Programming Interview Questions

dee707

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Comment hidden because of low score. Click to expand.

of 0 vote

a DP solution where it coputes results in O(nm) and BigOmega(nm). Using Java

package com.dinesh.cup.stringcheated;

//TODO ::
public class StringCheated {
	
	public static void main(String[] args) {
		String a = "asdsasad ds sada sd asd sad hello world im a cat";
		String b = "gfh hfgh h gfh gfhgh Yello world im not a cat gfhgfhgf ghgf gh";

		System.out.println(hasCopied(a,b, 10));
		System.out.println(hasCopied(a,b, 13));
		System.out.println(hasCopied(a,b, 17));
	}

	private static boolean hasCopied(String a, String b, int len) {
		int[][] copyMatrix = new int[a.length()][b.length()];
		for (int i=0; i<a.length(); i++) {
			for (int j=0; j<b.length(); j++) {
				if (a.charAt(i) == b.charAt(j)) {
					int prevValue = 0;
					if (i>0 && j>0) {
						prevValue = copyMatrix[i-1][j-1];
					}
					int newValue = prevValue + 1;
					copyMatrix[i][j] = newValue;
					if (newValue > len) {
						System.out.print(a.substring(i-len, i) + " : ");
						return true;
					}
				}
			}
		}
		return false;
	}

}

- dee707 October 18, 2016 | Flag Reply

Comment hidden because of low score. Click to expand.

of 0 vote

Java Code based on above discussion

import java.util.Arrays;
import java.util.BitSet;
import java.util.List;

// Given a list of 4 billion integers, 
// find an integer not in the list using 4MB of memory. (interview was in Java)
public class MissingIntFinder {
	
	// 4 MB can accommodate -> 4*8*1024*1024 = 32*(2^20) = 2^25
	// 32 bit integer can have possible -> 2 ^ 32 numbers
	
	// so if we assume the Integer range is 32 bit and we'll do the scan in 2 stages
	// (0 -> 16bit) -> (16 -> 32bit)
	
	// Following is a result of 1 stage scan

	public static void main(String[] args) {
	
		List<Integer> data = Arrays.asList(0,1,5,3,4,5,12,2,16,21);
		
		BitSet bitSet = new BitSet(2^16);
		for (Integer val : data) {
			bitSet.set(val);
		}
		
		// nextClearBit : Returns the index of the first bit that is set to false that occurs 
		//                on or after the specified starting index.
		int nextClearBit = bitSet.nextClearBit(0);
		System.out.println(nextClearBit);
	}
}

- dee707 October 18, 2016 | Flag Reply

Comment hidden because of low score. Click to expand.

of 0 vote

Here is a solution which is very similar to one provided here. But still this is a brute-force solution.

public class AnyDirectionWordFinder {
	
	public static void main(String[] args) {
		String[][] letters = new String[][]{
			{"a", "b", "c", "d", "e", "f"},
			{"z", "n", "a", "b", "c", "f"},
			{"f", "g", "f", "a", "b", "c"},
		};
		
		System.out.println(letters.length); // << NOTE THAT THIS IS "3"
		System.out.println(letters[0].length);  // << NOTE THAT THIS IS "6"
		
		System.out.println(isWordFound(letters, "fnz")); // true
		System.out.println(isWordFound(letters, "gng")); // false
		System.out.println(isWordFound(letters, "fbz")); // false
	}

	private static boolean isWordFound(String[][] letters, String word) {
		for (int x=0; x<letters.length; x++) {
			for (int y=0; y<letters[0].length; y++) {
				boolean[][] visited = new boolean[letters.length][letters[0].length];			
				if (traverse(word, letters, visited, x, y)) {
					return true;
				}
			}
		}
		return false;
	}

	private static boolean traverse(String wordToMatch, String[][] letters, boolean[][] visited, int x, int y) {
		if (wordToMatch.length() == 0) {
			return true;
		}
		if (x >= letters.length || y >= letters[0].length
				|| x < 0 || y < 0) {
			return false;
		}
		String firstLetter = wordToMatch.substring(0,1);
		if (!visited[x][y] && letters[x][y].equals(firstLetter)) {
			visited[x][y] = true;
			String newWordToMatch = wordToMatch.substring(1);
			return 
					traverse(newWordToMatch, letters, visited, x+1, y) ||
					traverse(newWordToMatch, letters, visited, x-1, y) ||
					traverse(newWordToMatch, letters, visited, x+1, y+1) ||
					traverse(newWordToMatch, letters, visited, x-1, y-1) ||
					traverse(newWordToMatch, letters, visited, x, y+1) ||
					traverse(newWordToMatch, letters, visited, x, y-1) ||
					traverse(newWordToMatch, letters, visited, x-1, y+1) ||
					traverse(newWordToMatch, letters, visited, x+1, y-1);
		}
		return false;
	}

}

- dee707 October 18, 2016 | Flag Reply

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