I'm currently building a tic tac toe in vanilla javascript. However the game is 'sort of' done but I'm trying to add levels of difficulty. So basically the thing I want to do is , on every player move , to get the the closest possible winning combination based on his moves and place computer's mark into the missing winning's combinations place.
Let's say I have multidimensional array with the winning combinations
winningCombinations: [
[0, 1, 2],
[3, 4, 5],
[6, 7, 8],
[0, 4, 8],
[0, 3, 6],
[1, 4, 7],
[2, 4, 6],
[2, 5, 8]
]
And the player X starts his moves. So his first move is 0, so saving player's current combination in array. So in first move the current comb is
currentPlayerCombintation: [0]
so I want to return [0,1,2], [0,4,8] and [0,3,6] from the winning combination's array.
However the player makes his second move , so he target's 4 so
currentPlayerCombination: [0,4]
and now I want to return the closest possible winning combination which is [0,4,8].
I've tried a lot of things including every() , some() , filter() but could not achieve the thing I want.
I've tried sort of
for(let i = 0; i < this.currentPlayerCombination.length ; i++) {
this.winningCombinations.some((arr) => {
if(arr.includes(this.currentPlayerCombination[i])) {
console.log(arr);
}
});
}
But this didnt work as expected :(
You could take a Set and map the count of the matching items, get the max count and filter the array.
function getWinningPositions(pos) {
var posS = new Set(pos),
temp = winningCombinations.map(a => [a, a.reduce((c, v) => c + posS.has(v), 0)]),
max = Math.max(...temp.map(({ 1: c }) => c))
return temp
.filter(({ 1: c }) => c === max)
.map(([a]) => a);
}
var winningCombinations = [[0, 1, 2], [3, 4, 5], [6, 7, 8], [0, 4, 8], [0, 3, 6], [1, 4, 7], [2, 4, 6], [2, 5, 8]];
console.log(getWinningPositions([0]).map(a => a.join(' ')));
console.log(getWinningPositions([0, 4]).map(a => a.join(' ')));
console.log(getWinningPositions([0, 4, 5]).map(a => a.join(' ')));
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First map the winningCombinations to an array of arrays whose numbers are only the numbers that have not been picked yet. Then, find the lowest length of those arrays, and you can identify the original winningCombinations which are closest to the currentPlayerCombination:
const winningCombinations = [
[0, 1, 2],
[3, 4, 5],
[6, 7, 8],
[0, 4, 8],
[0, 3, 6],
[1, 4, 7],
[2, 4, 6],
[2, 5, 8]
];
const currentPlayerCombination = [0, 4];
// eg: get [[1, 2], [3, 5,], [6, 7, 8], [8], ...]
const winningCombsWithoutCurrent = winningCombinations.map(arr => (
arr.filter(num => !currentPlayerCombination.includes(num))
));
// eg: here, lowestLength should be 1, because [8] has a length of 1
const lowestLength = winningCombsWithoutCurrent.reduce((a, { length }) => Math.min(a, length), 3);
const combosWithLowestLength = winningCombsWithoutCurrent
.reduce((a, { length }, i) => {
if (length === lowestLength) {
a.push(winningCombinations[i]);
}
return a;
}, []);
console.log(combosWithLowestLength);
Related
Can anyone tell me how to solv this problem please:
I tried doing this with array.map, array.filter, array.reduce but i did not got result:
Write a function putNum(arrayOfNum: number[], num: number),
which would find all possible combinations of numbers from arrayOfNum,
whose sum is equal to number. Wherein:
arrayOfNum contains only unique positive numbers (>0)
there should not be repetitions of numbers in the combination
all combinations must be unique
#param arrayOfNum: number[]
#param num: number[]
#return Array<Array<number>>
function putNum(arrayOfNum, num) {
***// write code only inside this function***
return [[1, 2], [3]];
}
// console.log(putNum([8, 2, 3, 4, 6, 7, 1], 99)); => []
// console.log(putNum([8, 2, 3, 4, 6, 7, 1], 5)); => [[2, 3], [4, 1]]
// console.log(putNum([1, 2, 3, 4, 5, 6, 7, 8], 8)); => [[1, 3, 4], [1, 2, 5], [3, 5], [2, 6], [1, 7], [8]]
let resultnum = result.filter(e => typeof e === 'number' && e > 0); // to make a new array with nums > 0
The best approach to solve this problem in optimized way is to use hash map
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([10,20,40,50,60,70,30],50));
Output:
[ [ 10, 40 ], [ 20, 30 ] ]
I have been trying to accomplish this since yesterday, though no luck yet. I have found solutions where there always is a slight difference in what I want to accomplish.
I am trying to get all possible combinations, slightly like this: combination_k, but I also want the same items to pair up with itself, so given the following:
input [1, 4, 5] and 2 (number of combinations) should return:
[1, 1], [1, 4], [1, 5], [4, 4], [4, 5], [5, 5]
input [1, 4, 5] and 3 should return:
[1, 1, 1], [1, 1, 4], [1, 1, 5], [1, 4, 4], [1, 4, 5], [4, 4, 4], [4, 4, 5], [5, 5, 5], [5, 5, 4], [5, 5, 1] (The order is not important).
I have been adjusting combination_k, it got me far enough that it worked with 2 but it didn't work when I provided 3 as a parameter.
const combinations = getAllCombinations([1, 4, 5], 2);
// combinations = [1, 1], [1, 4], [1, 5], [4, 4], [4, 5], [5, 5]
Any tips are welcome!
The problem is commonly referred to as k-combinations with repetitions.
Here's a solution that relies on recursion to get the desired result:
const combinations = (array, r) => {
const result = [];
const fn = (array, selected, c, r, start, end) => {
if (c == r) {
result.push([...selected]);
return;
}
for (let i = start; i <= end; i++) {
selected[c] = array[i];
fn(array, selected, c + 1, r, i, end);
}
}
fn(array, [], 0, r, 0, array.length - 1);
return result;
}
console.log(combinations([1, 4, 5], 3));
A modified version of the code you provided:
function getAllCombinations(arr, n) {
if (n <= 0) return [];
if (n === 1) return [...arr];
return arr.reduce((acc, cur, i) => {
const head = arr.slice(i, i + 1);
const combinations = getAllCombinations(arr.slice(i), n - 1)
.map(x => head.concat(x));
return [...acc, ...combinations];
}, []);
}
console.log(getAllCombinations([1, 4, 5], 2).join('|'));
console.log(getAllCombinations([1, 4, 5], 3).join('|'));
If I have one array:
let x= [0,1,2,3,5]
And I have an array with several subarrays:
let winningIndices = [[0, 1, 2], [6, 7, 8], [0, 3, 6], [2, 5, 8], [0, 4, 8], [2, 4, 6]]
How can I check if array x contains all of the elements of any one subarray.
In other words, how can I check if array x has combinations of either the numbers 0,1,2 or 6,7,8...
Thanks in advance
"How can I check if array x contains all of the elements of any one subarray."
Here's the most straightforward functional interpretation.
const won = winningIndices.some(indices=>
indices.every(
item=>x.includes(item)
)
)
This could be an option. If you'd like to remove xIncludesItem key from result, just map it again.
const winningIndices = [
[0, 1, 2],
[6, 7, 8],
[0, 3, 6],
[2, 5, 8],
[0, 4, 8],
[2, 4, 6]
];
const x = [0, 1, 2, 3, 5];
const result = winningIndices
.map((item) => {
const xIncludesItem = item
.map((i) => x.includes(i))
.every(includes => includes);
return { item, xIncludesItem };
})
.filter(result => result.xIncludesItem);
console.log(result);
Array.includes is much slower than Set.has. A slightly more performant solution based on Ted Brownlow's solution would be:
const setX = new Set(x);
const won = winningIndices.some(indices=>
indices.every(
item=>setX.has(item)
)
)
I have a multidimensional array like below and I want to shift column positions using javascript or ES6 with no jquery or any other plugins.
Eg: Initial array will look like this.
1|2|3|4
2|2|6|4
4|2|3|4
9|2|7|4
How can I shift the 4th column to 1st position so that it will look like this?
4|1|2|3
4|2|2|6
4|4|2|3
4|9|2|7
Could someone can help with logic to shift any columns like this?
You could assing a mapped outer array with new items by slicing the inner arrays with a given index.
For getting the original sort, you could shiftby the delta of length and index.
const shift = (array, index) => array.map(a => [...a.slice(index), ...a.slice(0, index)]);
var array = [[1, 2, 3, 4], [2, 2, 6, 4], [4, 2, 3, 4], [9, 2, 7, 4]],
index = 3;
array = shift(array, index);
console.log(array.map(a => a.join(' ')));
array = shift(array, array[0].length - index);
console.log(array.map(a => a.join(' ')));
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You can use array.map to re-arrange the values:
function rearrange(rows, pos) {
return rows.map(function(cols) {
return pos.map(function(i) {
return cols[i];
});
});
}
var old_arr;
var new_arr;
old_arr = [
[1, 2, 3, 4],
[2, 2, 6, 4],
[4, 2, 3, 4],
[9, 2, 7, 4]
];
new_arr = rearrange(old_arr, [3, 0, 1, 2]);
console.log(new_arr);
old_arr = [
[1, 2, 3, 4],
[2, 2, 6, 4],
[4, 2, 3, 4],
[9, 2, 7, 4]
];
new_arr = rearrange(old_arr, [3, 2, 1, 0]);
console.log(new_arr);
I am trying to solve this freecodecamp algorithm question where I had to collect the difference of two or more arrays. I used map to get the difference of array but the problem is I only get two elements;
function sym(args) {
args = [].slice.call(arguments);
var newArr = args.map(function(el, index, arr){
console.log(arr.indexOf(arr[index]));
if(arr.indexOf(arr[index] === -1 )){
// console.log(arr[index]);
return args.push(arr[index]);
}
});
return newArr; // my newArr returns [3, 4] instead of [3,4,5]
}
console.log(sym([1, 2, 3], [5, 2, 1, 4]));
//sym([1, 2, 3], [5, 2, 1, 4]) should return [3, 4, 5]
//sym([3, 3, 3, 2, 5], [2, 1, 5, 7], [3, 4, 6, 6], [1, 2, 3], [5, 3, 9, 8], [1]) should return [1, 2, 4, 5, 6, 7, 8, 9]
I think we could do also this way since we want them to be ordered at the end.
For more detail about the original problem please consult this link: FreecodeCamp Link: Symmetric Difference
const sym = (...args) => {
// Merge all the different arrays and remove duplicate elements it means elements that are present both on two related arrays
let tab = args.reduce((a, b) => [
...a.filter(i => !b.includes(i)),
...b.filter(j => !a.includes(j))
], []);
// Then remove the rest of duplicated values and sort the obtained array
return Array.from(new Set(tab)).sort((a, b) => a - b);
}
console.log(sym([1, 2, 3, 3], [5, 2, 1, 4])); // [3, 4, 5]
console.log(sym([1, 1, 2, 5], [2, 2, 3, 5], [3, 4, 5, 5])); // [1, 4, 5]
console.log(sym([3, 3, 3, 2, 5], [2, 1, 5, 7], [3, 4, 6, 6], [1, 2, 3], [5, 3, 9, 8], [1])); // [1, 2, 4, 5, 6, 7, 8, 9]
The Set data structure is used here to remove duplicated values thanks to its characteristics.
Well your function is a little more complex than only selecting the unique values, cause you want to filter them out... and also accept multiple arrays. This should work.
var sym = (...arrays)=>{
//Concat Items
const allItems = arrays.reduce((a,c)=>a.concat(c), []);
// Identify repeated items
const repeatedItems = allItems.filter((v,i,a)=>a.indexOf(v) !== i);
// Filter repeated items out
const diff = allItems.filter(item=>repeatedItems.indexOf(item) < 0);
console.log(diff);
};
sym([1, 2, 3], [5, 2, 1, 4]); // [3,5,4]
I don't think your approach will work; you're supposed to create an array with elementos from both arrays, so a single .map won't do the job. Filtering through both arrays should work, although it will probably leave enough room for optimization.
my newArr returns [3, 4] instead of [3,4,5]
You are using map which will only return one value per iteration (which is why you are getting only 2 values) and in your case you are checking if the index is found or not (not the item)
You need to concatenate all the arrays and then remove those which are repeated
Concatenate
var newArr = args.reduce( ( a, c ) => a.concat( c ) , []);
Create a map by number of occurrences
var map = newArr.reduce( (a,c) => ( a[c] = (a[c] || 0) + 1, a ) , {});
Iterate and filter through those keys whose value is 1
var output = Object.keys( map ).filter( s => map[s] === 1 ).map( Number );
Demo
function sym(args)
{
args = [].slice.call(arguments);
var newArr = args.reduce( ( a, c ) => a.concat( c ) , []);
var map = newArr.reduce( (a,c) => ( a[c] = (a[c] || 0) + 1, a ) , {});
return Object.keys( map ).filter( s => map[s] === 1 ).map( Number );
}
console.log(sym([1, 2, 3], [5, 2, 1, 4]));
You could take an Object for counting the items and return only the items which have a count.
function sym(array) {
return array.reduce((a, b) => {
var count = {};
a.forEach(v => count[v] = (count[v] || 0) + 1);
b.forEach(v => count[v] = (count[v] || 0) - 1);
return Object.keys(count).map(Number).filter(k => count[k]);
});
}
console.log(sym([[3, 3, 3, 2, 5], [2, 1, 5, 7], [3, 4, 6, 6], [1, 2, 3], [5, 3, 9, 8], [1]]));