Mask a String

Easy Strings

Mask all but the last 4 characters of a string with asterisks — like showing a saved card or account number in an app.

Sample input

1234567812345678

Sample output

************5678

Solution

s = "1234567812345678"
visible = 4
print("*" * (len(s) - visible) + s[-visible:])

const s = "1234567812345678";
const visible = 4;
console.log("*".repeat(s.length - visible) + s.slice(-visible));

public class Main {
    public static void main(String[] args) {
        String s = "1234567812345678";
        int visible = 4;
        System.out.println("*".repeat(s.length() - visible) + s.substring(s.length() - visible));
    }
}

fun main() {
    val s = "1234567812345678"
    val visible = 4
    println("*".repeat(s.length - visible) + s.takeLast(visible))
}

let s = "1234567812345678"
let visible = 4
let masked = String(repeating: "*", count: s.count - visible) + s.suffix(visible)
print(masked)

void main() {
  String s = '1234567812345678';
  int visible = 4;
  print('*' * (s.length - visible) + s.substring(s.length - visible));
}

#include <iostream>
#include <string>
using namespace std;

int main() {
    string s = "1234567812345678";
    int visible = 4;
    cout << string(s.size() - visible, '*') << s.substr(s.size() - visible) << endl;
    return 0;
}

#include <stdio.h>
#include <string.h>

int main() {
    char s[] = "1234567812345678";
    int len = strlen(s);
    int visible = 4;
    for (int i = 0; i < len - visible; i++) putchar('*');
    printf("%s\n", s + len - visible);
    return 0;
}