How to compare an array to an array of arrays?

Question

This is an attempt in a tic tac toe game app. I have two arrays playerMoves and winningCombinations. Like this.

var playerMoves= [0,1,4];
var winningCombinations = [
        [0,1,2],[3,4,5],[6,7,8],
        [0,3,6],[1,4,7],[2,5,8],
        [0,4,8],[2,4,6]
      ];

I need to filter the winningCombination array such that at-least and at-most two values of playerMoves array matches with each array in winningCombination.

findPossibleMove(playerMoves);
// should return [[0,1,2],[1,4,7], [0,4,8] ]

My attempt

function findPossibleMove(arr){
  var found = 0;
  return arr.forEach((item)=>{
    winningCombinations.map((obj)=>{
      if(obj.indexOf(item) !== -1) {
        found++;
      }
      if(found===2){
        return obj;
      }        
    })
  })      
}

kind user · Accepted Answer · 2017-07-18 12:05:01Z

12

Three simple steps:

Use indexOf function to check, if specified element from the subarray of winningCombinations array is present in the playerMoves array.
If so - filter it out with Array#filter function.
If the returned, filtered subarray has length equal to 2, it means that two (no more, nor less) elements have appeared - it fulfills our condition - filter it once again with yet another Array#filter.

let playerMoves = [0, 1, 4];
let winningCombinations = [
  [0, 1, 2],
  [3, 4, 5],
  [6, 7, 8],
  [0, 3, 6],
  [1, 4, 7],
  [2, 5, 8],
  [0, 4, 8],
  [2, 4, 6],
];

let res = winningCombinations.filter(v => v.filter(c => {
  return playerMoves.indexOf(c) > -1;
}).length == 2);
  
  console.log(JSON.stringify(res));

edited Jul 18, 2017 at 12:05

answered Jul 17, 2017 at 19:24

kind user

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1 Comment

Tim Bogdanov Over a year ago

how would you go about instead of filtering out the none matching once, replacing them. so if i have [['WA', 0], [OR, 0] and the other array is [['WA', 1], [OR, 0], I want WA in the first array to be replaces with the WA value in the second array.

Alberto Trindade Tavares · Accepted Answer · 2017-07-17 19:30:11Z

3

You can use filter and includes to achieve that:

var playerMoves= [0,1,4];
var winningCombinations = [
  [0,1,2],[3,4,5],[6,7,8],
  [0,3,6],[1,4,7],[2,5,8],
  [0,4,8],[2,4,6]
];

var filteredCombinations = winningCombinations.filter((combination) =>
  combination.filter(x => playerMoves.includes(x)).length === 2);

console.log(filteredCombinations);

answered Jul 17, 2017 at 19:30

Alberto Trindade Tavares

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Comments

Koushik Chatterjee · Accepted Answer · 2017-07-17 20:10:36Z

1

since we have to check with the length (matched item) in each filtered array, how about skipping creation of filtered array against array and reducing it to a number of matched element and check directly with that instead of the length?

let playerMoves = [0, 1, 4];
let winningCombinations = [
  [0, 1, 2],
  [3, 4, 5],
  [6, 7, 8],
  [0, 3, 6],
  [1, 4, 7],
  [2, 5, 8],
  [0, 4, 8],
  [2, 4, 6],
];
let res = winningCombinations.filter(a=> a.reduce((r, v) => r + playerMoves.includes(v), 0)==2);

console.log('matching array: ', res)

edited Jul 17, 2017 at 20:10

answered Jul 17, 2017 at 19:57

Koushik Chatterjee

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