LeetCode #1115 — MEDIUM

Print FooBar Alternately

Move from brute-force thinking to an efficient approach using core interview patterns strategy.

The Problem

Problem Statement

Suppose you are given the following code:

class FooBar {
  public void foo() {
    for (int i = 0; i < n; i++) {
      print("foo");
    }
  }

  public void bar() {
    for (int i = 0; i < n; i++) {
      print("bar");
    }
  }
}

The same instance of FooBar will be passed to two different threads:

thread A will call foo(), while
thread B will call bar().

Modify the given program to output "foobar" n times.

Example 1:

Input: n = 1
Output: "foobar"
Explanation: There are two threads being fired asynchronously. One of them calls foo(), while the other calls bar().
"foobar" is being output 1 time.

Example 2:

Input: n = 2
Output: "foobarfoobar"
Explanation: "foobar" is being output 2 times.

Constraints:

1 <= n <= 1000

Step 01

Brute Force Baseline

Problem summary: Suppose you are given the following code: class FooBar { public void foo() { for (int i = 0; i < n; i++) { print("foo"); } } public void bar() { for (int i = 0; i < n; i++) { print("bar"); } } } The same instance of FooBar will be passed to two different threads: thread A will call foo(), while thread B will call bar(). Modify the given program to output "foobar" n times.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

Example 2

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1115: Print FooBar Alternately
class FooBar {
    private int n;
    private Semaphore f = new Semaphore(1);
    private Semaphore b = new Semaphore(0);

    public FooBar(int n) {
        this.n = n;
    }

    public void foo(Runnable printFoo) throws InterruptedException {
        for (int i = 0; i < n; i++) {
            f.acquire(1);
            // printFoo.run() outputs "foo". Do not change or remove this line.
            printFoo.run();
            b.release(1);
        }
    }

    public void bar(Runnable printBar) throws InterruptedException {
        for (int i = 0; i < n; i++) {
            b.acquire(1);
            // printBar.run() outputs "bar". Do not change or remove this line.
            printBar.run();
            f.release(1);
        }
    }
}

// Accepted solution for LeetCode #1115: Print FooBar Alternately
// Auto-generated Go example from java.
func exampleSolution() {
}
// Reference (java):
// // Accepted solution for LeetCode #1115: Print FooBar Alternately
// class FooBar {
//     private int n;
//     private Semaphore f = new Semaphore(1);
//     private Semaphore b = new Semaphore(0);
// 
//     public FooBar(int n) {
//         this.n = n;
//     }
// 
//     public void foo(Runnable printFoo) throws InterruptedException {
//         for (int i = 0; i < n; i++) {
//             f.acquire(1);
//             // printFoo.run() outputs "foo". Do not change or remove this line.
//             printFoo.run();
//             b.release(1);
//         }
//     }
// 
//     public void bar(Runnable printBar) throws InterruptedException {
//         for (int i = 0; i < n; i++) {
//             b.acquire(1);
//             // printBar.run() outputs "bar". Do not change or remove this line.
//             printBar.run();
//             f.release(1);
//         }
//     }
// }

# Accepted solution for LeetCode #1115: Print FooBar Alternately
from threading import Semaphore


class FooBar:
    def __init__(self, n):
        self.n = n
        self.f = Semaphore(1)
        self.b = Semaphore(0)

    def foo(self, printFoo: "Callable[[], None]") -> None:
        for _ in range(self.n):
            self.f.acquire()
            # printFoo() outputs "foo". Do not change or remove this line.
            printFoo()
            self.b.release()

    def bar(self, printBar: "Callable[[], None]") -> None:
        for _ in range(self.n):
            self.b.acquire()
            # printBar() outputs "bar". Do not change or remove this line.
            printBar()
            self.f.release()

// Accepted solution for LeetCode #1115: Print FooBar Alternately
use std::sync::Condvar;
use std::sync::Mutex;

struct FooBar {
    flag: (Mutex<bool>, Condvar),
    n: usize,
}

impl FooBar {
    fn new(n: usize) -> Self {
        let flag = (Mutex::new(true), Condvar::new());
        FooBar { flag, n }
    }

    fn foo(&self, print_foo: impl Fn()) {
        for _ in 0..self.n {
            let (lock, cvar) = &self.flag;
            let mut g = cvar
                .wait_while(lock.lock().unwrap(), |flag| !*flag)
                .unwrap();
            *g = false;
            print_foo();
            cvar.notify_all();
        }
    }

    fn bar(&self, print_bar: impl Fn()) {
        for _ in 0..self.n {
            let (lock, cvar) = &self.flag;
            let mut g = cvar.wait_while(lock.lock().unwrap(), |flag| *flag).unwrap();
            *g = true;
            print_bar();
            cvar.notify_all();
        }
    }
}

#[test]
fn test() {
    use std::sync::mpsc::channel;
    use std::sync::Arc;
    use threadpool::ThreadPool;

    let foo_bar = Arc::new(FooBar::new(4));
    let (tx, rx) = channel();
    let pool = ThreadPool::new(4);
    {
        let tx = tx.clone();
        let foo_bar = foo_bar.clone();
        pool.execute(move || {
            foo_bar.foo(|| tx.send("foo".to_string()).unwrap());
        });
    }
    {
        let tx = tx;
        let foo_bar = foo_bar;
        pool.execute(move || {
            foo_bar.bar(|| tx.send("bar".to_string()).unwrap());
        });
    }
    pool.join();
    let ss: Vec<String> = rx.try_iter().collect();
    let res = ss.join("");
    let ans = "foobarfoobarfoobarfoobar".to_string();
    assert_eq!(ans, res);
}

// Accepted solution for LeetCode #1115: Print FooBar Alternately
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1115: Print FooBar Alternately
// class FooBar {
//     private int n;
//     private Semaphore f = new Semaphore(1);
//     private Semaphore b = new Semaphore(0);
// 
//     public FooBar(int n) {
//         this.n = n;
//     }
// 
//     public void foo(Runnable printFoo) throws InterruptedException {
//         for (int i = 0; i < n; i++) {
//             f.acquire(1);
//             // printFoo.run() outputs "foo". Do not change or remove this line.
//             printFoo.run();
//             b.release(1);
//         }
//     }
// 
//     public void bar(Runnable printBar) throws InterruptedException {
//         for (int i = 0; i < n; i++) {
//             b.acquire(1);
//             // printBar.run() outputs "bar". Do not change or remove this line.
//             printBar.run();
//             f.release(1);
//         }
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.