Javascript: How do I check if an array contains exactly another array?

Question

Suppose I have arrays parent and child. I want to check if a child exists inside a parent array. The ordering matters.
Example:

parent = ["x", "a", "b", "c", "d", "e", "f", "g"]
child = ["a", "b", "c"]
//returns true

Example:
When parent has a different order:

parent = ["x", "g", "b", "c", "d", "e", "f", "a"]
child = ["a", "b", "c"]
//It should return false

How would I achieve this in Javascript?
Edit: I have tried this How to check if an array contains another array? but it did not work for my case

before asking how to do something on stackoverflow, please tell us what you have already tried. We are not a code writing service, you have to try something, and then ask for help — Pierre
– Pierre, Commented Apr 7, 2018 at 11:33
You're asking us to di the work for you... What have you tried so far? — NielsNet
– NielsNet, Commented Apr 7, 2018 at 11:34
I have tried this stackoverflow.com/questions/41661287/…. It didn't work — J Duh
– J Duh, Commented Apr 7, 2018 at 11:35
@JDuh That’s not the same situation at all. You’re searching for sub-arrays, not for array elements that are arrays. Do you also need to check whether the items in the “child” array follow one another in the “parent” array? I.e. is ["a", "b", "x", "c"] a valid “parent” for ["a", "b", "c"]? — Sebastian Simon
– Sebastian Simon, Commented Apr 7, 2018 at 11:47

Ankit Agarwal · Accepted Answer · 2018-04-07 11:45:24Z

2

You can run a loop for child and change the index accordingly. You can also use a match variable to detect change in the sequence.

RETURN TRUE

var parent = ["x", "a", "b", "c", "d", "e", "f", "g"]
var child = ["a", "b", "c"];

var initIndex = parent.indexOf(child[0]);
var match = true;
for(var i=1; i<child.length; i++){
  var varIndex = parent.indexOf(child[i]);
  if( varIndex === initIndex+1){
    initIndex = varIndex;
    continue;
  }
  match = false;
}

console.log(match);

RETURN FALSE

var parent = ["x", "g", "b", "c", "d", "e", "f", "a"]
var child = ["a", "b", "c"];

var initIndex = parent.indexOf(child[0]);
var match = true;
for(var i=1; i<child.length; i++){
  var varIndex = parent.indexOf(child[i]);
  if( varIndex === initIndex+1){
    initIndex = varIndex;
    continue;
  }
  match = false;
}

//return false
console.log(match);

USING STRING OPERATION

You can also convert the array to string to avoid those loop:

var parent = ["x", "g", "b", "c", "d", "e", "f", "a"]
var child = ["a", "b", "c"]

var parentStr = parent.toString();
var match = parentStr.indexOf(child.toString()) !== -1;
//return false
console.log(match);

parent = ["x", "a", "b", "c", "d", "e", "f", "g"]
child = ["a", "b", "c"]

parentStr = parent.toString();
match = parentStr.indexOf(child.toString()) !== -1;
//return true
console.log(match);

edited Apr 7, 2018 at 11:45

answered Apr 7, 2018 at 11:38

Ankit Agarwal

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1 Comment

Sebastian Simon Over a year ago

The first approach doesn’t work for repeating elements. The second approach doesn’t work for elements that contain commas.

Sankar · Accepted Answer · 2018-04-07 11:42:53Z

1

Convert the array to string by using JSON.stringify() and remove the square brackets from child string.

Now check the indexOf child in parent to check if it contains the child.

let parent = ["x", "a", "b", "c", "d", "e", "f", "g"];
let child = ["a", "b", "c"];
var parStr = JSON.stringify(parent);
var chldStr = JSON.stringify(child).replace('[', '').replace(']', '')

console.log(parStr.indexOf(chldStr) !== -1);

edited Apr 7, 2018 at 11:42

answered Apr 7, 2018 at 11:37

Sankar

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2 Comments

Andreas Over a year ago

Instead of using JSON.stringify() and then replace the "array" part... use .join()

Sebastian Simon Over a year ago

@Andreas No. This will fail if the array elements contain commas. The JSON approach is more robust as it guarantees valid JSON delimiters and quote escaping.

S. Walker · Accepted Answer · 2018-04-07 12:43:49Z

I wrote a function for this a while back that takes some paramenters:

Array.prototype.containsArray = function (child, orderSensitivity, caseSensitivity, typeSensitivity) {
    var self = this;
    if (orderSensitivity) return orderSensitiveComparer();
    else return orderInsensitiveComparer();

    function orderSensitiveComparer() {
        var resultArry = [],
            placeholder = 0;
        if (child.length > self.length) return false;
        for (var i = 0; i < child.length; i++) {
            for (var k = placeholder; k < self.length; k++) {
                if (equalityComparer(self[k], child[i])) {
                    resultArry.push(true);
                    if (resultArry.length === child.length) return true;
                    placeholder = k + 1;
                    break;
                }
                else resultArry = [];
            }
        }
        return false;
    }
    function orderInsensitiveComparer() {
        for (var i = 0; i < child.length; i++) {
            var childHasParentElement = false;
            for (var k = 0; k < self.length; k++) {
                if (equalityComparer(child[i], self[k])) {
                    childHasParentElement = true;
                    break;
                }
            }
            if (!childHasParentElement) return false;
        }
        return true;
    }
    function equalityComparer(a, b) {

        if (caseSensitivity && typeSensitivity) return caseSensitiveEq(a, b) && typeSensitiveEq(a, b);
        else if (!caseSensitivity && typeSensitivity) return caseInsensitiveEq(a, b) && typeSensitiveEq(a, b);
        else if (caseSensitivity && !typeSensitivity) return caseSensitiveEq(a, b) && typeInsensitiveEq(a, b);
        else if (!caseSensitivity && !typeSensitivity) return caseInsensitiveEq(a, b) && typeInsensitiveEq(a, b);
        else throw "Unknown set of parameters";


        function caseSensitiveEq(a, b) {
            return a == b;
        }
        function caseInsensitiveEq(a, b) {
            return (a + "").toLowerCase() == (b + "").toLowerCase();
        }
        function typeSensitiveEq(a, b) {
            return typeof(a) === typeof(b);
        }
        function typeInsensitiveEq(a, b) {
            return true;
        }
    }
}

var parent = [1, 2, 3, "a", "b", "c"];
var child = [1, 2, 3];
var child2 = ["1", "2", "3"];
var child3 = ["A", "b", "C"];
var child4 = ["a", "b", "c"];
var child5 = ["c", "b", "a"];


// Tests:
console.log(parent.containsArray(parent));
console.log(parent.containsArray(child));
console.log(parent.containsArray(child2));

// parent to child 2, order sensitive, not case, not type. => true.
console.log(parent.containsArray(child2, true, false, false));

// parent to child 2, order, not case, type. => false. b/c of type.
console.log(parent.containsArray(child2, true, false, true));

// parent to child 3, order, not case, type. => true.
console.log(parent.containsArray(child3, true, false, true));

// parent to child 4, order, case and type => true.
console.log(parent.containsArray(child4, true, true, true));

// parent to child 4, not order, case and type. => true.
console.log(parent.containsArray(child4, false, true, true));

// parent to child 5, not order case or type => true.
console.log(parent.containsArray(child5));

Donghua Liu · Accepted Answer · 2018-04-07 11:46:31Z

0

I have a simple method for small sized arrays of this problem.

First join the array to a string, see Array/join
Search substring, see String/indexOf

parent = ["x", "a", "b", "c", "d", "e", "f", "g"];
child = ["a", "b", "c"];
function matchSubArray(parent, child) {
    parentStr = parent.join('');
    childStr = child.join('');
    return parentStr.indexOf(childStr) != -1;
}
matchSubArray(parent, child);

answered Apr 7, 2018 at 11:46

Donghua Liu

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2 Comments

Donghua Liu Over a year ago

If you want more good performance, see details like strstr of c, see stackoverflow.com/questions/12784766/…

Sebastian Simon Over a year ago

This doesn’t work. [ "abc" ] isn’t a subarray of [ "a", "b", "c", "d" ].

Ben Rivers · Accepted Answer · 2018-04-07 12:06:21Z

0

var parent = ["x", "g", "b", "c", "d", "e", "f", "a"]
var child = ["a", "b", "c"]

if(parent.join("").search(child.join("")) === -1) {
    console.log("Not found");
} else {
    console.log("found")
}

edited Apr 7, 2018 at 12:06

Ben Rivers

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answered Apr 7, 2018 at 11:36

Sagar Jajoriya

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3 Comments

wizzwizz4 Over a year ago

This will fail sometimes.

Sagar Jajoriya Over a year ago

In which case,?

wizzwizz4 Over a year ago

If I remember my JavaScript correctly, parent = ["a,b", "c"]; child = ["a", "b", "c"]; ought to do it.

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