JavaScript Return an array that contains all the strings appearing in all the subarrays

We have an array of arrays like this ?

const arr = [
   ['foo', 'bar', 'hey', 'oi'],
   ['foo', 'bar', 'hey'],
   ['foo', 'bar', 'anything'],
   ['bar', 'anything']
]

We are required to write a JavaScript function that takes in such array and returns an array that contains all the strings which appears in all the subarrays.

Let's write the code for this function

Example

const arr = [
   ['foo', 'bar', 'hey', 'oi'],
   ['foo', 'bar', 'hey'],
   ['foo', 'bar', 'anything'],
   ['bar', 'anything']
];

const commonArray = arr => {
   return arr.reduce((acc, val, index) => {
      return acc.filter(el => val.indexOf(el) !== -1);
   });
};

console.log(commonArray(arr));

Output

The output in the console will be ?

['bar']

How It Works

The function uses the reduce() method to iterate through each subarray. Starting with the first subarray as the accumulator, it filters elements that exist in the current subarray using indexOf(). Only elements present in all subarrays remain after the reduction.

Alternative Approach Using Set

const arr = [
   ['foo', 'bar', 'hey', 'oi'],
   ['foo', 'bar', 'hey'],
   ['foo', 'bar', 'anything'],
   ['bar', 'anything']
];

const findCommonStrings = arr => {
   if (arr.length === 0) return [];
   
   return arr[0].filter(element => 
      arr.every(subarray => subarray.includes(element))
   );
};

console.log(findCommonStrings(arr));

['bar']

Comparison

Conclusion

Both approaches effectively find common elements across all subarrays. The filter() + every() method is more readable, while reduce() offers a functional programming style for finding intersections.