Given N cards having positive integers printed on the front and the back of each card (possibly different). Any number of cards can be flipped, and after that we choose one card from the deck. If the number X written on the back of the chosen card is not in the front of any card, then we say number X is good. The task is to find the smallest number that is good. If no number is good, then print 0.
Note: A flip swaps the front and back numbers present on the card i.e. the value on the front is now on the back and vice versa.
Examples:
Input: fronts = [1, 2, 4, 4, 7, 8], backs = [1, 3, 4, 1, 3, 9]
Output: 2 If we flip the second card, the fronts are [1, 3, 4, 4, 7, 8] and the backs are [1, 2, 4, 1, 3, 9]. Now, we choose the second card, having number 2 on the back, and it isn’t on the front of any other card, so 2 is good.Input: fronts = [1, 2, 3, 4, 5], backs = [6, 7, 8, 9, 10]
Output: 1
Approach:
- If a card has the same value K written on the front and the back then K cannot be the answer as no matter how many times you flip the card the result will be the same.
- Every other card that has different numbers written on the front and backs are candidates for the answer as no matter how many times the number K repeats in the front array, just flip all the cards then there will no longer be a card that has K written on the front then we can simply choose any card (minimum) that has K written on it’s back and it is the answer.
- Now the problem reduces to finding all the numbers K1, K2, …, Kn such that they are not repeated on any card and then among all the numbers written on the backs of K1, K2, …, Kn find the minimum.
- If the result is not possible then print 0.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>#define MAX 9999using namespace std;int flipgame(int fronts[], int backs[], int fl, int bl){ set<int> same; for (int i = 0; i < fl; i++) if (fronts[i] == backs[i]) same.insert(fronts[i]); // Initialize answer to arbitrary value int ans = MAX; int arr[fl + bl]; for (int i = 0; i < fl + bl; i++) { if (i < fl) arr[i] = fronts[i]; else arr[i] = backs[i - fl]; } for (int k : arr) if (same.find(k) == same.end()) ans = min(ans, k); // Return final answer return ans % MAX;}// Driver Codeint main(){ int fronts[] = { 1, 2, 4, 4, 7 }; int backs[] = { 1, 3, 4, 1, 3 }; int fl = sizeof(fronts) / sizeof(fronts[0]); int bl = sizeof(backs) / sizeof(backs[0]); cout << flipgame(fronts, backs, fl, bl);}// This code is contributed by phasing17 |
Java
// Java implementation of the approachimport java.util.*;class GFG { static int MAX = 9999; static int flipgame(int[] fronts, int[] backs, int fl, int bl) { HashSet<Integer> same = new HashSet<Integer>(); for (int i = 0; i < fl; i++) if (fronts[i] == backs[i]) same.add(fronts[i]); // Initialize answer to arbitrary value int ans = MAX; int[] arr = new int[fl + bl]; for (int i = 0; i < fl + bl; i++) { if (i < fl) arr[i] = fronts[i]; else arr[i] = backs[i - fl]; } for (int k : arr) if (!same.contains(k)) ans = Math.min(ans, k); // Return final answer return ans % MAX; } // Driver Code public static void main(String[] args) { int[] fronts = { 1, 2, 4, 4, 7 }; int[] backs = { 1, 3, 4, 1, 3 }; int fl = fronts.length; int bl = backs.length; System.out.println(flipgame(fronts, backs, fl, bl)); }}// This code is contributed by phasing17 |
Python3
# Python3 iomplementation of the approachimport itertoolsMAX = 9999def flipgame(fronts, backs): same = {k for i, k in enumerate(fronts) if k == backs[i]} # Initialize answer to arbitrary value ans = MAX for k in itertools.chain(fronts, backs): if k not in same: ans = min(ans, k) # Return final answer return ans % MAX# Driver Codefronts = [1, 2, 4, 4, 7]backs = [1, 3, 4, 1, 3]print(flipgame(fronts, backs)) |
C#
// C# implementation of the approachusing System;using System.Collections.Generic;class GFG { static int MAX = 9999; static int flipgame(int[] fronts, int[] backs, int fl, int bl) { HashSet<int> same = new HashSet<int>(); for (int i = 0; i < fl; i++) if (fronts[i] == backs[i]) same.Add(fronts[i]); // Initialize answer to arbitrary value int ans = MAX; int[] arr = new int[fl + bl]; for (int i = 0; i < fl + bl; i++) { if (i < fl) arr[i] = fronts[i]; else arr[i] = backs[i - fl]; } foreach(int k in arr) if (!same.Contains(k)) ans = Math.Min(ans, k); // Return final answer return ans % MAX; } // Driver Code public static void Main(string[] args) { int[] fronts = { 1, 2, 4, 4, 7 }; int[] backs = { 1, 3, 4, 1, 3 }; int fl = fronts.Length; int bl = backs.Length; Console.WriteLine(flipgame(fronts, backs, fl, bl)); }}// This code is contributed by phasing17 |
Javascript
// JavaScript implementation of the approachlet MAX = 9999function flipgame(fronts, backs){ let same = new Set(); for (var i = 0; i < fronts.length; i++) if (fronts[i] == backs[i]) same.add(fronts[i]) // Initialize answer to arbitrary value let ans = MAX let arr = [...fronts] arr.push(...backs) for (var k of arr) if (!same.has(k)) ans = Math.min(ans, k) // Return final answer return ans % MAX}// Driver Codelet fronts = [1, 2, 4, 4, 7]let backs = [1, 3, 4, 1, 3]console.log(flipgame(fronts, backs))// This code is contributed by phasing17 |
Output
2
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