feat: Implement Levenshtein distance and BFS

algorithms/dp/levenshtein-distance/levenshteinDistance.test.ts

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+import {levenshteinDistance} from './levenshteinDistance';
+
+describe('levenshteinDistance', () => {
+  test('identical strings should have a distance of 0', () => {
+    expect(levenshteinDistance('abc', 'abc')).toBe(0);
+    expect(levenshteinDistance('', '')).toBe(0);
+    expect(levenshteinDistance('hello', 'hello')).toBe(0);
+  });
+  test('empty string vs non-empty string should have distance equal to length', () => {
+    expect(levenshteinDistance('', 'abc')).toBe(3);
+    expect(levenshteinDistance('abc', '')).toBe(3);
+  });
+  test('simple single operation cases', () => {
+    expect(levenshteinDistance('abc', 'abcd')).toBe(1);
+    expect(levenshteinDistance('abc', 'ab')).toBe(1);
+    expect(levenshteinDistance('abc', 'abd')).toBe(1);
+  });
+  test('complex edit distance cases', () => {
+    expect(levenshteinDistance('kitten', 'sitting')).toBe(3);
+    expect(levenshteinDistance('saturday', 'sunday')).toBe(3);
+    expect(levenshteinDistance('intention', 'execution')).toBe(5);
+  });
+  test('case sensitivity', () => {
+    expect(levenshteinDistance('abc', 'ABC')).toBe(3);
+    expect(levenshteinDistance('Hello', 'hello')).toBe(1);
+  });
+});

algorithms/dp/levenshtein-distance/levenshteinDistance.ts

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+/**
+ * Calculates the Levenshtein distance (edit distance) between two strings.
+ * The edit distance is the minimum number of single-character operations
+ * (insertions, deletions, or substitutions) required to change one string into the other.
+ *
+ * @param {string} s1 - The first string.
+ * @param {string} s2 - The second string.
+ * @returns {number} The minimum number of operations required to transform s1 into s2.
+ */
+export function levenshteinDistance(s1: string, s2: string): number {
+  const m = s1.length;
+  const n = s2.length;
+  // Create a matrix of size (m+1) x (n+1)
+  const dp: number[][] = Array.from({length: m + 1}, () => Array(n + 1).fill(0));
+  // Initialize the first column
+  for (let i = 0; i <= m; i++) {
+    dp[i][0] = i;
+  }
+  // Initialize the first row
+  for (let j = 0; j <= n; j++) {
+    dp[0][j] = j;
+  }
+  // Fill the matrix
+  for (let i = 1; i <= m; i++) {
+    for (let j = 1; j <= n; j++) {
+      if (s1[i - 1] === s2[j - 1]) {
+        dp[i][j] = dp[i - 1][j - 1]; // No operation needed
+      } else {
+        dp[i][j] = 1 + Math.min(dp[i - 1][j], dp[i][j - 1], dp[i - 1][j - 1]);
+      }
+    }
+  }
+  return dp[m][n];
+}

algorithms/search/bfs/bfs.test.ts

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+import {bfs, reconstructPath} from './bfs';
+
+describe('BFS Algorithm', () => {
+  test('should find the shortest paths in a simple graph', () => {
+    const graph = [
+      [1, 2], // Edges from vertex 0
+      [0, 3, 4], // Edges from vertex 1
+      [0, 5], // Edges from vertex 2
+      [1], // Edges from vertex 3
+      [1], // Edges from vertex 4
+      [2], // Edges from vertex 5
+    ];
+    const {distances, predecessors} = bfs(graph, 0);
+    expect(distances).toEqual([0, 1, 1, 2, 2, 2]);
+    expect(predecessors).toEqual([-1, 0, 0, 1, 1, 2]);
+  });
+  test('should handle disconnected graph', () => {
+    const graph = [
+      [1], // Edges from vertex 0
+      [0], // Edges from vertex 1
+      [3], // Edges from vertex 2
+      [2], // Edges from vertex 3
+      [], // Edges from vertex 4 (isolated)
+    ];
+    const {distances, predecessors} = bfs(graph, 0);
+    expect(distances).toEqual([0, 1, Infinity, Infinity, Infinity]);
+    expect(predecessors).toEqual([-1, 0, -1, -1, -1]);
+    expect(distances[2]).toBe(Infinity);
+    expect(distances[3]).toBe(Infinity);
+    expect(distances[4]).toBe(Infinity);
+    expect(predecessors[2]).toBe(-1);
+    expect(predecessors[3]).toBe(-1);
+    expect(predecessors[4]).toBe(-1);
+  });
+  test('should throw error for invalid start vertex', () => {
+    const graph = [[1, 2], [0], [0]];
+    expect(() => bfs(graph, -1)).toThrow('Start vertex is out of range');
+    expect(() => bfs(graph, 3)).toThrow('Start vertex is out of range');
+  });
+  test('should handle a graph with a single vertex', () => {
+    const graph = [[]];
+    const {distances, predecessors} = bfs(graph, 0);
+    expect(distances).toEqual([0]);
+    expect(predecessors).toEqual([-1]);
+  });
+  test('should handle BFS on a tree structure', () => {
+    const graph = [
+      [1, 2], // Root has two children
+      [3, 4], // Left child has two children
+      [5], // Right child has one child
+      [],
+      [],
+      [], // Leaf nodes
+    ];
+    const {distances, predecessors} = bfs(graph, 0);
+    expect(distances).toEqual([0, 1, 1, 2, 2, 2]);
+    expect(predecessors).toEqual([-1, 0, 0, 1, 1, 2]);
+  });
+  test('should handle cyclic graphs correctly', () => {
+    // A graph with cycles: 0-1-2-0 and 3-4-5-3
+    const graph = [
+      [1, 2], // Edges from vertex 0
+      [0, 2], // Edges from vertex 1
+      [0, 1], // Edges from vertex 2
+      [4, 5], // Edges from vertex 3
+      [3, 5], // Edges from vertex 4
+      [3, 4], // Edges from vertex 5
+    ];
+    const {distances} = bfs(graph, 0);
+    expect(distances[0]).toBe(0);
+    expect(distances[1]).toBe(1);
+    expect(distances[2]).toBe(1);
+    expect(distances[3]).toBe(Infinity);
+    expect(distances[4]).toBe(Infinity);
+    expect(distances[5]).toBe(Infinity);
+
+    const result2 = bfs(graph, 3);
+    expect(result2.distances[3]).toBe(0);
+    expect(result2.distances[4]).toBe(1);
+    expect(result2.distances[5]).toBe(1);
+    expect(result2.distances[0]).toBe(Infinity);
+    expect(result2.distances[1]).toBe(Infinity);
+    expect(result2.distances[2]).toBe(Infinity);
+  });
+
+  test('should handle large graphs efficiently', () => {
+    // Create a larger graph (path graph with 1000 vertices)
+    const largeGraph: number[][] = Array(1000)
+      .fill(0)
+      .map((_, i) => {
+        if (i === 0) return [1];
+        if (i === 999) return [998];
+        return [i - 1, i + 1];
+      });
+
+    const startTime = performance.now();
+    const {distances} = bfs(largeGraph, 0);
+    const endTime = performance.now();
+
+    expect(distances[0]).toBe(0);
+    expect(distances[1]).toBe(1);
+    expect(distances[10]).toBe(10);
+    expect(distances[100]).toBe(100);
+    expect(distances[999]).toBe(999);
+    expect(endTime - startTime).toBeLessThan(1000); // Should complete in less than 1 second
+  });
+});
+
+describe('reconstructPath', () => {
+  test('should reconstruct the correct path', () => {
+    const graph = [[1, 2], [3], [4], [], []];
+    const {predecessors} = bfs(graph, 0);
+    const path1 = reconstructPath(0, 3, predecessors);
+    expect(path1).toEqual([0, 1, 3]);
+
+    const path2 = reconstructPath(0, 4, predecessors);
+    expect(path2).toEqual([0, 2, 4]);
+  });
+
+  test('should handle path from vertex to itself', () => {
+    const graph = [[1], [2], []];
+    const {predecessors} = bfs(graph, 0);
+
+    const path = reconstructPath(0, 0, predecessors);
+    expect(path).toEqual([0]);
+  });
+
+  test('should return null for unreachable vertices', () => {
+    const graph = [[1], [0], [3], [2]];
+    const {predecessors, distances} = bfs(graph, 0);
+
+    expect(distances[2]).toBe(Infinity);
+    expect(distances[3]).toBe(Infinity);
+
+    const path = reconstructPath(0, 2, predecessors);
+    expect(path).toBeNull();
+  });
+
+  test('should handle invalid target vertex', () => {
+    const graph = [[1], [0]];
+    const {predecessors} = bfs(graph, 0);
+
+    const path = reconstructPath(0, 2, predecessors);
+    expect(path).toBeNull();
+  });
+});

algorithms/search/bfs/bfs.ts

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+/**
+ * Performs a breadth-first search (BFS) traversal on a graph.
+ * BFS explores all vertices at the current level before moving to vertices at the next level.
+ *
+ * @param {number[][]} graph - An adjacency list representation of the graph.
+ * @param {number} start - The starting vertex.
+ * @returns {{ distances: number[], predecessors: number[] }}
+ *          An object containing distances from the start vertex to all other vertices,
+ *          and predecessors array for path reconstruction.
+ * @throws {Error} If the start vertex is out of range.
+ */
+export function bfs(
+  graph: number[][],
+  start: number,
+): {
+  distances: number[];
+  predecessors: number[];
+} {
+  if (start < 0 || start >= graph.length) {
+    throw new Error('Start vertex is out of range');
+  }
+  const n = graph.length;
+  const distances: number[] = Array(n).fill(Infinity);
+  const predecessors: number[] = Array(n).fill(-1);
+  const visited: boolean[] = Array(n).fill(false);
+  // Initialize the queue with the start vertex
+  const queue: number[] = [start];
+  distances[start] = 0;
+  visited[start] = true;
+
+  while (queue.length > 0) {
+    const current = queue.shift()!;
+    // Visit all adjacent vertices
+    for (const neighbor of graph[current]) {
+      if (!visited[neighbor]) {
+        visited[neighbor] = true;
+        distances[neighbor] = distances[current] + 1;
+        predecessors[neighbor] = current;
+        queue.push(neighbor);
+      }
+    }
+  }
+
+  return {distances, predecessors};
+}
+
+/**
+ * Reconstructs the shortest path from a source vertex to a target vertex
+ * using the predecessors array obtained from BFS.
+ *
+ * @param {number} source - The source vertex.
+ * @param {number} target - The target vertex.
+ * @param {number[]} predecessors - The predecessors array from BFS.
+ * @returns {number[] | null} The path from source to target, or null if no path exists.
+ */
+export function reconstructPath(
+  source: number,
+  target: number,
+  predecessors: number[],
+): number[] | null {
+  if (
+    target < 0 ||
+    target >= predecessors.length ||
+    (predecessors[target] === -1 && source !== target)
+  ) {
+    return null;
+  }
+
+  const path: number[] = [];
+  let current = target;
+
+  while (current !== -1) {
+    path.unshift(current);
+    if (current === source) {
+      break;
+    }
+    current = predecessors[current];
+  }
+  // Check if the path starts with the source vertex
+  return path[0] === source ? path : null;
+}