#StackBounty: #python #algorithm #python-2.7 #sudoku Sudoku solver recursive solution

Bounty: 50

Here is my code in Python 2.7 for a Sudoku resolver. Any advice on performance improvement, code bugs or general code style advice is appreciated.

My major idea is:

  1. Using method generate some random data, then using check_invalid to see if it is a valid Sudoku
  2. If from 1, it is a valid Sudoku, then I will call resolve_sudoku to fill valid values

I find my code sometimes run a long time without completion. Are there any code bugs you can find?

import random
from collections import defaultdict
found = False
def check_invalid(matrix):
    # check for each row
    for row in range(len(matrix)):
        cur_row = set()
        for col in range(len(matrix[0])):
            if matrix[row][col] == 0:
                continue
            elif 1 <= matrix[row][col] <= 9:
                if matrix[row][col] in cur_row:
                    return False
                else:
                    cur_row.add(matrix[row][col])
            else:
                return False # invalid number
    # check each col
    for col in range(len(matrix[0])):
        cur_col = set()
        for row in range(len(matrix)):
            if matrix[row][col] == 0:
                continue
            elif 1 <= matrix[row][col] <= 9:
                if matrix[row][col] in cur_col:
                    return False
                else:
                    cur_col.add(matrix[row][col])
            else:
                return False # invalid number
    # check each 3*3 square
    for start_row in [0,3,6]:
        for start_col in [0,3,6]:
            cur_square = set()
            for row in range(start_row, start_row+3):
                for col in range(start_col, start_col + 3):
                    if matrix[row][col] == 0:
                        continue
                    elif 1 <= matrix[row][col] <= 9:
                        if matrix[row][col] not in cur_square:
                            cur_square.add(matrix[row][col])
                        else:
                            return False
                    else:
                        return False # invalid value
    return True

def resolve_sudoku(matrix, row_map, col_map, square_map, cur_row, cur_col):
    global found
    if found:
        return
    if cur_row == len(matrix):
        found = True
        for r in matrix:
            print r
    elif cur_col == len(matrix[0]):
        resolve_sudoku(matrix, row_map, col_map, square_map, cur_row+1, 0)
    elif matrix[cur_row][cur_col] != 0:
        resolve_sudoku(matrix, row_map, col_map, square_map, cur_row, cur_col+1)
    else:
        for val in range(1,10):
            square_x = cur_row / 3
            square_y = cur_col / 3
            if val in row_map[cur_row] or val in col_map[cur_col] or val in square_map[(square_x, square_y)]:
                continue
            else:
                row_map[cur_row].add(val)
                col_map[cur_col].add(val)
                square_map[(square_x, square_y)].add(val)
                matrix[cur_row][cur_col] = val
                resolve_sudoku(matrix, row_map, col_map, square_map, cur_row, cur_col+1)
                row_map[cur_row].remove(val)
                col_map[cur_col].remove(val)
                square_map[(square_x, square_y)].remove(val)
                matrix[cur_row][cur_col] = 0
                if found:
                    return
if __name__ == "__main__":
    matrix = []
    for row in range(9):
        cur_row = []
        for col in range(9):
            if random.random() < 0.1:
                cur_row.append(random.randint(1,9))
            else:
                cur_row.append(0)
        matrix.append(cur_row)
    for r in matrix:
        print r
    re = check_invalid(matrix)
    print re
    if re:
        # init for row map and col map
        row_map = defaultdict(set)
        col_map = defaultdict(set)
        square_map = defaultdict(set)
        for row in range(len(matrix)):
            for col in range(len(matrix[0])):
                square_x = row / 3
                square_y = row / 3
                if matrix[row][col] != 0:
                    row_map[row].add(matrix[row][col])
                    col_map[col].add(matrix[row][col])
                    square_map[(row, col)].add(matrix[row][col])
        resolve_sudoku(matrix, row_map, col_map, square_map, 0, 0)


Get this bounty!!!

K random combinations of N elements in List in Java

Given a List of N Strings, generate and print all possible combinations of R elements in array and return X random combinations from the result. Following is the code for implementing it:

HackerRank: Repeated String

Problem

Lilah has a string, s, of lowercase English letters that she repeated infinitely many times.

Given an integer, n, find and print the number of letter a‘s in the first letters of Lilah’s infinite string.

Input Format

The first line contains a single string, s.
The second line contains an integer, n.

Constraints

  • 1<=|s|<=100
  • 1<=|n|<=10^12
  • For 25% of the test cases, n <= 10^6

Output Format

Print a single integer denoting the number of letter a’s in the first letters of the infinite string created by repeating infinitely many times.

Sample Input 0

aba
10

Sample Output 0

7

Explanation 0

The first n = 10 letters of the infinite string are abaabaabaa. Because there are 7 a‘s, we print on a new line.

Sample Input 1

a
1000000000000

Sample Output 1

1000000000000

Explanation 1

Because all of the first n=1000000000000 letters of the infinite string are a, we print 1000000000000 on a new line.

Solution

Apache Commons DbUtils Mini Wrapper

This is a very small DB Connector code in Java as a wrapper class to Apache DBUtils.

The Commons DbUtils library is a small set of classes designed to make working with JDBC easier. JDBC resource cleanup code is mundane, error prone work so these classes abstract out all of the cleanup tasks from your code leaving you with what you really wanted to do with JDBC in the first place: query and update data.

Some of the advantages of using DbUtils are:

  • No possibility for resource leaks. Correct JDBC coding isn’t difficult but it is time-consuming and tedious. This often leads to connection leaks that may be difficult to track down.
  • Cleaner, clearer persistence code. The amount of code needed to persist data in a database is drastically reduced. The remaining code clearly expresses your intention without being cluttered with resource cleanup.
  • Automatically populate Java Bean properties from Result Sets. You don’t need to manually copy column values into bean instances by calling setter methods. Each row of the Result Set can be represented by one fully populated bean instance.

DbUtils is designed to be:

  • Small – you should be able to understand the whole package in a short amount of time.
  • Transparent – DbUtils doesn’t do any magic behind the scenes. You give it a query, it executes it and cleans up for you.
  • Fast – You don’t need to create a million temporary objects to work with DbUtils.

DbUtils is not:

  • An Object/Relational bridge – there are plenty of good O/R tools already. DbUtils is for developers looking to use JDBC without all the mundane pieces.
  • A Data Access Object (DAO) framework – DbUtils can be used to build a DAO framework though.
  • An object oriented abstraction of general database objects like a Table, Column, or Primary Key.
  • A heavyweight framework of any kind – the goal here is to be a straightforward and easy to use JDBC helper library.

Wrapper:

HackerRank: Circular Array Rotation

Problem

John Watson performs an operation called a right circular rotation on an array of integers, [a(0),a(1).a(2)...a(n-2),a(n-1)]. After performing one right circular rotation operation, the array is transformed from

[a(0),a(1).a(2)...a(n-2),a(n-1)]

to

[a(n-1),a(0),a(1).a(2)...a(n-2)].

Watson performs this operation k times. To test Sherlock’s ability to identify the current element at a particular position in the rotated array, Watson asks q queries, where each query consists of a single integer, m, for which you must print the element at index in the rotated array (i.e., the value of a(m)).

Input Format

The first line contains space-separated integers, n, k, and q, respectively.
The second line contains space-separated integers, where each integer i describes array element a(i)(where 0 <= i <= n).
Each of the q subsequent lines contains a single integer denoting m.

Constraints

  • 0 <= i <= 10^5
  • 0 <= a(i) <= 10^5
  • 0 <= k <= 10^5
  • 0 <= q <= 500
  • 0 <= m <= N-1

Output Format

For each query, print the value of the element at index m of the rotated array on a new line.

Sample Input
3 2 3
1 2 3
0
1
2
Sample Output
2
3
1

Explanation

After the first rotation, the array becomes [3,1,2].
After the second (and final) rotation, the array becomes [2,3,1].

Let’s refer to the array’s final state as array b. For each query, we just have to print the value of b(m) on a new line:

  • m=0 , so we print 2 on a new line.
  • m=1 , so we print 3 on a new line.
  • m=2 , so we print 1 on a new line.

Soluton

HackerEarth: Battle Of Bots 6: Draughts

Problem:

Sample Game

Draughts is a two player board game which is played on a 8X8 grid of cells and is played on opposite sides of the game-board. Each player has an allocated color, Red ( First Player ) or White ( Second Player ) being conventional. Players take turns involving diagonal moves of uniform game pieces in the forward direction only and mandatory captures by jumping over opponent pieces.

Rules:

  • Player can only move diagonally to the adjacent cell and in forward direction, if the diagonally adjacent cell is vacant.
  • A player may not move an opponent’s piece.
  • If the diagonally adjacent cell contains an opponent’s piece, and the cell immediately beyond it is vacant, the opponent’s piece may be captured (and removed from the game) by jumping over it in the forward direction only.
  • If a player made a jump, then its mandatory to keep on jumping as long as the jump is possible.
  • Player cannot move to the diagonally adjacent cell once the player made a jump.

The game will end when any of the players don’t have any move left. At the end of the game the player with majority of pieces will win.

We will play it on an 8X8 grid. The top left of the grid is [0,0] and the bottom right is [7,7].

Input:
The input will be a 8X8 matrix consisting only of 0o2. Then another line will follow which will contain a number –  1 or 2 which is your player id. In the given matrix, top-left is [0,0] and bottom-right is [7,7]. The x-coordinate increases from left to right, and y-coordinate increases from top to bottom.

The cell marked 0 means it doesn’t contain any stones. The cell marked 1 means it contains first player’s stone which is Red in color. The cell marked 2 means it contains second player’s stone which is white in color.

Output:
In the first line print the coordinates of the cell separated by space, the piece he / she wants to move.
In second line print an integer N, number of steps or jumps the piece will make in one move.
In the next N lines print the coordinates of the cells in which the piece will make jump.
You must take care that you don’t print invalid coordinates. For example, [1,1] might be a valid coordinate in the game play if [1,1] in diagonal to the piece in which is going to jump, but [9,10] will never be. Also if you play an invalid move or your code exceeds the time/memory limit while determining the move, you lose the game.

Starting state
The starting state of the game is the state of the board before the game starts.

0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 2 0 2 0 2 0 2
2 0 2 0 2 0 2 0

First Input
This is the input give to the first player at the start of the game.

0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 2 0 2 0 2 0 2
2 0 2 0 2 0 2 0
1
SAMPLE INPUT
0 1 0 1 0 1 0 1
1 0 1 0 1 0 0 0
0 0 0 0 0 1 0 0
0 0 0 0 2 0 0 0
0 0 0 0 0 0 0 0
0 0 2 0 0 0 0 0
0 0 0 2 0 0 0 2
2 0 2 0 2 0 2 0
1
SAMPLE OUTPUT
2 5
2
4 3
6 1

Explanation

This is player 1’s turn, and the player will move the piece at [2,5] and will make two jumps. First jump will be at [4,3and second jump will be at [6,1]

After his/her move the state of game becomes:

0 1 0 1 0 1 0 1
1 0 1 0 1 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 1 0 2 0 2 0 2
2 0 2 0 2 0 2 0

This state will be fed as input to program of player 2.

Other valid move for the first player is

2 5
1
3 6

But the following are invalid moves.
Case 1:

2 5
1
4 3

Because after making a jump its possible to jump again and its mandatory to jump as long as its possible to jump.

Case 2:

2 5
2
4 3
5 4

Because after making a jump its invalid to move to diagonally adjacent cell.

Here is the code of the Random Bot.

Time Limit:1.0 sec(s) for each input file.
Memory Limit:256 MB
Source Limit:1024 KB

Solution

This is the solution submitted by me

HackerEarth: Battle Of Bots #5: Reversi

Problem:

Reversi is a two player board game which is played on a 10 x 10 grid of cells. Each player has an allocated color, Black ( First Player ) or White ( Second Player ) being conventional. Players take turns placing a stone of their color on a single cell. A player must place a stone on the board, in such a position that there exists at least one straight (horizontal, vertical, or diagonal) occupied line between the new stone and another stone of same color, with one or more contiguous other color stone between them.

During a game, any stone of the opponent’s color that are in a straight line and bounded by the stone just placed and another stone of the current player’s color are turned over to the current player’s color. The game will end when the board is completely filled or both the players don’t have any move left. At the end of the game the player with majority of stone will win.

We will play it on an 10 x 10 grid. The top left of the grid is [0,0] and the bottom right is [9,9]. The rule is that a cell[i,j] is connected to any of top, left, right, or bottom cell.

Input:
The input will be a 10 x 10 matrix consisting only of 0,1,2 or 3. Then another line will follow which will contain a number – 1 or 2 which is your player id.

In the given matrix, top-left is [0,0] and bottom-right is [9,9].

In cell[row,column], row increases from top to bottom and column increases from left to right.

The cell marked 0 means it doesn’t contain any stones. The cell marked 1 means it contains first player’s stone which is Black in color. The cell marked 2 means it contains second player’s stone which is white in color. The cell marked 3 means it is a valid place for player whose turn it is.

Output:
Print the coordinates of the cell separated by space, where you want to play your move. You must take care that you don’t print invalid coordinates. For example, [1] might be a valid coordinate in the game play if cell[i,j]=3, but [10] will never be. Also if you play an invalid move or your code exceeds the time/memory limit while determining the move, you lose the game.

Starting state
The starting state of the game is the state of the board before the game starts.

0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 2 1 0 0 0 0
0 0 0 0 1 2 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0

First Input
This is the input give to the first player at the start of the game.

0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 3 0 0 0 0 0
0 0 0 3 2 1 0 0 0 0
0 0 0 0 1 2 3 0 0 0
0 0 0 0 0 3 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
1

Scoring
The scores are calculated by running tournament of all submissions. Your latest submission will be taken into tournament. Scores are assigned according to the Glicko-2 rating system. For more information and questions, see Bot problem judge.

SAMPLE INPUT

0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 3 0 0 0 0 0
0 0 0 3 2 1 3 0 0 0
0 0 0 0 2 2 0 0 0 0
0 0 0 3 1 1 2 3 0 0
0 0 0 1 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
1
SAMPLE OUTPUT
4 3

Explanation

This is player 1’s turn, and the player puts his/her stone in cell[4,3].
After his/her move the state of game becomes:

0 0 0 0 0 0 0 0 0 0  
0 0 0 0 0 0 0 0 0 0  
0 0 0 0 0 0 0 0 0 0  
0 0 0 3 0 3 0 0 0 0  
0 0 0 1 1 1 0 0 0 0  
0 0 0 3 1 2 0 0 0 0  
0 0 0 3 1 1 2 0 0 0  
0 0 0 1 0 3 0 0 0 0  
0 0 3 0 0 0 0 0 0 0  
0 0 0 0 0 0 0 0 0 0  

This state will be fed as input to program of player 2.


Time Limit:1.0 sec(s) for each input file.
Memory Limit: 256 MB
Source Limit: 1024 KB
Marking Scheme:Marks are awarded if any testcase passes.
Allowed Languages:C, CPP, CLOJURE, CSHARP, D, ERLANG, FSHARP, GO, GROOVY, HASKELL, JAVA, JAVA8, JAVASCRIPT, JAVASCRIPT_NODE, LISP, LISP_SBCL, LUA, OBJECTIVEC, OCAML, OCTAVE, PASCAL, PERL, PHP, PYTHON, PYTHON3, R, RACKET, RUBY, RUST, SCALA, SWIFT, VB

My Solution:

CodeEval: Penultimate Word

Challenge Description:

Write a program which finds the next-to-last word in a string.

Input Sample:

Your program should accept as its first argument a path to a filename. Input example is the following

some line with text
another line

Each line has more than one word.

Output Sample:

Print the next-to-last word in the following way.

with
another

Solution:

 

CodeEval: Shortest Repetition

Challenge Description:

Write a program to determine the shortest repetition in a string.
A string is said to have period p if it can be formed by concatenating one or more repetitions of another string of length p. For example, the string “xyzxyzxyzxyz” has period 3, since it is formed by 4 repetitions of the string “xyz”. It also has periods 6 (two repetitions of “xyzxyz”) and 12 (one repetition of “xyzxyzxyzxyz”).

Input Sample:

Your program should accept as its first argument a path to a filename. Each line will contain a string of up to 80 non-blank characters. E.g.

abcabcabcabc
bcbcbcbcbcbcbcbcbcbcbcbcbcbc
dddddddddddddddddddd
adcdefg

Output Sample:

Print out the smallest period of the input string. E.g.

3
2
1
7

Solution:

 

CodeEval: PASS TRIANGLE

CHALLENGE DESCRIPTION:

By starting at the top of the triangle and moving to adjacent numbers on the row below, the maximum total from top to bottom is 27.

   5
  9 6
 4 6 8
0 7 1 5

5 + 9 + 6 + 7 = 27

INPUT SAMPLE:

Your program should accept as its first argument a path to a filename. Input example is the following:

5
9 6
4 6 8
0 7 1 5

You make also check full input file which will be used for your code evaluation.

OUTPUT SAMPLE:

The correct output is the maximum sum for the triangle. So for the given example the correct answer would be 27

Solution