Answer:
Omega Software is attempting to make a change in the area of people.
Explanation:
Types of Organizational Change:
Most organizations have to go through a change to keep up with the changing market dynamics, trends and technologies. There are four major types of organizational change:
- Structural
- Strategic
- People
- Process
People change:
One of the organizational change is people change where organizations strive towards the improvement of their employee's skills set and productivity. This is usually achieved by launching various specialized and general training programs for employees to enhance their knowledge and widen their skills set so that they can work more efficiently for the organization.
Therefore, it can be concluded that Omega Software is attempting to make a change in the area of People.
Answer:
{"double", "char", "char", "double"} will be stored in word.
Explanation:
Given the word is an array of four strings, {"algorithm", "boolean", "char", "double"}. Hence, the length of the word array is 4.
The for-loop will only run for two iterations due to i < word.length/2, with i = 0 (first loop) and i = 1 (second loop).
In the first loop,
- word[word.length - 1 - i] = word[4 - 1 - 0] = word[3] = "double"
- Hence, the string "double" will be assigned to word[0] and overwrite "algorithm"
In the second loop,
- word[word.length - 1 - i] = word[4 - 1 - 1] = word[2] = "char"
- Hence, the string "char" will be assigned to word[1] and overwrite "boolean"
At last, the word array will hold {"double", "char", "char", "double"}
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: well it all depends on what is wrong with your computer
plz mark as brainliest
Answer:
1. Speaking or texting on the cell phone while driving.
= It cause an accident.
2. Using technology for bullying others.
Hope this answer will help you.
