Question

What is the worst-case Big-O runtime for the following method? Assume that n is the length of the array. 1. public static void divdeAndConquer(int[] input) { 2. if (input.length <= 0) 3. return; 4. else { 5. int[] leftHalf = new int[input.length/2]; 6. for (int i = 0; i < leftHalf.length; i++) { 7. leftHalf[i] = input[i]; 8. } 9. divideAndConquer(leftHalf); 10. int[] rightHalf = new int[input.length - leftHalf.length]; 11. for (int i = 0; i < rightHalf.length; i++) { 12. rightHalf[i] = input[leftHalf.length + i]; 13. } 14. divideAndConquer(right); 15. } 16.}

          What is the worst-case Big-O runtime for the following method? Assume that n is the
length of the array.
1. public static void
divdeAndConquer(int[] input) {
2. 
if (input.length <= 0)
3. 	return;
4. 	else {
5. 	int[] leftHalf = new
int[input.length/2];
6. 	for (int i = 0; i < leftHalf.length;
i++) {
7. 		leftHalf[i] = input[i];
8. 	}
9. 	divideAndConquer(leftHalf);
10. 	int[] rightHalf = new
int[input.length - leftHalf.length];
11. 	for (int i = 0; i < rightHalf.length;
i++) {
12. 		rightHalf[i] = 
input[leftHalf.length + i];
13. 	}
14. 	divideAndConquer(right);
15. 	}
16.}

What is the worst-case Big-O runtime for the following method? Assume that n is the
length of the array.
1. public static void
divdeAndConquer(int[] input)
2.
if (input.length <= 0)
3. return;
4. else
5. int[] leftHalf = new
int[input.length/2];
6. for (int i = 0; i < leftHalf.length;
i++)
7. leftHalf[i] = input[i];
8.
9. divideAndConquer(leftHalf);
10. int[] rightHalf = new
int[input.length - leftHalf.length];
11. for (int i = 0; i < rightHalf.length;
i++)
12. rightHalf[i] =
input[leftHalf.length + i];
13.
14. divideAndConquer(right);
15.
16.

Added by Jackson M.

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