Write a function that returns all the Barcode that can be created in an array when M must be an even sequence of natural numbers from 1 to N and the length must be M.*
condition
function test (N, M) {
};
outputs should go like below
const output1 = test(2, 1); console.log(output1); // --> [1, 2]
const output2 = test(3, 2); console.log(output2); // --> [12, 13, 21, 23, 31, 32]
const output3 = test(3, 3); console.log(output3); // --> [123, 132, 213, 231, 312, 321]
I've tried solve this for hours , but I still can't get it. ... It's hard to even start. I would appreciate it if you could help me solve the code.
try this
const test = (n, m) => {
if(m > n){
return []
}
const numbers = Array(n).fill(1).map((n, i) => (n + i).toString())
const loop = (arr, res, elements) => {
if (elements === 0) {
return res.join('')
}
return arr.flatMap((a, i) => loop(arr.filter(b => b !== a), [...res, a], elements - 1))
}
return loop(numbers, [], m)
}
console.log(test(1, 1))
console.log(test(2, 1))
console.log(test(3, 2))
console.log(test(5, 3))Explanation
first there is a check to ensure that m is lesser than n because in that case there are no possible combination
Internally this function uses a recursive function loop to calculate all possible combination of length m
the arguments are
at first the loop function is called with numbers as array , empty array as res and m as number of elements
First it checks if elements is zero (exit condition)
Then there it take the arr and do a flatMap over it
Foreach element it call the loop function recursively with argument
example (2, 2)
first run loop(['1', '2'], [], 2)
elements > 0
loop(['2'], ['1'], 1)
loop(['1'], ['2'], 1)