I want to see if two arrays of strings are equal.
Eg:
compare(["abc", "def"], ["def", "abc"])
should return true
and similarly,
compare(["abc", "def"], ["def", "ghi"])
should return false
.
What is the best way to do this?
I want to see if two arrays of strings are equal.
Eg:
compare(["abc", "def"], ["def", "abc"])
should return true
and similarly,
compare(["abc", "def"], ["def", "ghi"])
should return false
.
What is the best way to do this?
JavaScript doesn't have a Set or Multiset data structure (at least, not one with wide browser support), which is what you'd normally use for testing that two sets of items are the same regardless of order. So I recommend sorting the arrays and checking that their contents are equal. If you know the arrays contain only strings, you can check the items with simple equality:
function compare(array1, array2) {
if (array1.length != array2.length) {
return false;
}
array1 = array1.slice();
array1.sort();
array2 = array2.slice();
array2.sort();
for (var i = 0; i < array1.length; i++) {
if (array1[i] != array2[i]) {
return false;
}
}
return true;
}
console.log(compare(["abc", "def"], ["def", "abc"])); // true
console.log(compare(["abc", "def"], ["def", "ghi"])); // false
For more general cases, you'll need a more complex definition of equality, and I recommend browsing the answers to this question.
Naive algorithm: O(N^2)
function compare(array1, array2){
for (var i = 0; i < array1.length; i++){
if (array2.indexOf(array1[i]) < 0) return false;
}
return true;
}
Better one: (uses sorting, O(NLOGN))
function compare(array1, array2){
array1.sort();
array2.sort();
for (var i = 0; i < array1.length; i++){
if (array1[i] !== array2[i]) return false;
}
return true;
}
JSON.stringify(array1.sort()) === JSON.stringify(array2.sort())
I would suggest following ES6-oneliner
const compare = (a1, a2) =>
(a1 = new Set(a1)) &&
(a2 = new Set(a2)) &&
a1.size === a2.size &&
[...a1].every(v => a2.has(v));
compare(['a', 'a'], ['a', 'b'])
should return false
).compare(['a', 'b'], ['a', 'b', 'c'])
should return false
).Unified solution:
function compare(a, b){
var isEqual = false;
if (Array.isArray(a) && Array.isArray(b) && a.length == b.length){
a.sort();
b.sort();
var i;
for (i = 0; i < arr1.length; i++){
if (a[i] === b[i]){
isEqual = true;
} else{
isEqual = false;
break;
}
}
}
return isEqual;
}
var arr1 = ["def", "abc"], arr2 = ["abc", "def"];
console.log(compare(arr2,arr1)); // gives 'true'
console.log(compare(["abc", "def"], ["def", "ghi"])); // gives 'false'
function compare(arr1, arr2){
var match = true
if(arr1.length != arr2.length){
match = false
}
arr1 = arr1.slice();
arr1.sort();
arr2.slice();
arr2.sort();
for(var i = 0; i < arr1.length; i++){
if(arr1[i] != arr2[i]){
match = false;
}
}
return match;
}
console.log(compare(["abc", "def"], ["def", "abc"])); // it will return true
console.log(compare(["abc", "def"], ["def", "ghi"])); // it will return false
Verbose typescript example for ordered or unordered
(obviously inspired by @dhilts answer above)
const VERBOSE = console.log // ()=>void
const areDeduplicatedArraysEqualUnordered = (a1: any[] | Set<any>, a2: any[] | Set<any>) => {
(a1 = new Set(a1)); (a2 = (new Set(a2))); VERBOSE('checking arrays as Sets', { a1, a2 })
return !!(a1.size === a2.size && [...a1].every(v => (a2 as Set<any>).has(v)))
}
const areDeduplicatedArraysEqualOrdered = (a1: any[], a2: any[]) => {
(a1 = [...(new Set(a1))] as any[]) && (a2 = [...(new Set(a2))] as any[]) && VERBOSE('checking deduplicated arrays', { a1, a2 })
return !!(a1.length === a2.length && a1.every((v: any, i) => a2[i] === v))
}
const a2 = [3,5,8]
const a1 = [8,5,3]
const notEqual = areDeduplicatedArraysEqualOrdered(a1,a2)
const yesEqual = areDeduplicatedArraysEqualUnordered(a1,a2)
VERBOSE('results', { notEqual, yesEqual})