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, Apr 07 '18 at 11:33
You're asking us to di the work for you... What have you tried so far? — NielsNet, Apr 07 '18 at 11:34
I have tried this https://stackoverflow.com/questions/41661287/how-to-check-if-an-array-contains-another-array. It didn't work — J Duh, Apr 07 '18 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, Apr 07 '18 at 11:47
@Tan nope, that's not really right. Consider https://jsfiddle.net/x1f6x6jn/ (logs "found"). — nicael, Apr 07 '18 at 13:14
@JDuh You can tick the answer if it was really helpful to you :) — Ankit Agarwal, Aug 11 '18 at 05:59

Ankit Agarwal · Answer 1 · 2018-04-07T11:45:24.320

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);

The first approach doesn’t work for repeating elements. The second approach doesn’t work for elements that contain commas. — Sebastian Simon, Oct 15 '22 at 14:26

Sankar · Answer 2 · 2018-04-07T11:42:53.643

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 07 '18 at 11:42

answered Apr 07 '18 at 11:37

Sankar

6,908
2
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Instead of using `JSON.stringify()` and then replace the "array" part... use `.join()` – Andreas Apr 07 '18 at 11:48
@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. – Sebastian Simon Oct 15 '22 at 14:33

score 1 · Answer 3 · answered Apr 07 '18 at 12:43

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));

score 0 · Answer 4 · edited Apr 07 '18 at 12:06

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 07 '18 at 12:06

Ben Rivers

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answered Apr 07 '18 at 11:36

Sagar Jajoriya

2,377
1
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2

This will fail sometimes. – wizzwizz4 Apr 07 '18 at 11:46
1

In which case,? – Sagar Jajoriya Apr 07 '18 at 11:58
1

If I remember my JavaScript correctly, `parent = ["a,b", "c"]; child = ["a", "b", "c"];` ought to do it. – wizzwizz4 Apr 07 '18 at 12:12

score 0 · Answer 5 · answered Apr 07 '18 at 11:46

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 07 '18 at 11:46

Donghua Liu

1,776
2
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If you want more good performance, see details like `strstr` of c, see https://stackoverflow.com/questions/12784766/check-substring-exists-in-a-string-in-c – Donghua Liu Apr 07 '18 at 11:50
This doesn’t work. `[ "abc" ]` isn’t a subarray of `[ "a", "b", "c", "d" ]`. – Sebastian Simon Oct 15 '22 at 14:28

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

5 Answers5