Answer:
The new root will be 2.
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Explanation:
The binary tree is not properly presented (See attachment)
To answer this; first, we need to order the nodes of the tree in a pre-order traversal.
We use pre-order because the question says if something is removed from the left child.
So, the nodes in pre-order form is: 14, 2, 1, 5, 4, 16.
The root of the binary tree is 14 and if 14 is removed, the next is 2.
<em>Hence, the new root will be 2.</em>
Answer:
Grace Hopper.
Explanation:
Grace Hopper was a US Naval Rear Admiral and an American computer scientist who was born on the 9th of December, 1906 in New York city, United States of America. She worked on the first commercial computer known as universal automatic computer (UNIVAC), after the second World War II.
In 1953 at the Remington Rand, Grace Hopper invented the first high-level programming language for UNIVAC 1 by using words and expressions.
Hence, Grace Hopper was the Navy Admiral who invented a high-level programming language FLOW-MATIC in 1953.
Additionally, FLOW-MATIC paved the way for the development of common business-oriented language (COBOL).
Since both arrays are already sorted, that means that the first int of one of the arrays will be smaller than all the ints that come after it in the same array. We also know that if the first int of arr1 is smaller than the first int of arr2, then by the same logic, the first int of arr1 is smaller than all the ints in arr2 since arr2 is also sorted.
public static int[] merge(int[] arr1, int[] arr2) {
int i = 0; //current index of arr1
int j = 0; //current index of arr2
int[] result = new int[arr1.length+arr2.length]
while(i < arr1.length && j < arr2.length) {
result[i+j] = Math.min(arr1[i], arr2[j]);
if(arr1[i] < arr2[j]) {
i++;
} else {
j++;
}
}
boolean isArr1 = i+1 < arr1.length;
for(int index = isArr1 ? i : j; index < isArr1 ? arr1.length : arr2.length; index++) {
result[i+j+index] = isArr1 ? arr1[index] : arr2[index]
}
return result;
}
So this implementation is kind of confusing, but it's the first way I thought to do it so I ran with it. There is probably an easier way, but that's the beauty of programming.
A quick explanation:
We first loop through the arrays comparing the first elements of each array, adding whichever is the smallest to the result array. Each time we do so, we increment the index value (i or j) for the array that had the smaller number. Now the next time we are comparing the NEXT element in that array to the PREVIOUS element of the other array. We do this until we reach the end of either arr1 or arr2 so that we don't get an out of bounds exception.
The second step in our method is to tack on the remaining integers to the resulting array. We need to do this because when we reach the end of one array, there will still be at least one more integer in the other array. The boolean isArr1 is telling us whether arr1 is the array with leftovers. If so, we loop through the remaining indices of arr1 and add them to the result. Otherwise, we do the same for arr2. All of this is done using ternary operations to determine which array to use, but if we wanted to we could split the code into two for loops using an if statement.
Answer:
import random
a = random.randint(1,10)
b = random.randint(1,10)
answer = a * b
print (str(a) + " X " + str(b) + " = " + str(answer))
Explanation:
Happy to help you mate
