {
return 0
} else {
let largestInt = Math.max.apply(null, array)
let smallestInt = Math.min.apply(null, array)
let indexSmallest = array.indexOf(largestInt)
let indexLargest = array.indexOf(smallestInt)
array.splice(indexSmallest, 1)
array.splice(indexLargest, 1)
let sum = 0
for (let i = 0; i < array.length; i++) {
sum += array[I]
}
return sum
}
}
The tests:
const {
assert
} = require("chai");
it("example tests", () => {
assert.strictEqual(sumArray(null), 0);
assert.strictEqual(sumArray([]), 0);
assert.strictEqual(sumArray([3]), 0);
assert.strictEqual(sumArray([3, 5]), 0);
assert.strictEqual(sumArray([6, 2, 1, 8, 10]), 16);
assert.strictEqual(sumArray([0, 1, 6, 10, 10]), 17);
assert.strictEqual(sumArray([-6, -20, -1, -10, -12]), -28);
assert.strictEqual(sumArray([-6, 20, -1, 10, -13]), 3);
});
The output:
Test Results:
example tests
expected -10 to equal 3
function total(array) {
// always assure at least an empty array.
array = Array.from(array ?? []);
// sort array values ascending.
array.sort((a, b) => a - b);
array.pop(); // remove last/higest value.
array.shift(); // remove first/lowest value.
// for any to be reduced/summed-up (empty) array
// the initial value of zero always assures the
// minimum expected result of zero.
return array
.reduce((total, value) => total + value, 0);
}
const testEntries = [
[ null, 0 ],
[ [ ], 0 ],
[ [ 3 ], 0 ],
[ [ 3, 5 ], 0 ],
[ [ 6, 2, 1, 8, 10 ], 16 ],
[ [ 0, 1, 6, 10, 10 ], 17 ],
[ [ -6, -20, -1, -10, -12 ], -28 ],
[ [ -6, 20, -1, 10, -13 ], 3 ],
];
console.log(
testEntries
.map(([value, result]) =>
`(total([${ value }]) === ${ result }) ... ${ total(value) === result }`
)
);
.as-console-wrapper { min-height: 100%!important; top: 0; }
I'm having some issues translating a depth list into a nested object.
For example, i have a list like so:
"depth": [ 0, 1, 2, 3, 3, 2, 3, 3, 3 ],
And i need to come up with a recursive function to produce an object like this:
"depth": [
{
"type": 0,
"children": [
{
"type": 1,
"children": [
{
"type": 2,
"children":[
{ "type": 3, "children": []},
{ "type": 3, "children": []},
]
},
{
"type:": 2,
"children":[
{ "type": 3, "children": []},
{ "type": 3, "children": []},
{ "type": 3, "children": []},
]
}
]
}
]
}
]
}
So the rule here, the lower numbers are parents, the higher numbers are siblings to the previous lower number.
So far what i have come up with is:
const depth = [0, 1, 2, 3, 3, 2, 3, 3, 3]
// Start looping through all the numbers in the depth
for (let i = 0; i < depth.length; i++) { //depth.length
// As i loop i want to only look at the array i have explored
// So that i can find the parent that is 1 lower in the array
let getParent = depth.slice(0, i).lastIndexOf(depth[i] - 1) // position of last lower array
// Here i check if the current depth item is bigger than the currently lower Item in the array
if (depth[i] > depth[getParent]) {
console.log(depth[i] + " Nesting into " + depth[getParent]) // Is child of that item
}
}
I think this successfully maps the child to the parent. But now I'm stuck on a way to produce my desired result.
If anyone has an suggestions I would appreciate it a lot.
Thanks
It's easier to create a tree when using an Map or a dictionary (object) to retrieve parents. After reducing the entire tree to a Map, retrieve the root, which holds the entire tree.
const createItem = (type) => ({ type, children: [] })
const fn = (arr, root) => arr.reduce((acc, d) => {
const parent = d - 1
const item = createItem(d)
if(d !== root && !acc.has(parent)) acc.set(parent, createItem(parent))
if(d !== root) acc.get(parent).children.push(item)
return acc.set(d, item);
}, new Map()).get(root);
const depth = [0, 1, 2, 3, 3, 2, 3, 3, 3 ]
const result = fn(depth, 0)
console.log(result)
As you get the data in pre-order sequence, you don't actually need a map:
function convert(data) {
let i = 0;
function getChildren(type) {
let children = [];
while (data[i] === type) {
i++;
children.push({
type,
children: getChildren(type + 1)
});
}
return children;
}
return getChildren(0);
}
let data = [0, 1, 2, 3, 3, 2, 3, 3, 3];
console.log(convert(data));
I have an array:
array = {
"data": [
{ "value": [ 100, 13, 16 ] },
{ "value": [ 101, 14, 17 ] },
{ "value": [ 12, 15, 18 ] }
]
}
Which I am reformatting into a new array of just the columns:
const columnArray = jsonData.map( (current, index, arr) => {
let out = [];
for( let i = 0; i < current.value.length; i++ ) {
out.push( arr[ i ].value[ index ] );
}
return out;
});
// output
[
[ 100, 101, 12 ],
[ 13, 14, 15 ],
[ 16, 17, 18 ]
]
How would I re-write the columnArray mapping to do the column array and be able to sum from the previous value?
So the intended output from the original array would be:
[
[ 100, 201, 213 ],
[ 13, 27, 42 ],
[ 16, 33, 51 ]
]
I would also like the summing to be scalable (though it will always be in a 1:1 ratio). So if the data has 20 items, then each value will have 20 integers in that array too.
I have tried looping through but that didn't work as I only sum from the previous, not all the previous. And this wouldn't scale either:
const columnArray = jsonData.map( (current, index, arr) => {
let out = [];
for( let i = 0; i < current.value.length; i++ ) {
// dont touch first
if( i < 1 ) {
out.push( arr[ i ].value[ index ] );
} else {
out.push( arr[ i ].value[ index ] + arr[ i - 1 ].value[ index ] )
}
}
return out;
});
Instead of pushing the array element, add it to a variable accumulating the running totals, and push that.
const jsonData = [{
"value": [100, 13, 16]
},
{
"value": [101, 14, 17]
},
{
"value": [12, 15, 18]
}
];
const columnArray = jsonData.map((current, index, arr) => {
let out = [];
let total = 0;
for (let i = 0; i < current.value.length; i++) {
total += arr[i].value[index]
out.push(total);
}
return out;
});
console.log(columnArray);
or with a nested map():
const jsonData = [{
"value": [100, 13, 16]
},
{
"value": [101, 14, 17]
},
{
"value": [12, 15, 18]
}
];
const columnArray = jsonData.map((current, index, arr) => {
let total = 0;
return arr.map(el => total += el.value[index])
});
console.log(columnArray);
You're thinking this in the wrong way. You're storing the sum in the list, not anywhere else. So even tho your index is increasing, the resulting sum resides in the list, so to achieve your goal you have to save it in some variable then push the variable into the final list. Follow this code below:
const columnArray = array.data.map((current, index, arr) => {
let out = [];
let temp;
for (let i = 0; i < current.value.length; i++) {
// dont touch first
if (i < 1) {
temp = arr[i].value[index];
out.push(arr[i].value[index]);
} else {
temp = arr[i].value[index] + temp;
out.push(temp);
}
}
return out;
});
something like that...
const array0 = {
"data": [
{ "value": [ 100, 13, 16 ] },
{ "value": [ 101, 14, 17 ] },
{ "value": [ 12, 15, 18 ] }
]
}
const
rowCount = array0.data.reduce((c,{value})=>Math.max(c,value.length) ,0)
, arrResult = Array(rowCount).fill(0).map(x=>Array(array0.data.length).fill(0))
;
arrResult.forEach((_,i,arr)=>
{
array0.data[i].value.forEach((v,j)=>
{
arr[j][i] = v + (i? arr[j][i-1] : 0 )
})
})
console.log( arrResult)
.as-console-wrapper {max-height: 100%!important;top:0}
I need to check if an array contains another array. The order of the subarray is important but the actual offset it not important. It looks something like this:
var master = [12, 44, 22, 66, 222, 777, 22, 22, 22, 6, 77, 3];
var sub = [777, 22, 22];
So I want to know if master contains sub something like:
if(master.arrayContains(sub) > -1){
//Do awesome stuff
}
So how can this be done in an elegant/efficient way?
With a little help from fromIndex parameter
This solution features a closure over the index for starting the position for searching the element if the array. If the element of the sub array is found, the search for the next element starts with an incremented index.
function hasSubArray(master, sub) {
return sub.every((i => v => i = master.indexOf(v, i) + 1)(0));
}
var array = [12, 44, 22, 66, 222, 777, 22, 22, 22, 6, 77, 3];
console.log(hasSubArray(array, [777, 22, 22]));
console.log(hasSubArray(array, [777, 22, 3]));
console.log(hasSubArray(array, [777, 777, 777]));
console.log(hasSubArray(array, [42]));
var master = [12, 44, 22, 66, 222, 777, 22, 22, 22, 6, 77, 3];
var sub = [777, 22, 22];
console.log(master.join(',').includes(sub.join(',')))
//true
You can do this by simple console.log(master.join(',').includes(sub.join(','))) this line of code using include method
The simplest way to match subset/sub-array
const master = [12, 44, 22, 66, 222, 777, 22, 22, 22, 6, 77, 3];
const sub1 = [777, 44, 222];
const sub2 = [777, 18, 66];
sub1.every(el => master.includes(el)); // reture true
sub2.every(el => master.includes(el)); // return false
Just came up with quick thought , but efficiency depends on size of the array
var master = [12, 44, 22, 66, 222, 777, 22, 22, 22, 6, 77, 3];
var sub = [777, 22, 22];
if ((master.toString()).indexOf(sub.toString()) > -1 ){
//body here
}
It’s surprising how often this is implemented incorrectly.
What we’re looking for is a substring in the mathematical sense.
In mathematics, a sequence is an enumerated collection of objects in which repetitions are allowed and order matters.
In mathematics, a subsequence of a given sequence is a sequence that can be derived from the given sequence by deleting some or no elements without changing the order of the remaining elements.
A subsequence which consists of a consecutive run of elements from the original sequence, such as ⟨ B, C, D ⟩ from ⟨ A, B, C, D, E, F ⟩ is a substring.
Note that a “string”, here, can consist of any element and is not limited to Unicode code-point sequences.
Effectively all previous answers have one of many possible flaws:
The string concatenation approach (array1.toString().includes(array2.toString())) fails when your array elements have commas. (Example: [ "a", "b" ] does not contain [ "a,b" ]).
Some implementations check beyond array bounds. (Example: [ "3" ] does not contain [ "3", undefined ], just because array[1] reports undefined for both).
Some implementations fail to handle repetition correctly.
Some implementations aren’t checking for substrings (in the mathematical sense) correctly, but for subsets or subsequences or something else.
Some implementations don’t account for the empty array. The empty string is the substring of every string.
Check if an array constitutes a “substring” of another array
Right off the bat, this handles the empty array correctly.
Then, it builds a list of candidate starting indexes by matching against the first element of the potential subarray.
Find the first candidate where every element of the slice matches index by index with the full array, offset by the candidate starting index.
The checked index also has to exist within the full array, hence Object.hasOwn.
const isSubArray = (full, slice) => {
if(slice.length === 0){
return true;
}
const candidateIndexes = full
.map((element, fullIndex) => ({
matched: element === slice[0],
fullIndex
}))
.filter(({ matched }) => matched),
found = candidateIndexes
.find(({ fullIndex }) => slice.every((element, sliceIndex) => Object.hasOwn(full, fullIndex + sliceIndex) && element === full[fullIndex + sliceIndex]));
return Boolean(found);
};
console.log(isSubArray([], []) === true);
console.log(isSubArray([ 0 ], []) === true);
console.log(isSubArray([ 0, 1, 2 ], [ 1, 2 ]) === true);
console.log(isSubArray([ 0, 1, 1, 2 ], [ 0, 1, 2 ]) === false);
console.log(isSubArray([ 2, 1 ], [ 1, 2 ]) === false);
console.log(isSubArray([ 1, 2, 3 ], [ 2, 3, undefined ]) === false);
console.log(isSubArray([ 0, 1, 1, 2, 3 ], [ 1, 1, 2 ]) === true);
console.log(isSubArray([ 0, 1, 1, 2, 3 ], [ 1, 2 ]) === true);
console.log(isSubArray([ 0, 1, 1, 2, 3 ], [ 0, 1, 1, 1 ]) === false);
console.log(isSubArray([ "a", "b" ], [ "a,b" ]) === false);
.as-console-wrapper { max-height: 100% !important; top: 0; }
This has quadratic complexity, yes.
There might be more efficient implementations using Trees or Ropes.
You might also want to research some efficient substring search algorithms and try to apply them to this problem.
Get the index of the found “substring”, or -1 if not found
It’s basically the same code, but with return true; replaced by return 0;, and return Boolean(found); replaced by return found?.fullIndex ?? -1;.
const findSubArrayIndex = (full, slice) => {
if(slice.length === 0){
return 0;
}
const candidateIndexes = full
.map((element, fullIndex) => ({
matched: element === slice[0],
fullIndex
}))
.filter(({ matched }) => matched),
found = candidateIndexes
.find(({ fullIndex }) => slice.every((element, sliceIndex) => Object.hasOwn(full, fullIndex + sliceIndex) && element === full[fullIndex + sliceIndex]));
return found?.fullIndex ?? -1;
};
console.log(findSubArrayIndex([], []) === 0);
console.log(findSubArrayIndex([ 0 ], []) === 0);
console.log(findSubArrayIndex([ 0, 1, 2 ], [ 1, 2 ]) === 1);
console.log(findSubArrayIndex([ 0, 1, 1, 2 ], [ 0, 1, 2 ]) === -1);
console.log(findSubArrayIndex([ 2, 1 ], [ 1, 2 ]) === -1);
console.log(findSubArrayIndex([ 1, 2, 3 ], [ 2, 3, undefined ]) === -1);
console.log(findSubArrayIndex([ 0, 1, 1, 2, 3 ], [ 1, 1, 2 ]) === 1);
console.log(findSubArrayIndex([ 0, 1, 1, 2, 3 ], [ 1, 2 ]) === 2);
console.log(findSubArrayIndex([ 0, 1, 1, 2, 3 ], [ 0, 1, 1, 1 ]) === -1);
console.log(findSubArrayIndex([ "a", "b" ], [ "a,b" ]) === -1);
.as-console-wrapper { max-height: 100% !important; top: 0; }
Semi-acceptable alternative: JSON
JSON-encoding both arrays might be a viable strategy as well.
Here, the surrounding […] of the potential subarray need to be removed, then an includes will tell you if the JSON string is included in the other JSON string.
This works — as opposed to the simple string concatenation or join approach — because JSON has delimiters that cannot appear verbatim in the encoded elements; if they do appear in the original elements, they’d be correctly escaped.
The caveat is that this won’t work for values that are not encodable in JSON.
const isSubArray = (full, slice) => JSON.stringify(full)
.includes(JSON.stringify(slice).replaceAll(/^\[|\]$/g, ""));
console.log(isSubArray([], []) === true);
console.log(isSubArray([ 0 ], []) === true);
console.log(isSubArray([ 0, 1, 2 ], [ 1, 2 ]) === true);
console.log(isSubArray([ 0, 1, 1, 2 ], [ 0, 1, 2 ]) === false);
console.log(isSubArray([ 2, 1 ], [ 1, 2 ]) === false);
console.log(isSubArray([ 1, 2, 3 ], [ 2, 3, undefined ]) === false);
console.log(isSubArray([ 0, 1, 1, 2, 3 ], [ 1, 1, 2 ]) === true);
console.log(isSubArray([ 0, 1, 1, 2, 3 ], [ 1, 2 ]) === true);
console.log(isSubArray([ 0, 1, 1, 2, 3 ], [ 0, 1, 1, 1 ]) === false);
console.log(isSubArray([ "a", "b" ], [ "a,b" ]) === false);
.as-console-wrapper { max-height: 100% !important; top: 0; }
If the order is important, it has to be an actually sub-array (and not the subset of array) and if the values are strictly integers then try this
console.log ( master.join(",").indexOf( subarray.join( "," ) ) == -1 )
for checking only values check this fiddle (uses no third party libraries)
var master = [12, 44, 22, 66, 222, 777, 22, 22, 22, 6, 77, 3];
var sub = [777, 22, 22];
function isSubset( arr1, arr2 )
{
for (var i=0; i<arr2.length; i++)
{
if ( arr1.indexOf( arr2[i] ) == -1 )
{
return false;
}
}
return true;
}
console.log( isSubset( master, sub ) );
There are faster options explained here as well.
EDIT
Misunderstood question initially.
function arrayContainsSub(arr, sub) {
var first = sub[0],
i = 0,
starts = [];
while (arr.indexOf(first, i) >= 0) {
starts.push(arr.indexOf(first, i));
i = arr.indexOf(first, i) + 1;
}
return !!starts
.map(function(start) {
for (var i = start, j = 0; j < sub.length; i++, j++) {
if (arr[i] !== sub[j]) {
return false;
}
if (j === sub.length - 1 && arr[i] === sub[j]) {
return true;
}
};
}).filter(function(res) {
return res;
}).length;
}
This solution will recursively check all available start points, so points where the first index of the sub has a match in the array
Old Answer Kept in case useful for someone searching.
if(master.indexOf(sub) > -1){
//Do awesome stuff
}
Important to remember that this will only match of master literally references sub. If it just contains an array with the same contents, but references a different specific object, it will not match.
You can try with filter and indexOf like this:
Note: This code works in case we do not cover the order in sub array.
Array.prototype.arrayContains = function (sub) {
var self = this;
var result = sub.filter(function(item) {
return self.indexOf(item) > -1;
});
return sub.length === result.length;
}
Example here.
UPDATED: Return index of sub array inside master (cover order in sub array)
Array.prototype.arrayContains = function(sub) {
var first;
var prev;
for (var i = 0; i < sub.length; i++) {
var current = this.indexOf(sub[i]);
if (current > -1) {
if (i === 0) {
first = prev = current;
continue;
} else {
if (++prev === current) {
continue;
} else {
return -1;
}
}
} else {
return -1;
}
}
return first;
}
Demo: here
For this answer, I am preserving the order of sub-array. Means, the elements of sub-array should be in Consecutive order. If there is any extra element while comparing with the master, it will be false.
I am doing it in 3 steps:
Find the index of the first element of sub in the master and store it an array matched_index[].
for each entry in matched_index[] check if each element of sub is same as master starting from the s_index. If it doesn't match then return false and break the for loop of sub and start next for-loop for next element in matched_index[]
At any point, if the same sub array is found in master, the loop will break and return true.
function hasSubArray(master,sub){
//collect all master indexes matching first element of sub-array
let matched_index = []
let start_index = master.indexOf(master.find(e=>e==sub[0]))
while(master.indexOf(sub[0], start_index)>0){
matched_index.push(start_index)
let index = master.indexOf(sub[0], start_index)
start_index = index+1
}
let has_array //flag
for(let [i,s_index] of matched_index.entries()){
for(let [j,element] of sub.entries()){
if(element != master[j+s_index]) {
has_array = false
break
}else has_array = true
}
if (has_array) break
}
return has_array
}
var master = [12, 44, 22, 66, 222, 777, 22, 22, 22, 6, 77, 3];
console.log(hasSubArray(master, [777, 22, 22]));
console.log(hasSubArray(master, [777, 22, 3]));
console.log(hasSubArray(master, [777, 777, 777]));
console.log(hasSubArray(master, [44]));
console.log(hasSubArray(master, [22, 66]));
I had a similar problem and resolved it using sets.
function _hasSubArray( mainArray, subArray )
{
mainArray = new Set( mainArray );
subArray = new Set( subArray );
for ( var element of subArray )
{
if ( !mainArray.has( element ) )
{
return false;
}
}
return true;
}
If run this snippet below it should work
x = [34, 2, 4];
y = [2, 4];
y.reduce((included, num) => included && x.includes(num), true);
EDIT:
#AlexanderGromnitsky You are right this code is incorrect and thank you for the catch! The above code doesn't actually do what the op asked for. I didn't read the question close enough and this code ignores order. One year later here is what I came up with and hopefully this may help someone.
var master = [12, 44, 22, 66, 222, 777, 22, 22, 22, 6, 77, 3];
var sub = [777, 22, 22];
var is_ordered_subset = master.join('|').includes(sub.join('|'))
This code is somewhat elegant and does what op asks for. The separator doesn't matter as long as its not an int.
async function findSelector(a: Uint8Array, selector: number[]): Promise<number> {
let i = 0;
let j = 0;
while (i < a.length) {
if (a[i] === selector[j]) {
j++;
if (j === selector.length) {
return i - j + 1;
}
} else {
j = 0;
}
i++;
}
return -1;
}
Try using every and indexOf
var mainArr = [1, 2, 3, 4, 5]
var subArr = [1, 2, 3]
function isSubArray(main, sub) {
return sub.every((eachEle) => {
return (main.indexOf(eachEle) + 1);
});
}
isSubArray(mainArr, subArr);