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

/**
 * @author Aman Shekhar
 * @link https://codingbat.com/java/Array-1
 */
public class ArraysOne {

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

	}

	// Given an array of ints, return true if 6 appears as either the first or last
	// element in the array. The array will be length 1 or more.
	//
	//
	// firstLast6([1, 2, 6]) → true
	// firstLast6([6, 1, 2, 3]) → true
	// firstLast6([13, 6, 1, 2, 3]) → false

	public boolean firstLast6(int[] nums) {
		return nums[0] == 6 || nums[nums.length - 1] == 6;
	}

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

	// Given an array of ints, return true if the array is length 1 or more, and the
	// first element and the last element are equal.

	// sameFirstLast([1, 2, 3]) → false
	// sameFirstLast([1, 2, 3, 1]) → true
	// sameFirstLast([1, 2, 1]) → true

	public boolean sameFirstLast(int[] nums) {
		if (nums.length >= 1)
			return nums[0] == nums[nums.length - 1];
		return false;
	}

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

	// Return an int array length 3 containing the first 3 digits of pi, {3, 1, 4}.
	//
	//
	// makePi() → [3, 1, 4]

	public int[] makePi() {
		int[] pi = { 3, 1, 4 };
		return pi;
	}

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

	// Given 2 arrays of ints, a and b, return true if they have the same first
	// element or they have the same last element. Both arrays will be length 1 or
	// more.
	//
	//
	// commonEnd([1, 2, 3], [7, 3]) → true
	// commonEnd([1, 2, 3], [7, 3, 2]) → false
	// commonEnd([1, 2, 3], [1, 3]) → true

	public boolean commonEnd(int[] a, int[] b) {
		int aLen = a.length;
		int bLen = b.length;
		if (aLen >= 1 && bLen >= 1)
			return (a[0] == b[0]) || (a[aLen - 1]) == b[bLen - 1];
		return false;
	}

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

	// Given an array of ints length 3, return the sum of all the elements.
	//
	//
	// sum3([1, 2, 3]) → 6
	// sum3([5, 11, 2]) → 18
	// sum3([7, 0, 0]) → 7
	//
	public int sum3(int[] nums) {
		return (nums[0] + nums[1] + nums[2]);
	}

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

	// Given an array of ints length 3, return an array with the elements "rotated
	// left" so {1, 2, 3} yields {2, 3, 1}.
	//
	//
	// rotateLeft3([1, 2, 3]) → [2, 3, 1]
	// rotateLeft3([5, 11, 9]) → [11, 9, 5]
	// rotateLeft3([7, 0, 0]) → [0, 0, 7]

	public int[] rotateLeft3(int[] nums) {
		return new int[] { nums[1], nums[2], nums[0] };
	}

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

	// Given an array of ints length 3, return a new array with the elements in
	// reverse order, so {1, 2, 3} becomes {3, 2, 1}.
	//
	//
	// reverse3([1, 2, 3]) → [3, 2, 1]
	// reverse3([5, 11, 9]) → [9, 11, 5]
	// reverse3([7, 0, 0]) → [0, 0, 7]

	public int[] reverse3(int[] nums) {
		return new int[] { nums[2], nums[1], nums[0] };
	}

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

	// Given an array of ints length 3, figure out which is larger, the first or
	// last element in the array, and set all the other elements to be that value.
	// Return the changed array.
	//
	//
	// maxEnd3([1, 2, 3]) → [3, 3, 3]
	// maxEnd3([11, 5, 9]) → [11, 11, 11]
	// maxEnd3([2, 11, 3]) → [3, 3, 3]

	public int[] maxEnd3(int[] nums) {
		int a = nums[0];
		int c = nums[2];

		if (a > c)
			return new int[] { nums[0], nums[0], nums[0] };
		return new int[] { nums[2], nums[2], nums[2] };
	}

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

	// Given an array of ints, return the sum of the first 2 elements in the array.
	// If the array length is less than 2, just sum up the elements that exist,
	// returning 0 if the array is length 0.
	//
	//
	// sum2([1, 2, 3]) → 3
	// sum2([1, 1]) → 2
	// sum2([1, 1, 1, 1]) → 2

	public int sum2(int[] nums) {
		if (nums.length >= 2)
			return nums[0] + nums[1];
		if (nums.length == 1)
			return nums[0];
		return 0;
	}

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

	// Given 2 int arrays, a and b, each length 3, return a new array length 2
	// containing their middle elements.
	//
	//
	// middleWay([1, 2, 3], [4, 5, 6]) → [2, 5]
	// middleWay([7, 7, 7], [3, 8, 0]) → [7, 8]
	// middleWay([5, 2, 9], [1, 4, 5]) → [2, 4]
	//

	public int[] middleWay(int[] a, int[] b) {
		return new int[] { a[1], b[1] };
	}

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

	// Given an array of ints, return a new array length 2 containing the first and
	// last elements from the original array. The original array will be length 1 or
	// more.
	//
	//
	// makeEnds([1, 2, 3]) → [1, 3]
	// makeEnds([1, 2, 3, 4]) → [1, 4]
	// makeEnds([7, 4, 6, 2]) → [7, 2]
	//

	public int[] makeEnds(int[] nums) {
		return new int[] { nums[0], nums[nums.length - 1] };
	}

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

	// Given an int array length 2, return true if it contains a 2 or a 3.
	//
	//
	// has23([2, 5]) → true
	// has23([4, 3]) → true
	// has23([4, 5]) → false

	public boolean has23(int[] nums) {
		return ((nums[0] == 2) || (nums[0] == 3) || (nums[1] == 2) || nums[1] == 3);
	}

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

	// Given an int array length 2, return true if it does not contain a 2 or 3.
	//
	//
	// no23([4, 5]) → true
	// no23([4, 2]) → false
	// no23([3, 5]) → false

	public boolean no23(int[] nums) {
		if (nums[0] == 2 || nums[0] == 3)
			return false;
		return !(nums[1] == 2 || nums[1] == 3);
	}

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

	// Given an int array, return a new array with double the length where its last
	// element is the same as the original array, and all the other elements are 0.
	// The original array will be length 1 or more. Note: by default, a new int
	// array contains all 0's.
	//
	//
	// makeLast([4, 5, 6]) → [0, 0, 0, 0, 0, 6]
	// makeLast([1, 2]) → [0, 0, 0, 2]
	// makeLast([3]) → [0, 3]

	public int[] makeLast(int[] nums) {
		int len = nums.length * 2;
		int[] dubsArr = new int[len];
		dubsArr[len - 1] = nums[nums.length - 1];
		return dubsArr;
	}

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

	// Given an int array, return true if the array contains 2 twice, or 3 twice.
	// The array will be length 0, 1, or 2.
	//
	//
	// double23([2, 2]) → true
	// double23([3, 3]) → true
	// double23([2, 3]) → false

	public boolean double23(int[] nums) {
		if (nums.length == 2) {
			if (nums[0] == 2 && nums[1] == 2)
				return true;
			return (nums[0] == 3 && nums[1] == 3);
		}
		return false;
	}

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

	// Given an int array length 3, if there is a 2 in the array immediately
	// followed by a 3, set the 3 element to 0. Return the changed array.
	//
	//
	// fix23([1, 2, 3]) → [1, 2, 0]
	// fix23([2, 3, 5]) → [2, 0, 5]
	// fix23([1, 2, 1]) → [1, 2, 1]

	public int[] fix23(int[] nums) {
		int[] fxArr = { nums[0], nums[1], nums[2] };
		if (nums[0] == 2 && nums[1] == 3)
			fxArr[1] = 0;
		if (nums[1] == 2 && nums[2] == 3)
			fxArr[2] = 0;
		return fxArr;
	}

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

	// Start with 2 int arrays, a and b, of any length. Return how many of the
	// arrays have 1 as their first element.
	//
	//
	// start1([1, 2, 3], [1, 3]) → 2
	// start1([7, 2, 3], [1]) → 1
	// start1([1, 2], []) → 1

	public int start1(int[] a, int[] b) {
		int ones = 0;
		if (a.length >= 1 && a[0] == 1)
			ones += 1;
		if (b.length >= 1 && b[0] == 1)
			ones += 1;
		return ones;
	}

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

	// Start with 2 int arrays, a and b, each length 2. Consider the sum of the
	// values in each array. Return the array which has the largest sum. In event of
	// a tie, return a.
	//
	//
	// biggerTwo([1, 2], [3, 4]) → [3, 4]
	// biggerTwo([3, 4], [1, 2]) → [3, 4]
	// biggerTwo([1, 1], [1, 2]) → [1, 2]

	public int[] biggerTwo(int[] a, int[] b) {
		int a0 = a[0];
		int a1 = a[1];
		int b0 = b[0];
		int b1 = b[1];

		if ((a0 + a1) == (b0 + b1))
			return a;
		if ((a0 + a1) > (b0 + b1))
			return a;
		return b;

	}

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

	// Given an array of ints of even length, return a new array length 2 containing
	// the middle two elements from the original array. The original array will be
	// length 2 or more.
	//
	//
	// makeMiddle([1, 2, 3, 4]) → [2, 3]
	// makeMiddle([7, 1, 2, 3, 4, 9]) → [2, 3]
	// makeMiddle([1, 2]) → [1, 2]

	public int[] makeMiddle(int[] nums) {
		int mLength = nums.length;
		int mHalfLength = mLength / 2;
		if (mLength > 2) {
			int[] mUpdatedArray = new int[2];
			mUpdatedArray[0] = nums[mHalfLength - 1];
			mUpdatedArray[1] = nums[mHalfLength];
			return mUpdatedArray;
		}
		return nums;
	}

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

	// Given 2 int arrays, each length 2, return a new array length 4 containing all
	// their elements.
	//
	//
	// plusTwo([1, 2], [3, 4]) → [1, 2, 3, 4]
	// plusTwo([4, 4], [2, 2]) → [4, 4, 2, 2]
	// plusTwo([9, 2], [3, 4]) → [9, 2, 3, 4]

	public int[] plusTwo(int[] a, int[] b) {
		return new int[] { a[0], a[1], b[0], b[1] };
	}

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

	// Given an array of ints, swap the first and last elements in the array. Return
	// the modified array. The array length will be at least 1.
	//
	//
	// swapEnds([1, 2, 3, 4]) → [4, 2, 3, 1]
	// swapEnds([1, 2, 3]) → [3, 2, 1]
	// swapEnds([8, 6, 7, 9, 5]) → [5, 6, 7, 9, 8]

	public int[] swapEnds(int[] nums) {
		int length = nums.length - 1;
		int temp = nums[0];
		nums[0] = nums[length];
		nums[length] = temp;
		return nums;
	}

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

	// Given an array of ints of odd length, return a new array length 3 containing
	// the elements from the middle of the array. The array length will be at least
	// 3.
	//
	//
	// midThree([1, 2, 3, 4, 5]) → [2, 3, 4]
	// midThree([8, 6, 7, 5, 3, 0, 9]) → [7, 5, 3]
	// midThree([1, 2, 3]) → [1, 2, 3]
	//

	public int[] midThree(int[] nums) {
		int[] halfArr = new int[3];
		int half = nums.length / 2;
		halfArr[0] = nums[half - 1];
		halfArr[1] = nums[half];
		halfArr[2] = nums[half + 1];
		return halfArr;
	}

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

	// Given an array of ints of odd length, look at the first, last, and middle
	// values in the array and return the largest. The array length will be a least
	// 1.
	//
	//
	// maxTriple([1, 2, 3]) → 3
	// maxTriple([1, 5, 3]) → 5
	// maxTriple([5, 2, 3]) → 5
	//

	public int maxTriple(int[] nums) {
		int max = nums[0];
		if (max <= nums[nums.length - 1])
			max = nums[nums.length - 1];
		if (max <= nums[nums.length / 2])
			max = nums[nums.length / 2];
		return max;
	}

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

	// Given an int array of any length, return a new array of its first 2 elements.
	// If the array is smaller than length 2, use whatever elements are present.
	//
	//
	// frontPiece([1, 2, 3]) → [1, 2]
	// frontPiece([1, 2]) → [1, 2]
	// frontPiece([1]) → [1]
	//
	public int[] frontPiece(int[] nums) {
		int[] front;
		if (nums.length >= 2) {
			front = new int[2];
			front[0] = nums[0];
			front[1] = nums[1];
		} else if (nums.length == 1) {
			front = new int[1];
			front[0] = nums[0];
		} else
			front = new int[0];
		return front;
	}

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

	// We'll say that a 1 immediately followed by a 3 in an array is an "unlucky" 1.
	// Return true if the given array contains an unlucky 1 in the first 2 or last 2
	// positions in the array.
	//
	//
	// unlucky1([1, 3, 4, 5]) → true
	// unlucky1([2, 1, 3, 4, 5]) → true
	// unlucky1([1, 1, 1]) → false

	public boolean unlucky1(int[] nums) {
		int lastP = nums.length - 1;
		if (lastP >= 2) {
			if ((nums[0] == 1 && nums[1] == 3) || (nums[1] == 1 && nums[2] == 3))
				return true;
			return (nums[lastP - 1] == 1 && nums[lastP] == 3);
		}
		if (lastP == 1)
			return ((nums[0] == 1 && nums[1] == 3) || (nums[1] == 1 && nums[2] == 3));
		return false;
	}

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

	// Given 2 int arrays, a and b, return a new array length 2 containing, as much
	// as will fit, the elements from a followed by the elements from b. The arrays
	// may be any length, including 0, but there will be 2 or more elements
	// available between the 2 arrays.
	//
	//
	// make2([4, 5], [1, 2, 3]) → [4, 5]
	// make2([4], [1, 2, 3]) → [4, 1]
	// make2([], [1, 2]) → [1, 2]
	//

	public int[] make2(int[] a, int[] b) {
		int[] comb = new int[2];
		if (a.length >= 2) {
			comb[0] = a[0];
			comb[1] = a[1];
		} else if (a.length == 1) {
			comb[0] = a[0];
			comb[1] = b[0];
		} else {
			comb[0] = b[0];
			comb[1] = b[1];
		}
		return comb;
	}

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

	// Given 2 int arrays, a and b, of any length, return a new array with the first
	// element of each array. If either array is length 0, ignore that array.
	//
	//
	// front11([1, 2, 3], [7, 9, 8]) → [1, 7]
	// front11([1], [2]) → [1, 2]
	// front11([1, 7], []) → [1]
	//

	public int[] front11(int[] a, int[] b) {
		int[] front;
		if (a.length >= 1) {
			if (b.length >= 1) {
				front = new int[2];
				front[0] = a[0];
				front[1] = b[0];
			} else {
				front = new int[1];
				front[0] = a[0];
			}
		} else if (b.length >= 1) {
			front = new int[1];
			front[0] = b[0];
		} else
			front = new int[0];
		return front;
	}

}