Basic Algorithm Scripting: Mutations help me pls

JavaScript
1

Can anyone explaine why it counts twice?
Screenshot_1

here is my code

function mutation(arr) {
  let z = arr[0].split("")
  let x = arr[1].split("")
  for (let i = 0; i<z.length; i++){
    z[i] = z[i].toLowerCase()
  }
  for (let i = 0; i<x.length; i++){
    x[i] = x[i].toLowerCase()
  }
  let c = 0;
  for (let i = 0; i<z.length; i++){
    for (let k = 0; k<x.length; k++){
      if (z[i] == x[k]){
        c++
        console.log(c, z[i], x[k])
      }
    }
  }
  if (c == x.length){
    return true
  } else {
    return false
  }
}

console.log(mutation(["floor", "for"]));

this lesson https://learn.freecodecamp.org/javascript-algorithms-and-data-structures/basic-algorithm-scripting/mutations/

2

In floor the o appear twice

3

Yes, i knew but although thank you. I think i understood how to fix this