Consider this 2D array:
[
["x","y","z"]
["m","n","o"]
["p","q","r"]
]
The provided word is :"znm" the desired result should be string : "xyzopq"
Consider this 2D array:
[
["x","y","z"]
["m","n","o"]
["p","q","r"]
]
The provided word is :"znm" the desired result should be string : "xyzopq"
Here is an example of using a function for filtering characters in a multidimensional array:
let arr = [
["a","b","c"],
["d","e","f"],
["g","h","i"]
];
function filterCharacters(arr,word) {
return arr // Return the character list,
.flat(1) // converted into a 1D array,
.filter(e => !word.includes(e)) // filtered to elements not in the word,
.join(''); // joined together in a string.
}
console.log(filterCharacters(arr,"iea"))It seems unnecessary to store the characters in a 2D array for this task. To efficiently work with them, you should consider using a 1D array instead.
const targetWord = "iea";
const charactersArray = [
["a", "b", "c"],
["d", "e", "f"],
["g", "h", "i"]
];
// Initialize all characters as unused
let unusedCharacters = charactersArray.flat(1);
// Loop through each row of the 2D array
for (let row = 0; row < charactersArray.length; row++) {
// Iterate over each element in the row
for (let col = 0; col < charactersArray[row].length; col++) {
let currentChar = charactersArray[row][col];
if (targetWord.includes(currentChar)) {
// If character is used, remove from unused list
let indexToRemove = unusedCharacters.indexOf(currentChar);
unusedCharacters.splice(indexToRemove, 1);
}
}
}
console.log(unusedCharacters); // Display remaining unused characters
console.log(unusedCharacters.join("")); // Concatenate unused charactersThis code snippet checks each character in the 2D array against the word and eliminates it from the list of unused characters if found in the word.
const letters = [
["x","y","z"],
["q","r","s"],
["u","v","w"]
];
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