Javascript array contains/includes sub array

Question

I need to check if an array contains another array. The order of the subarray is important but the actual offset it not important. It looks something like this:

var master = [12, 44, 22, 66, 222, 777, 22, 22, 22, 6, 77, 3]; 

var sub = [777, 22, 22]; 

So I want to know if master contains sub something like:

if(master.arrayContains(sub) > -1){
    //Do awesome stuff
}

So how can this be done in an elegant/efficient way?

Answer 1

With a little help from fromIndex parameter

This solution features a closure over the index for starting the position for searching the element if the array. If the element of the sub array is found, the search for the next element starts with an incremented index.

function hasSubArray(master, sub) {
    return sub.every((i => v => i = master.indexOf(v, i) + 1)(0));
}

var array = [12, 44, 22, 66, 222, 777, 22, 22, 22, 6, 77, 3];

console.log(hasSubArray(array, [777, 22, 22]));
console.log(hasSubArray(array, [777, 22, 3]));
console.log(hasSubArray(array, [777, 777, 777]));
console.log(hasSubArray(array, [42]));

Answer 2

var master = [12, 44, 22, 66, 222, 777, 22, 22, 22, 6, 77, 3]; 

var sub = [777, 22, 22]; 

console.log(master.join(',').includes(sub.join(',')))

//true

You can do this by simple console.log(master.join(',').includes(sub.join(','))) this line of code using include method

Answer 3

The simplest way to match subset/sub-array

const master = [12, 44, 22, 66, 222, 777, 22, 22, 22, 6, 77, 3]; 

const sub1 = [777, 44, 222]; 
const sub2 = [777, 18, 66]; 

sub1.every(el => master.includes(el));  // reture true
sub2.every(el => master.includes(el));  // return false

Answer 4

It’s surprising how often this is implemented incorrectly.

What we’re looking for is a substring in the mathematical sense.

In mathematics, a sequence is an enumerated collection of objects in which repetitions are allowed and order matters.

In mathematics, a subsequence of a given sequence is a sequence that can be derived from the given sequence by deleting some or no elements without changing the order of the remaining elements.

A subsequence which consists of a consecutive run of elements from the original sequence, such as ⟨ B, C, D ⟩ from ⟨ A, B, C, D, E, F ⟩ is a substring.

Note that a “string”, here, can consist of any element and is not limited to Unicode code-point sequences.

Effectively all previous answers have one of many possible flaws:

• The string concatenation approach (array1.toString().includes(array2.toString())) fails when your array elements have commas. (Example: [ "a", "b" ] does not contain [ "a,b" ]).
• Some implementations check beyond array bounds. (Example: [ "3" ] does not contain [ "3", undefined ], just because array[1] reports undefined for both).
• Some implementations fail to handle repetition correctly.
• Some implementations aren’t checking for substrings (in the mathematical sense) correctly, but for subsets or subsequences or something else.
• Some implementations don’t account for the empty array. The empty string is the substring of every string.

Check if an array constitutes a “substring” of another array

Right off the bat, this handles the empty array correctly.

Then, it builds a list of candidate starting indexes by matching against the first element of the potential subarray.

Find the first candidate where every element of the slice matches index by index with the full array, offset by the candidate starting index. The checked index also has to exist within the full array, hence Object.hasOwn.

const isSubArray = (full, slice) => {
    if(slice.length === 0){
      return true;
    }

    const candidateIndexes = full
        .map((element, fullIndex) => ({
          matched: element === slice[0],
          fullIndex
        }))
        .filter(({ matched }) => matched),
      found = candidateIndexes
        .find(({ fullIndex }) => slice.every((element, sliceIndex) => Object.hasOwn(full, fullIndex + sliceIndex) && element === full[fullIndex + sliceIndex]));

    return Boolean(found);
  };

console.log(isSubArray([], []) === true);
console.log(isSubArray([ 0 ], []) === true);
console.log(isSubArray([ 0, 1, 2 ], [ 1, 2 ]) === true);
console.log(isSubArray([ 0, 1, 1, 2 ], [ 0, 1, 2 ]) === false);
console.log(isSubArray([ 2, 1 ], [ 1, 2 ]) === false);
console.log(isSubArray([ 1, 2, 3 ], [ 2, 3, undefined ]) === false);
console.log(isSubArray([ 0, 1, 1, 2, 3 ], [ 1, 1, 2 ]) === true);
console.log(isSubArray([ 0, 1, 1, 2, 3 ], [ 1, 2 ]) === true);
console.log(isSubArray([ 0, 1, 1, 2, 3 ], [ 0, 1, 1, 1 ]) === false);
console.log(isSubArray([ "a", "b" ], [ "a,b" ]) === false);

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This has quadratic complexity, yes. There might be more efficient implementations using Trees or Ropes. You might also want to research some efficient substring search algorithms and try to apply them to this problem.

Get the index of the found “substring”, or -1 if not found

It’s basically the same code, but with return true; replaced by return 0;, and return Boolean(found); replaced by return found?.fullIndex ?? -1;.

const findSubArrayIndex = (full, slice) => {
    if(slice.length === 0){
      return 0;
    }

    const candidateIndexes = full
        .map((element, fullIndex) => ({
          matched: element === slice[0],
          fullIndex
        }))
        .filter(({ matched }) => matched),
      found = candidateIndexes
        .find(({ fullIndex }) => slice.every((element, sliceIndex) => Object.hasOwn(full, fullIndex + sliceIndex) && element === full[fullIndex + sliceIndex]));

    return found?.fullIndex ?? -1;
  };

console.log(findSubArrayIndex([], []) === 0);
console.log(findSubArrayIndex([ 0 ], []) === 0);
console.log(findSubArrayIndex([ 0, 1, 2 ], [ 1, 2 ]) === 1);
console.log(findSubArrayIndex([ 0, 1, 1, 2 ], [ 0, 1, 2 ]) === -1);
console.log(findSubArrayIndex([ 2, 1 ], [ 1, 2 ]) === -1);
console.log(findSubArrayIndex([ 1, 2, 3 ], [ 2, 3, undefined ]) === -1);
console.log(findSubArrayIndex([ 0, 1, 1, 2, 3 ], [ 1, 1, 2 ]) === 1);
console.log(findSubArrayIndex([ 0, 1, 1, 2, 3 ], [ 1, 2 ]) === 2);
console.log(findSubArrayIndex([ 0, 1, 1, 2, 3 ], [ 0, 1, 1, 1 ]) === -1);
console.log(findSubArrayIndex([ "a", "b" ], [ "a,b" ]) === -1);

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Semi-acceptable alternative: JSON

JSON-encoding both arrays might be a viable strategy as well. Here, the surrounding […] of the potential subarray need to be removed, then an includes will tell you if the JSON string is included in the other JSON string. This works — as opposed to the simple string concatenation or join approach — because JSON has delimiters that cannot appear verbatim in the encoded elements; if they do appear in the original elements, they’d be correctly escaped.

The caveat is that this won’t work for values that are not encodable in JSON.

const isSubArray = (full, slice) => JSON.stringify(full)
    .includes(JSON.stringify(slice).replaceAll(/^\[|\]$/g, ""));

console.log(isSubArray([], []) === true);
console.log(isSubArray([ 0 ], []) === true);
console.log(isSubArray([ 0, 1, 2 ], [ 1, 2 ]) === true);
console.log(isSubArray([ 0, 1, 1, 2 ], [ 0, 1, 2 ]) === false);
console.log(isSubArray([ 2, 1 ], [ 1, 2 ]) === false);
console.log(isSubArray([ 1, 2, 3 ], [ 2, 3, undefined ]) === false);
console.log(isSubArray([ 0, 1, 1, 2, 3 ], [ 1, 1, 2 ]) === true);
console.log(isSubArray([ 0, 1, 1, 2, 3 ], [ 1, 2 ]) === true);
console.log(isSubArray([ 0, 1, 1, 2, 3 ], [ 0, 1, 1, 1 ]) === false);
console.log(isSubArray([ "a", "b" ], [ "a,b" ]) === false);

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Answer 5

Just came up with quick thought , but efficiency depends on size of the array

var master = [12, 44, 22, 66, 222, 777, 22, 22, 22, 6, 77, 3];
var sub = [777, 22, 22];

if ((master.toString()).indexOf(sub.toString()) > -1 ){
    //body here
}

Answer 6

If the order is important, it has to be an actually sub-array (and not the subset of array) and if the values are strictly integers then try this

console.log ( master.join(",").indexOf( subarray.join( "," ) ) == -1 )

for checking only values check this fiddle (uses no third party libraries)

var master = [12, 44, 22, 66, 222, 777, 22, 22, 22, 6, 77, 3]; 

var sub = [777, 22, 22]; 

function isSubset( arr1, arr2 )
{
    for (var i=0; i<arr2.length; i++)
    {
        if ( arr1.indexOf( arr2[i] ) == -1 )
        {
          return false;
        }
    }
    return true;
}
console.log( isSubset( master, sub ) );

There are faster options explained here as well.

Answer 7

EDIT

Misunderstood question initially.

function arrayContainsSub(arr, sub) {
        var first = sub[0],
            i = 0,
            starts = [];

        while (arr.indexOf(first, i) >= 0) {
            starts.push(arr.indexOf(first, i));
            i = arr.indexOf(first, i) + 1;
        }

        return !!starts
                    .map(function(start) {
                        for (var i = start, j = 0; j < sub.length; i++, j++) {
                            if (arr[i] !== sub[j]) {
                                return false;
                            }
                            if (j === sub.length - 1 && arr[i] === sub[j]) {
                                return true;
                            }
                        };

                    }).filter(function(res) {
                        return res;
                    }).length;
    }

This solution will recursively check all available start points, so points where the first index of the sub has a match in the array

---

Old Answer Kept in case useful for someone searching.

if(master.indexOf(sub) > -1){ //Do awesome stuff }

Important to remember that this will only match of master literally references sub. If it just contains an array with the same contents, but references a different specific object, it will not match.

Answer 8

You can try with filter and indexOf like this:

Note: This code works in case we do not cover the order in sub array.

Array.prototype.arrayContains = function (sub) {
  var self = this;
  var result = sub.filter(function(item) {
    return self.indexOf(item) > -1;
  });
  return sub.length === result.length;
}

Example here.

UPDATED: Return index of sub array inside master (cover order in sub array)

Array.prototype.arrayContains = function(sub) {
  var first;
  var prev;
  for (var i = 0; i < sub.length; i++) {
    var current = this.indexOf(sub[i]);
    if (current > -1) {
      if (i === 0) {
        first = prev = current;
        continue;
      } else {
        if (++prev === current) {
          continue;
        } else {
          return -1;
        }
      }
    } else {
      return -1;
    }
  }
  return first;
}

Demo: here

Answer 9

For this answer, I am preserving the order of sub-array. Means, the elements of sub-array should be in Consecutive order. If there is any extra element while comparing with the master, it will be false.

I am doing it in 3 steps:

1. Find the index of the first element of sub in the master and store it an array matched_index[].
2. for each entry in matched_index[] check if each element of sub is same as master starting from the s_index. If it doesn't match then return false and break the for loop of sub and start next for-loop for next element in matched_index[]
3. At any point, if the same sub array is found in master, the loop will break and return true.

function hasSubArray(master,sub){

    //collect all master indexes matching first element of sub-array
    let matched_index = [] 
    let start_index = master.indexOf(master.find(e=>e==sub[0]))
    
    while(master.indexOf(sub[0], start_index)>0){
        matched_index.push(start_index)
        let index = master.indexOf(sub[0], start_index)
        start_index = index+1
    } 

    let has_array //flag
    
    for(let [i,s_index] of matched_index.entries()){
        for(let [j,element] of sub.entries()){
            if(element != master[j+s_index]) {
                has_array = false
                break
            }else has_array = true
        }
        if (has_array) break
    }
    return has_array
}

var master = [12, 44, 22, 66, 222, 777, 22, 22, 22, 6, 77, 3];

console.log(hasSubArray(master, [777, 22, 22]));
console.log(hasSubArray(master, [777, 22, 3]));
console.log(hasSubArray(master, [777, 777, 777]));
console.log(hasSubArray(master, [44]));
console.log(hasSubArray(master, [22, 66]));

Answer 10

If run this snippet below it should work

x = [34, 2, 4];
y = [2, 4];
y.reduce((included, num) => included && x.includes(num), true);

EDIT: @AlexanderGromnitsky You are right this code is incorrect and thank you for the catch! The above code doesn't actually do what the op asked for. I didn't read the question close enough and this code ignores order. One year later here is what I came up with and hopefully this may help someone.

var master = [12, 44, 22, 66, 222, 777, 22, 22, 22, 6, 77, 3]; 
var sub = [777, 22, 22]; 
var is_ordered_subset = master.join('|').includes(sub.join('|'))

This code is somewhat elegant and does what op asks for. The separator doesn't matter as long as its not an int.

Answer 11

  async function findSelector(a: Uint8Array, selector: number[]): Promise<number> {
    let i = 0;
    let j = 0;
    while (i < a.length) {
      if (a[i] === selector[j]) {
        j++;
        if (j === selector.length) {
          return i - j + 1;
        }
      } else {
        j = 0;
      }
      i++;
    }
    return -1;
  }

Answer 12

I had a similar problem and resolved it using sets.

function _hasSubArray( mainArray, subArray )
{
    mainArray = new Set( mainArray );
    subArray = new Set( subArray );

    for ( var element of subArray )
    {
        if ( !mainArray.has( element ) )
        {
            return false;
        }
    }
    return true;
}

Answer 13

Try using every and indexOf

var mainArr = [1, 2, 3, 4, 5]
var subArr = [1, 2, 3]

function isSubArray(main, sub) {
    return sub.every((eachEle) => {
        return (main.indexOf(eachEle) + 1);
    });
}
isSubArray(mainArr, subArr);