Merge branch 'master' into aesencryption_add_tests

Author: siriak

DIRECTORY.md

@@ -58,7 +58,7 @@ public void setCoeffs(double[] aCoeffs, double[] bCoeffs) throws IllegalArgument
             throw new IllegalArgumentException("bCoeffs must be of size " + order + ", got " + bCoeffs.length);
         }
 
-        for (int i = 0; i <= order; i++) {
+        for (int i = 0; i < order; i++) {
             coeffsA[i] = aCoeffs[i];
             coeffsB[i] = bCoeffs[i];
         }

src/main/java/com/thealgorithms/audiofilters/IIRFilter.java

@@ -4,33 +4,26 @@
 import java.util.Comparator;
 import java.util.List;
 
-/*
-    * Problem Statement: -
-
-    Given a N*N board with the Knight placed on the first block of an empty board. Moving according
-   to the rules of chess knight must visit each square exactly once. Print the order of each cell in
-   which they are visited.
-
-    Example: -
-
-    Input : N = 8
-
-    Output:
-        0  59  38  33  30  17   8  63
-        37  34  31  60   9  62  29  16
-        58   1  36  39  32  27  18   7
-        35  48  41  26  61  10  15  28
-        42  57   2  49  40  23   6  19
-        47  50  45  54  25  20  11  14
-        56  43  52   3  22  13  24   5
-        51  46  55  44  53   4  21  12
-
+/**
+ * The KnightsTour class solves the Knight's Tour problem using backtracking.
+ *
+ * Problem Statement:
+ * Given an N*N board with a knight placed on the first block, the knight must
+ * move according to chess rules and visit each square on the board exactly once.
+ * The class outputs the sequence of moves for the knight.
+ *
+ * Example:
+ * Input: N = 8 (8x8 chess board)
+ * Output: The sequence of numbers representing the order in which the knight visits each square.
  */
 public final class KnightsTour {
     private KnightsTour() {
     }
 
+    // The size of the chess board (12x12 grid, with 2 extra rows/columns as a buffer around a 8x8 area)
     private static final int BASE = 12;
+
+    // Possible moves for a knight in chess
     private static final int[][] MOVES = {
         {1, -2},
         {2, -1},
@@ -40,36 +33,40 @@ private KnightsTour() {
         {-2, 1},
         {-2, -1},
         {-1, -2},
-    }; // Possible moves by knight on chess
-    private static int[][] grid; // chess grid
-    private static int total; // total squares in chess
+    };
+
+    // Chess grid representing the board
+    static int[][] grid;
+
+    // Total number of cells the knight needs to visit
+    static int total;
 
-    public static void main(String[] args) {
+    /**
+     * Resets the chess board to its initial state.
+     * Initializes the grid with boundary cells marked as -1 and internal cells as 0.
+     * Sets the total number of cells the knight needs to visit.
+     */
+    public static void resetBoard() {
         grid = new int[BASE][BASE];
         total = (BASE - 4) * (BASE - 4);
-
         for (int r = 0; r < BASE; r++) {
             for (int c = 0; c < BASE; c++) {
                 if (r < 2 || r > BASE - 3 || c < 2 || c > BASE - 3) {
-                    grid[r][c] = -1;
+                    grid[r][c] = -1; // Mark boundary cells
                 }
             }
         }
-
-        int row = 2 + (int) (Math.random() * (BASE - 4));
-        int col = 2 + (int) (Math.random() * (BASE - 4));
-
-        grid[row][col] = 1;
-
-        if (solve(row, col, 2)) {
-            printResult();
-        } else {
-            System.out.println("no result");
-        }
     }
 
-    // Return True when solvable
-    private static boolean solve(int row, int column, int count) {
+    /**
+     * Recursive method to solve the Knight's Tour problem.
+     *
+     * @param row   The current row of the knight
+     * @param column The current column of the knight
+     * @param count  The current move number
+     * @return True if a solution is found, False otherwise
+     */
+    static boolean solve(int row, int column, int count) {
         if (count > total) {
             return true;
         }
@@ -80,49 +77,72 @@ private static boolean solve(int row, int column, int count) {
             return false;
         }
 
+        // Sort neighbors by Warnsdorff's rule (fewest onward moves)
         neighbor.sort(Comparator.comparingInt(a -> a[2]));
 
         for (int[] nb : neighbor) {
-            row = nb[0];
-            column = nb[1];
-            grid[row][column] = count;
-            if (!orphanDetected(count, row, column) && solve(row, column, count + 1)) {
+            int nextRow = nb[0];
+            int nextCol = nb[1];
+            grid[nextRow][nextCol] = count;
+            if (!orphanDetected(count, nextRow, nextCol) && solve(nextRow, nextCol, count + 1)) {
                 return true;
             }
-            grid[row][column] = 0;
+            grid[nextRow][nextCol] = 0; // Backtrack
         }
 
         return false;
     }
 
-    // Returns List of neighbours
-    private static List<int[]> neighbors(int row, int column) {
+    /**
+     * Returns a list of valid neighboring cells where the knight can move.
+     *
+     * @param row   The current row of the knight
+     * @param column The current column of the knight
+     * @return A list of arrays representing valid moves, where each array contains:
+     *         {nextRow, nextCol, numberOfPossibleNextMoves}
+     */
+    static List<int[]> neighbors(int row, int column) {
         List<int[]> neighbour = new ArrayList<>();
 
         for (int[] m : MOVES) {
             int x = m[0];
             int y = m[1];
-            if (grid[row + y][column + x] == 0) {
+            if (row + y >= 0 && row + y < BASE && column + x >= 0 && column + x < BASE && grid[row + y][column + x] == 0) {
                 int num = countNeighbors(row + y, column + x);
                 neighbour.add(new int[] {row + y, column + x, num});
             }
         }
         return neighbour;
     }
 
-    // Returns the total count of neighbors
-    private static int countNeighbors(int row, int column) {
+    /**
+     * Counts the number of possible valid moves for a knight from a given position.
+     *
+     * @param row    The row of the current position
+     * @param column The column of the current position
+     * @return The number of valid neighboring moves
+     */
+    static int countNeighbors(int row, int column) {
         int num = 0;
         for (int[] m : MOVES) {
-            if (grid[row + m[1]][column + m[0]] == 0) {
+            int x = m[0];
+            int y = m[1];
+            if (row + y >= 0 && row + y < BASE && column + x >= 0 && column + x < BASE && grid[row + y][column + x] == 0) {
                 num++;
             }
         }
         return num;
     }
 
-    // Returns true if it is orphan
-    private static boolean orphanDetected(int count, int row, int column) {
+    /**
+     * Detects if moving to a given position will create an orphan (a position with no further valid moves).
+     *
+     * @param count   The current move number
+     * @param row     The row of the current position
+     * @param column  The column of the current position
+     * @return True if an orphan is detected, False otherwise
+     */
+    static boolean orphanDetected(int count, int row, int column) {
         if (count < total - 1) {
             List<int[]> neighbor = neighbors(row, column);
             for (int[] nb : neighbor) {
@@ -133,17 +153,4 @@ private static boolean orphanDetected(int count, int row, int column) {
         }
         return false;
     }
-
-    // Prints the result grid
-    private static void printResult() {
-        for (int[] row : grid) {
-            for (int i : row) {
-                if (i == -1) {
-                    continue;
-                }
-                System.out.printf("%2d ", i);
-            }
-            System.out.println();
-        }
-    }
 }

src/main/java/com/thealgorithms/backtracking/KnightsTour.java

@@ -1,51 +1,72 @@
 package com.thealgorithms.backtracking;
 
-/*
-Word Search Problem (https://en.wikipedia.org/wiki/Word_search)
-
-Given an m x n grid of characters board and a string word, return true if word exists in the grid.
-
-The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are
-those horizontally or vertically neighboring. The same letter cell may not be used more than once.
-
-For example,
-Given board =
-
-[
- ['A','B','C','E'],
- ['S','F','C','S'],
- ['A','D','E','E']
-]
-word = "ABCCED", -> returns true,
-word = "SEE", -> returns true,
-word = "ABCB", -> returns false.
-*/
-
-/*
-   Solution
-   Depth First Search in matrix (as multiple sources possible) with backtracking
-   like finding cycle in a directed graph. Maintain a record of path
-
-   Tx = O(m * n * 3^L): for each cell, we look at 3 options (not 4 as that one will be visited), we
-   do it L times Sx = O(L) : stack size is max L
-*/
-
+/**
+ * Word Search Problem
+ *
+ * This class solves the word search problem where given an m x n grid of characters (board)
+ * and a target word, the task is to check if the word exists in the grid.
+ * The word can be constructed from sequentially adjacent cells (horizontally or vertically),
+ * and the same cell may not be used more than once in constructing the word.
+ *
+ * Example:
+ * - For board =
+ *     [
+ *       ['A','B','C','E'],
+ *       ['S','F','C','S'],
+ *       ['A','D','E','E']
+ *     ]
+ *   and word = "ABCCED", -> returns true
+ *   and word = "SEE",    -> returns true
+ *   and word = "ABCB",   -> returns false
+ *
+ * Solution:
+ * - Depth First Search (DFS) with backtracking is used to explore possible paths from any cell
+ *   matching the first letter of the word. DFS ensures that we search all valid paths, while
+ *   backtracking helps in reverting decisions when a path fails to lead to a solution.
+ *
+ * Time Complexity: O(m * n * 3^L)
+ *  - m = number of rows in the board
+ *  - n = number of columns in the board
+ *  - L = length of the word
+ *  - For each cell, we look at 3 possible directions (since we exclude the previously visited direction),
+ *    and we do this for L letters.
+ *
+ * Space Complexity: O(L)
+ *  - Stack space for the recursive DFS function, where L is the maximum depth of recursion (length of the word).
+ */
 public class WordSearch {
     private final int[] dx = {0, 0, 1, -1};
     private final int[] dy = {1, -1, 0, 0};
     private boolean[][] visited;
     private char[][] board;
     private String word;
 
+    /**
+     * Checks if the given (x, y) coordinates are valid positions in the board.
+     *
+     * @param x The row index.
+     * @param y The column index.
+     * @return True if the coordinates are within the bounds of the board; false otherwise.
+     */
     private boolean isValid(int x, int y) {
         return x >= 0 && x < board.length && y >= 0 && y < board[0].length;
     }
 
+    /**
+     * Performs Depth First Search (DFS) from the cell (x, y)
+     * to search for the next character in the word.
+     *
+     * @param x       The current row index.
+     * @param y       The current column index.
+     * @param nextIdx The index of the next character in the word to be matched.
+     * @return True if a valid path is found to match the remaining characters of the word; false otherwise.
+     */
     private boolean doDFS(int x, int y, int nextIdx) {
         visited[x][y] = true;
         if (nextIdx == word.length()) {
             return true;
         }
+
         for (int i = 0; i < 4; ++i) {
             int xi = x + dx[i];
             int yi = y + dy[i];
@@ -56,10 +77,19 @@ private boolean doDFS(int x, int y, int nextIdx) {
                 }
             }
         }
-        visited[x][y] = false;
+
+        visited[x][y] = false; // Backtrack
         return false;
     }
 
+    /**
+     * Main function to check if the word exists in the board. It initiates DFS from any
+     * cell that matches the first character of the word.
+     *
+     * @param board The 2D grid of characters (the board).
+     * @param word  The target word to search for in the board.
+     * @return True if the word exists in the board; false otherwise.
+     */
     public boolean exist(char[][] board, String word) {
         this.board = board;
         this.word = word;

src/main/java/com/thealgorithms/backtracking/WordSearch.java

@@ -48,4 +48,32 @@ public long countSetBits(long num) {
         }
         return cnt;
     }
+
+    /**
+     * This approach takes O(1) running time to count the set bits, but requires a pre-processing.
+     *
+     * So, we divide our 32-bit input into 8-bit chunks, with four chunks. We have 8 bits in each chunk.
+     *
+     * Then the range is from 0-255 (0 to 2^7).
+     * So, we may need to count set bits from 0 to 255 in individual chunks.
+     *
+     * @param num takes a long number
+     * @return the count of set bits in the binary equivalent
+     */
+    public int lookupApproach(int num) {
+        int[] table = new int[256];
+        table[0] = 0;
+
+        for (int i = 1; i < 256; i++) {
+            table[i] = (i & 1) + table[i >> 1]; // i >> 1 equals to i/2
+        }
+
+        int res = 0;
+        for (int i = 0; i < 4; i++) {
+            res += table[num & 0xff];
+            num >>= 8;
+        }
+
+        return res;
+    }
 }