// Helene Martin, CSE 142
// I mentioned that array problems are my favorite final exam problems because
// they bring all we've learned together and often include interesting
// algorithmic complexity.

// These are all examples of 15-point array questions (not the 10 point harder one)
import java.util.*;

public class ArrayPractice {
	public static void main(String[] args) {
		int[] a = {6, 3, 2};
		System.out.println(Arrays.toString(delta(a)));
		
		System.out.println(isUnique(a));
		
		String[] b = {"a", "b", "c", "d"};
		swapPairs(b);
		System.out.println(Arrays.toString(b));
		
		System.out.println(Arrays.toString(runningSum(a)));
	}
	
	// Write a static method named delta that accepts an array of integers as a 
	// parameter and returns a new array formed by inserting between each pair of 
	// values the difference between those values.  For example, given this array: 
	// int[] numbers = {3, 8, 6}; The call of delta(numbers) should return the following 
	// array: {3, 5, 8, -2, 6}
	// 6 3 2 3
	// 6 -3 3 -1 2 1 3
	// i   i*2
	// 0 -> 0
	// 1 -> 2
	// 2 -> 4 
	public static int[] delta(int[] a) {
		int[] result = new int[a.length * 2 - 1];
		for (int i = 0; i < a.length - 1; i++) {
			int diff = a[i + 1] - a[i];
			result[i * 2] = a[i];
			result[i * 2 + 1] = diff;			
		}
		
		result[result.length - 1] = a[a.length - 1];
		
		return result;
	}
	
	// Write a static method called runningSum that takes an array of integers
	// as a parameter and that returns a new array that contains the running
	// sum of the numbers in the array (where the i-th value of the new array
	// is the sum of the first i values of the original array)
		public static int[] runningSum(int[] a) {
		int[] sums = new int[a.length];
		if (a.length > 0) { // careful about empty cases!
			sums[0] = a[0]; // fencepost
			for (int i = 1; i < a.length; i++) {
				sums[i] = a[i] + sums[i - 1];
			}
		}
		return sums;
	}
	
	//	Write a static method called isUnique that takes an array of integers as
	// a parameter and that returns true if the values in the list are unique
	// and that returns false otherwise.  The values in the list are considered
	// unique if there is no pair of values that are equal.
	// 4 2 45 1 23 1 2 67 --> false
	public static boolean isUnique(int[] a) {
		for (int i = 0; i < a.length - 1; i++) {
			for (int j = i + 1; j < a.length; j++) {
				if (a[i] == a[j]) {
					// as soon as two match, can't all be unique
					return false;
				} 
			}
		}
		// we ONLY know that everything was unique once we've checked all 
		// values against all others.
		return true;
	}
	
	// Write a method named swapPairs that accepts an array of strings as a parameter 
	// and switches the order of values in a pairwise fashion. Your method should switch 
	// the order of the first two values, then switch the order of the next two, switch 
	// the order of the next two, and so on. If there are an odd number of values, the final 
	// element is not moved.
	// a b c d --> b a d c
	public static void swapPairs(String[] a) {
		// be sure to simulate the code in your mind!
		// we need to go up by twos to make sure we're swapping pairwise
		// we need to stop 1 short of the end to work for the odd-length case
		for (int i = 0; i < a.length - 1; i = i + 2) { // also ok: i += 2
			String temp = a[i];
			a[i] = a[i + 1];
			a[i + 1] = temp;
		}
	}
}