LogicTwo.java

/**
 * 
 */
package coding.bat.solutions;

/**
 * @author Aman Shekhar
 *
 */
public class LogicTwo {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub

	}

	// --------------------------------------------------------------------------------------------

	// We want to make a row of bricks that is goal inches long. We have a number of
	// small bricks (1 inch each) and big bricks (5 inches each). Return true if it
	// is possible to make the goal by choosing from the given bricks. This is a
	// little harder than it looks and can be done without any loops. See also:
	// Introduction to MakeBricks
	//
	//
	// makeBricks(3, 1, 8) → true
	// makeBricks(3, 1, 9) → false
	// makeBricks(3, 2, 10) → true

	public boolean makeBricks(int small, int big, int goal) {
		return goal - big * 5 <= small && goal % 5 <= small;
	}

	// --------------------------------------------------------------------------------------------

	// Given 3 int values, a b c, return their sum. However, if one of the values is
	// the same as another of the values, it does not count towards the sum.
	//
	//
	// loneSum(1, 2, 3) → 6
	// loneSum(3, 2, 3) → 2
	// loneSum(3, 3, 3) → 0
	//
	public int loneSum(int a, int b, int c) {

		if ((a == b) && (b == c))
			return 0;
		else if ((a == c))
			return b;
		else if ((a == b))
			return c;
		else if ((b == c))
			return a;
		return a + b + c;
	}

	// --------------------------------------------------------------------------------------------

	// Given 3 int values, a b c, return their sum. However, if one of the values is
	// 13 then it does not count towards the sum and values to its right do not
	// count. So for example, if b is 13, then both b and c do not count.
	//
	//
	// luckySum(1, 2, 3) → 6
	// luckySum(1, 2, 13) → 3
	// luckySum(1, 13, 3) → 1

	public int luckySum(int a, int b, int c) {
		if (a == 13)
			return 0;
		if (b == 13)
			return a;
		if (c == 13)
			return a + b;
		return a + b + c;
	}

	// --------------------------------------------------------------------------------------------

	// Given 3 int values, a b c, return their sum. However, if any of the values is
	// a teen -- in the range 13..19 inclusive -- then that value counts as 0,
	// except 15 and 16 do not count as a teens. Write a separate helper "public int
	// fixTeen(int n) {"that takes in an int value and returns that value fixed for
	// the teen rule. In this way, you avoid repeating the teen code 3 times (i.e.
	// "decomposition"). Define the helper below and at the same indent level as the
	// main noTeenSum().
	//
	//
	// noTeenSum(1, 2, 3) → 6
	// noTeenSum(2, 13, 1) → 3
	// noTeenSum(2, 1, 14) → 3

	public int noTeenSum(int a, int b, int c) {
		int aT = ((a >= 13 && a < 15 || a > 16 && a <= 19) ? 0 : a);
		int bT = ((b >= 13 && b < 15 || b > 16 && b <= 19) ? 0 : b);
		int cT = ((c >= 13 && c < 15 || c > 16 && c <= 19) ? 0 : c);
		return aT + bT + cT;
	}

	// --------------------------------------------------------------------------------------------

	// For this problem, we'll round an int value up to the next multiple of 10 if
	// its rightmost digit is 5 or more, so 15 rounds up to 20. Alternately, round
	// down to the previous multiple of 10 if its rightmost digit is less than 5, so
	// 12 rounds down to 10. Given 3 ints, a b c, return the sum of their rounded
	// values. To avoid code repetition, write a separate helper "public int
	// round10(int num) {" and call it 3 times. Write the helper entirely below and
	// at the same indent level as roundSum().
	//
	//
	// roundSum(16, 17, 18) → 60
	// roundSum(12, 13, 14) → 30
	// roundSum(6, 4, 4) → 10

	public int roundSum(int a, int b, int c) {
		int aT = (a % 10 < 5) ? (a / 10 * 10) : (a / 10 * 10 + 10);
		int bT = (b % 10 < 5) ? (b / 10 * 10) : (b / 10 * 10 + 10);
		int cT = (c % 10 < 5) ? (c / 10 * 10) : (c / 10 * 10 + 10);
		return aT + bT + cT;
	}

	// --------------------------------------------------------------------------------------------

	// Given three ints, a b c, return true if one of b or c is "close" (differing
	// from a by at most 1), while the other is "far", differing from both other
	// values by 2 or more. Note: Math.abs(num) computes the absolute value of a
	// number.
	//
	//
	// closeFar(1, 2, 10) → true
	// closeFar(1, 2, 3) → false
	// closeFar(4, 1, 3) → true

	public boolean closeFar(int a, int b, int c) {
		return Math.abs(b - c) >= 2
				&& (Math.abs(a - b) <= 1 && Math.abs(a - c) >= 2 || Math.abs(a - c) <= 1 && Math.abs(a - b) >= 2);
	}

	// Given 2 int values greater than 0, return whichever value is nearest to 21
	// without going over. Return 0 if they both go over.
	//
	//
	// blackjack(19, 21) → 21
	// blackjack(21, 19) → 21
	// blackjack(19, 22) → 19
	//

	// --------------------------------------------------------------------------------------------

	public int blackjack(int a, int b) {
		int aUpdated = Math.abs(21 - a);
		int bUpdated = Math.abs(21 - b);
		if (a > 21 && b > 21)
			return 0;
		if (a > 21)
			return b;
		if (b > 21)
			return a;
		if (aUpdated > bUpdated)
			return b;
		return a;
	}

	// --------------------------------------------------------------------------------------------

	// Given three ints, a b c, one of them is small, one is medium and one is
	// large. Return true if the three values are evenly spaced, so the difference
	// between small and medium is the same as the difference between medium and
	// large.
	//
	//
	// evenlySpaced(2, 4, 6) → true
	// evenlySpaced(4, 6, 2) → true
	// evenlySpaced(4, 6, 3) → false

	public boolean evenlySpaced(int a, int b, int c) {
		int min = Math.min(Math.min(a, b), c);
		int mid = Math.max(Math.min(a, b), c);
		int mid2 = Math.min(Math.max(a, b), c);
		int max = Math.max(Math.max(a, b), c);
		return Math.abs(mid - min) == Math.abs(mid - max) || Math.abs(mid2 - min) == Math.abs(mid2 - max);
	}

	// --------------------------------------------------------------------------------------------

	// We want make a package of goal kilos of chocolate. We have small bars (1 kilo
	// each) and big bars (5 kilos each). Return the number of small bars to use,
	// assuming we always use big bars before small bars. Return -1 if it can't be
	// done.
	//
	//
	// makeChocolate(4, 1, 9) → 4
	// makeChocolate(4, 1, 10) → -1
	// makeChocolate(4, 1, 7) → 2
	//

	public int makeChocolate(int small, int big, int goal) {
		int maxBig = goal / 5;
		if (big > maxBig)
			return (goal <= 5 * maxBig + small) ? (goal - 5 * maxBig) : -1;
		return (goal <= 5 * big + small) ? (goal - 5 * big) : -1;
	}

}