const array1 = [[1, 2, 3], [1,3,4], [1,2,5]];
let b = []
let c = [2,3]
array1.forEach(e => {
c.some(r => {
if(e.includes(r))
b.push(e)
})
})
console.log(b)Upon running the code, the output was [ [ 1, 2, 3 ], [ 1, 2, 3 ], [ 1, 3, 4 ], [ 1, 3, 5 ] ]
However, the expected result should have been [ [ 1, 2, 3 ]]
Utilize filtering on the original array and leverage Array.every() along with Array.includes() to validate that all elements in c are present in the subArray:
const array1 = [[1,2,3], [1,3,4], [1,3,5], [1,2,5]];
const c = [2,3]
const result = array1.filter(subArray =>
c.every(n => subArray.includes(n))
)
console.log(result)You can also employ the latest Set functionalities to verify if c is a subset of subArray:
const array1 = [[1,2,3], [1,3,4], [1,3,5], [1,2,5]];
const c = new Set([2,3])
const result = array1.filter(subArray =>
c.isSubsetOf(new Set(subArray))
)
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