LeetCode #822 — MEDIUM

Card Flipping Game

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given two 0-indexed integer arrays fronts and backs of length n, where the i^th card has the positive integer fronts[i] printed on the front and backs[i] printed on the back. Initially, each card is placed on a table such that the front number is facing up and the other is facing down. You may flip over any number of cards (possibly zero).

After flipping the cards, an integer is considered good if it is facing down on some card and not facing up on any card.

Return the minimum possible good integer after flipping the cards. If there are no good integers, return 0.

Example 1:

Input: fronts = [1,2,4,4,7], backs = [1,3,4,1,3]
Output: 2
Explanation:
If we flip the second card, the face up numbers are [1,3,4,4,7] and the face down are [1,2,4,1,3].
2 is the minimum good integer as it appears facing down but not facing up.
It can be shown that 2 is the minimum possible good integer obtainable after flipping some cards.

Example 2:

Input: fronts = [1], backs = [1]
Output: 0
Explanation:
There are no good integers no matter how we flip the cards, so we return 0.

Constraints:

n == fronts.length == backs.length
1 <= n <= 1000
1 <= fronts[i], backs[i] <= 2000

Step 01

Brute Force Baseline

Problem summary: You are given two 0-indexed integer arrays fronts and backs of length n, where the ith card has the positive integer fronts[i] printed on the front and backs[i] printed on the back. Initially, each card is placed on a table such that the front number is facing up and the other is facing down. You may flip over any number of cards (possibly zero). After flipping the cards, an integer is considered good if it is facing down on some card and not facing up on any card. Return the minimum possible good integer after flipping the cards. If there are no good integers, return 0.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

[1,2,4,4,7]
[1,3,4,1,3]

Example 2

[1]
[1]

Step 02

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 #822: Card Flipping Game
class Solution {
    public int flipgame(int[] fronts, int[] backs) {
        Set<Integer> s = new HashSet<>();
        int n = fronts.length;
        for (int i = 0; i < n; ++i) {
            if (fronts[i] == backs[i]) {
                s.add(fronts[i]);
            }
        }
        int ans = 9999;
        for (int v : fronts) {
            if (!s.contains(v)) {
                ans = Math.min(ans, v);
            }
        }
        for (int v : backs) {
            if (!s.contains(v)) {
                ans = Math.min(ans, v);
            }
        }
        return ans % 9999;
    }
}

// Accepted solution for LeetCode #822: Card Flipping Game
func flipgame(fronts []int, backs []int) int {
	s := map[int]struct{}{}
	for i, a := range fronts {
		if a == backs[i] {
			s[a] = struct{}{}
		}
	}
	ans := 9999
	for _, v := range fronts {
		if _, ok := s[v]; !ok {
			ans = min(ans, v)
		}
	}
	for _, v := range backs {
		if _, ok := s[v]; !ok {
			ans = min(ans, v)
		}
	}
	return ans % 9999
}

# Accepted solution for LeetCode #822: Card Flipping Game
class Solution:
    def flipgame(self, fronts: List[int], backs: List[int]) -> int:
        s = {a for a, b in zip(fronts, backs) if a == b}
        return min((x for x in chain(fronts, backs) if x not in s), default=0)

// Accepted solution for LeetCode #822: Card Flipping Game
use std::collections::HashSet;

impl Solution {
    pub fn flipgame(fronts: Vec<i32>, backs: Vec<i32>) -> i32 {
        let n = fronts.len();
        let mut s: HashSet<i32> = HashSet::new();

        for i in 0..n {
            if fronts[i] == backs[i] {
                s.insert(fronts[i]);
            }
        }

        let mut ans = 9999;
        for &v in fronts.iter() {
            if !s.contains(&v) {
                ans = ans.min(v);
            }
        }
        for &v in backs.iter() {
            if !s.contains(&v) {
                ans = ans.min(v);
            }
        }

        if ans == 9999 {
            0
        } else {
            ans
        }
    }
}

// Accepted solution for LeetCode #822: Card Flipping Game
function flipgame(fronts: number[], backs: number[]): number {
    const s: Set<number> = new Set();
    const n = fronts.length;
    for (let i = 0; i < n; ++i) {
        if (fronts[i] === backs[i]) {
            s.add(fronts[i]);
        }
    }
    let ans = 9999;
    for (const v of fronts) {
        if (!s.has(v)) {
            ans = Math.min(ans, v);
        }
    }
    for (const v of backs) {
        if (!s.has(v)) {
            ans = Math.min(ans, v);
        }
    }
    return ans % 9999;
}

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(n)

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.