Project Euler > Problem 54 > Poker hands (Java Solution)


In the card game poker, a hand consists of five cards and are ranked, from lowest to highest, in the following way:

* High Card: Highest value card.
* One Pair: Two cards of the same value.
* Two Pairs: Two different pairs.
* Three of a Kind: Three cards of the same value.
* Straight: All cards are consecutive values.
* Flush: All cards of the same suit.
* Full House: Three of a kind and a pair.
* Four of a Kind: Four cards of the same value.
* Straight Flush: All cards are consecutive values of same suit.
* Royal Flush: Ten, Jack, Queen, King, Ace, in same suit.

The cards are valued in the order:
2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, Ace.

If two players have the same ranked hands then the rank made up of the highest value wins; for example, a pair of eights beats a pair of fives (see example 1 below). But if two ranks tie, for example, both players have a pair of queens, then highest cards in each hand are compared (see example 4 below); if the highest cards tie then the next highest cards are compared, and so on.

Consider the following five hands dealt to two players:

Hand Player 1 Player 2 Winner
1 5H 5C 6S 7S KD
Pair of Fives
2C 3S 8S 8D TD
Pair of Eights
Player 2
2 5D 8C 9S JS AC
Highest card Ace
2C 5C 7D 8S QH
Highest card Queen
Player 1
3 2D 9C AS AH AC
Three Aces
3D 6D 7D TD QD
Flush with Diamonds
Player 2
4 4D 6S 9H QH QC
Pair of Queens
Highest card Nine
3D 6D 7H QD QS
Pair of Queens
Highest card Seven
Player 1
5 2H 2D 4C 4D 4S
Full House
With Three Fours
3C 3D 3S 9S 9D
Full House
with Three Threes
Player 1

The file, poker.txt, contains one-thousand random hands dealt to two players. Each line of the file contains ten cards (separated by a single space): the first five are Player 1's cards and the last five are Player 2's cards. You can assume that all hands are valid (no invalid characters or repeated cards), each player's hand is in no specific order, and in each hand there is a clear winner.

How many hands does Player 1 win?



The solution may include methods that will be found here: .

public interface EulerSolution{

public String run();

* Solution to Project Euler problem 4
* By Nayuki Minase

public final class p004 implements EulerSolution {

public static void main(String[] args) {
System.out.println(new p004().run());

public String run() {
int maxPalin = -1;
for (int i = 100; i < 1000; i++) {
for (int j = 100; j < 1000; j++) {
int prod = i * j;
if (Library.isPalindrome(prod) && prod > maxPalin)
maxPalin = prod;
return Integer.toString(maxPalin);


1 comment :

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