CODEWITHSUNDEEP
javascriptarrayshashhashtable
Adam262
Adam262

Reputation: 39

loop over hash table of arrays

I have a hash table of arrays, ie:

function (findPairs) {

    var arrMaster = [];

    var hash = {

    a: [1,2,3,4,5],
    b: [2,3,5,7,9],
    c: [7,2,3,8],
    d: [1,2]

    }

   return arrMaster

}

My goal is to return an array that shows all pairs of keys who have three instances of a common element in their value array. eg, in above example:

I sketched out code in JSFiddle. My main roadblock now is comparing a given value array to all other value arrays - I have code that compares any array to the immediate successive array. The code is getting quite complicated, with multiple nests, and frankly my head is spinning.

Upvotes: 0

Views: 118

Answers (2)

Esteban Felix
Esteban Felix

Reputation: 1561

You could create an intermediate function to find the pairs like so:

function isPair(first, second) {
    var values = {},
        pairsFound = 0;

    for (var i = 0; i < first.length; i++) {
        var str = first[i];
        values[str] = true;
    }

    for (var j = 0; j < second.length; j++) {
        if (values[second[j]] === true) {
            pairsFound++;
        }
    }

    return pairsFound > 3;
}

Upvotes: 0

Alnitak
Alnitak

Reputation: 340045

Think in smaller code chunks, i.e. how could you count the number of common elements in a pair of arrays:

function countCommon(a, b) {
    return a.reduce(function(p, c, e) {
        return p + (b.indexOf(c) >= 0 ? 1 : 0);
    }, 0);
}

Your problem is then reduced (no pun intended) to creating all possible pairs of the keys of hash and evaluating the above function for each key's data:

function makePairs(hash) {
    var result = [];
    var keys = Object.keys(hash);
    for (var i = 0, n = keys.length; i < n; ++i) {
        for (var j = i + 1; j < n; ++j) {
            result.push([keys[i], keys[j]]);
        }
    }
    return result;
}

and then you just need the pairs that satisfy the criteria:

var result = makePairs(hash).filter(function(pair) {
    return countCommon(hash[pair[0]], hash[pair[1]]) >= 3;
});

Hey presto, only one nested loop! Demo at http://jsfiddle.net/alnitak/kxdngy7n/

Upvotes: 1

Related Questions