Bit manipulation

src/TechInterview/BitManipulation.java

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+package TechInterview;
+
+import java.util.List;
+
+/**
+ * This class has solution for Bit Manipulation questions
+ * from book Cracking the Coding Interview
+ * @author JayOnCaffeine
+ *
+ */
+public class BitManipulation {
+
+	public boolean getBitFromNthPosition(int binum, int position) {
+		return ((binum & (1 << position)) != 0);
+	}
+
+	public int updateBitOnNthPosition(int binum, int value, int position) {
+		int mask = ~(1 << position);
+		binum = binum & mask;
+		return binum | (value << position);
+	}
+
+	public void printBinary(final int num) {
+		System.out.println(Integer.toBinaryString(num));
+	}
+
+	public void printMessage(final Object msg) {
+		System.out.println(msg);
+	}
+
+	public boolean isBinaryString(final String num) {
+		for(Character digit : num.toCharArray()) {
+			if(Integer.parseInt(digit.toString()) > 1) {
+				return false;
+			}
+		}
+		return true;
+	}
+
+	/*
+	 * You are given two 32-bit numbers, N and M, and two bit positions, i and j.
+	 * Write a method to insert M into N such that M starts at bit j and ends at bit i. 
+	 * You can assume that the bits j through i have enough space to fit all of M.
+	 */
+	public void insertion(String N, String M, int i, int j) {
+		int Nint = Integer.parseInt(N, 2);
+		int Mint = Integer.parseInt(M, 2);
+		if ( j > i && (j-i+1) == M.length()) {
+			for(int k = M.length(); k > 0 && j >= i; k--) {
+				if(getBitFromNthPosition(Mint, k-1)) {
+					Nint = updateBitOnNthPosition(Nint, 1, j);
+				} else {
+					Nint = updateBitOnNthPosition(Nint, 0, j);
+				}
+				j--;
+			}
+			printBinary(Nint);
+		} else {
+			printMessage(" j should be greater than i ");
+		}
+	}
+
+	/*
+	 * Given a real number between O and 1 (e.g., 0.72) that is passed in as a double, 
+	 * print the binary representation. If the number cannot be represented accurately in binary 
+	 * with at most 32 characters, print "ERROR:'
+	 */
+	public void binaryToString(double num) {
+		StringBuffer result = new StringBuffer();
+		result.append("0.");
+		while(num > 0) {
+			num = num * 2;
+			byte bit = (byte) (num / 1);
+			result.append(bit);
+			if(bit != 0) {
+				num = num - 1;
+			}
+			if(result.length() > 32) {
+				printMessage("ERROR");
+				return;
+			}
+		}
+		printMessage(result.toString());
+	}
+
+	/*
+	 * You have an integer and you can flip exactly one bit from a 0 to a 1. 
+	 * Write code to find the length of the longest sequence of ls you could create.
+	 */
+	public void flipToWin(String num) {
+		if(!isBinaryString(num)) {
+			num = Integer.toBinaryString(Integer.parseInt(num));
+		}
+
+		int binum = Integer.parseInt(num, 2);
+
+		List<String> flippedNumList = new java.util.ArrayList<String>();
+		int flipped = 0;
+		int iteration = 0;
+		while(iteration < num.length()){
+			int check0Flipped = 0;
+			for(int i = 0; i < num.length(); i++) {
+				if(num.charAt(i) == '0') {
+					check0Flipped++;
+					if(check0Flipped > flipped) {
+						flippedNumList.add(Integer.toBinaryString(updateBitOnNthPosition(binum, 1, num.length() - i - 1)));
+						flipped++;
+						break;
+					}
+				}
+			}
+			iteration++;
+		}
+
+		int highestCount = 0;
+
+		for(String flippedStr : flippedNumList) {
+			int counter = 0;
+			for(int i = 0; i < flippedStr.length(); i++) {
+				if(flippedStr.charAt(i) == '1') {
+					counter++;
+				} else if(flippedStr.charAt(i) == '0') {
+					if(counter > highestCount) {
+						highestCount = counter;
+					}
+					counter = 0;
+				}
+				if(i == flippedStr.length()-1) {
+					if(counter > highestCount) {
+						highestCount = counter;
+					}
+					counter = 0;
+				}
+			}
+		}
+
+		printMessage(highestCount);
+	}
+
+	/*
+	 * Given a positive integer, print the next smallest and the next largest number that
+	 * have the same number of 1 bits in their binary representation.
+	 */
+	public void nextNumber(String num) {
+		if(!isBinaryString(num)) {
+			num = Integer.toBinaryString(Integer.parseInt(num));
+		}
+
+		int binum = Integer.parseInt(num, 2);
+
+		int position = -1;
+
+		// Find position of 0 bit which is next to 1 bit
+		for(int i = num.length()-1; i >= 0; i--) {
+			if(num.charAt(i) == '0') {
+				position = num.length() - i - 1;	
+			} else if(position != -1 && num.charAt(i) == '1') {
+				break;
+			}
+		}
+
+		if(position == -1) {
+			printMessage("Unable to find next numbers");
+			return;
+		}
+
+		int nxtSmallNum = updateBitOnNthPosition(binum, 1, position);
+		nxtSmallNum = updateBitOnNthPosition(nxtSmallNum, 0, position + 1);
+
+		if(position == 0) {
+			// Find position of 0 bit which is next to 1 bit
+			for(int i = num.length()-2; i >= 0; i--) {
+				if(num.charAt(i) == '0') {
+					position = num.length() - i - 1;	
+				} else if(position != 0 && num.charAt(i) == '1') {
+					break;
+				}
+			}
+		}
+		int nxtLargeNum = updateBitOnNthPosition(binum, 1, position);
+		nxtLargeNum = updateBitOnNthPosition(nxtLargeNum, 0, position-1);
+
+		printMessage(num + " small number " + nxtSmallNum);
+		printMessage(num + " large number " + nxtLargeNum);
+	}
+
+	public static void main(String args[]) {
+		final BitManipulation bm = new BitManipulation();
+		bm.printMessage("==== Insertion ====");
+		bm.insertion("10000000000", "10011", 2, 6);
+		bm.insertion("10000000000000", "10010011", 1, 8);
+		bm.insertion("1000000000011", "110010011", 2, 10);
+
+		bm.printMessage("==== Binary To String ====");
+		bm.binaryToString(.625);
+		bm.binaryToString(.78);
+
+		bm.printMessage("==== Flip To Win ====");
+		bm.flipToWin("11011101111");
+		bm.flipToWin("0000000000000000001");
+		bm.flipToWin("1010101010101010101010");
+
+		bm.printMessage("==== Next Number ====");
+		bm.nextNumber("203");
+		bm.nextNumber("999");
+		bm.nextNumber("26");
+	}
+}