React-Native | Generate a non-repeating (name) generator in React-Native

if I understand you want to select without return

let arr1 = ['a', 'b', 'c'];
let value;

Option1: Copy the array to temp array

let arr2 = [...arr1];
let random_index = Math.floor(Math.random() * arr2.length);
value = arr2[random_index];
arr2 = arr2.filter((val, index) => index !== random_index);

if (arr2.length === 0)
    arr2 = [...arr1];

Option2: Save the indexes of the array

let arr2 = Array.from(Array(arr1.length).keys());
let random_index = Math.floor(Math.random() * arr2.length);
value = arr1[arr2[random_index]];
arr2 = arr2.filter((val, index) => index !== random_index);

if (arr2.length === 0)
    arr2 = Array.from(Array(arr1.length).keys());

Option 1: Quick and Easy

Create an object in state to track the used name IDs.

const [usedIds, setUsedIds] = useState([]);

Then update the findLinkByName function to use this array. You should also invoke the random number generator inside the function.

function findLinkByName() {
  // clear array if full
  if(usedIds.length === differentNames.namesContainer.names.length) {
    setUsedIds([]);
  }

  // find unique ID
  let randomId;
  do {
    randomId = RandomNumber(1,5);
  } while(usedIds.includes(randomId));
  
  // add used ID to array
  setUsedIds(prev => [...prev, randomId]);

  // return random name
  return differentNames.namesContainer.names.find(n => n.id === randomId);
}

Option 2: Move Names to State

You can also simply append a used property to each name object in the name container and change it to be stored in state so that we can mutate it. The ugliest part of this is that the names are kept 3 levels deep in the object. If that can be lifted up, then the following statements could be much shorter.

const [names, setNames] = useState({
  namesContainer: {
    names: [
      { name: "(1) Sophie", id: 1, used: false },
      { name: "(2) Emma", id: 2, used: false },
      { name: "(3) Lisa", id: 3, used: false },
      { name: "(4) Esmée", id: 4, used: false },
      { name: "(5) Zoe", id: 5, used: false },
    ],
  },
});

I'd also recommend using const and let over var for various reasons

Then your findLinkByName function can be updated to work much more efficiently like this:

function findLinkByName() {
  // clear array if full
  if(names.namesContainer.names.filter(n => !n.used).length === 0) {
    let newNames = {...names};
    newNames.namesContainer.names.map(n => {...n, used: false});
    setNames(newNames);
  }

  // find random ID
  const unusedNames = names.filter(n => !n.unsed);
  const randId = Math.floor(Math.random() * unusedNames.length);
  
  // update state
  let newNames = {...names};
  newNames.namesContainer.names.map(n => {
    return (n.id === randId) ? {...n, used: true} : {...n}
  });
  setNames(newNames);

  // return random name
  return names.namesContainer.names.find(n => n.id === randId);
}

You could keep track of two states. One holds already selectedNames and the other one holds still availableNames as follows.

  const [selectedNames, setSelectedNames] = useState([])
  const [availableNames, setAvailableNames] = useState([
        { name: "(1) Sophie", id: 1 },
        { name: "(2) Emma", id: 2 },
        { name: "(3) Lisa", id: 3 },
        { name: "(4) Esmée", id: 4 },
        { name: "(5) Zoe", id: 5 },
      ])

Then, we choose a random number between 0 and the length of avialableNames which represents the index we want to pick from avialableNames.

const random = RandomNumber(0, availableNames.length - 1);

Then, your onPress function looks as follows.

const onPress = () => {
    setAvailableNames(availableNames.filter(n => n !== availableNames[random]))
    setSelectedNames([...selectedNames, availableNames[random]])  
};

We add the new randomly picked name to selectedNames and remove it from availableNames at the same time.

Your findLinkByName function could look as follows.

function findLinkByName(random) {
   
    if (availableNames.length === 0) {
      setAvailableNames(selectedNames.sort((a, b) => a.id - b.id))
      setSelectedNames([])
      return availableNames[0] 
    }
    return availableNames[random].name
 }

As long as there are names in availableNames, that is its length is not equal to 0, we just pick it and return its name. If all avialable names have been selected, we reset the states, sort the selectedNames by their id prop and return the first name of the list again.

Here is a working snack.