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                // Array & Set

// test cases
const a = [1, 2, 3, 1];
const b = [2, 3, 4, 2];
const c = ['x', 1, 'z'];

// union (unique)
const union = (a1, a2) => [ Set(a1.concat(a2))];
// Note: [ Set(a1, a2))] works for same type, failed with [1], ['x']
// Use Set only to dedupe, and use Array for manipulation

// intersection (unique)
const intersection = (a1, a2) => [ Set(a1.filter(x => a2.includes(x)))];

// minus: a1 - a2 (unique)
const minus = (a1, a2) => [... new Set(a1.filter(x => !a2.includes(x)))];

console.log('a: ', a);
console.log('b: ', b);
console.log('c: ', c);
console.log('union(a, b): ', union(a, b));
console.log('union(a, c): ', union(a, c));
// union multiple arrays with the union function
console.log('union a b & c: ', [a, b, c].reduce((total, arr) => union(total, arr), []));

console.log('intersection(a, b): ', intersection(a, b));
console.log('minus(a, b): ', minus(a, b));
console.log('minus(b, a): ', minus(b, a));

// How to accept an arbitrage number of sets?
// Can implement with [arr1, arr2, arr3].reduce((result, item) => union(result, item), []);

// Flattening arrays
const nested1 = [ ['a'], [b, [1, 2]]];

const flatten = (arr) => arr.reduce((a, value) => a.concat(
  Array.isArray(value) ? flatten(value) : value
), []);

console.log('flattening: ', nested1);
console.log('result: ', flatten(nested1));