Answer:
There is also an attachment below
Explanation:
Since we are talking about binary search, let's assume that the items are sorted according to some criteria.
Time complexity of binary search is O(logN) in worst case, best case and average case as well. That means it can search for an item in Log N time where N is size of the input. Here problem talks about the item not getting found. So, this is a worst case scenario. Even in this case, binary search runs in O(logN) time.
N = 700000000.
So, number of comparisions can be log(N) = 29.3 = 29.
So, in the worst case it does comparisions 29 times
Answer:
if(soldYesterday > soldToday){
salesTrend = -1;
} else if(soldToday > soldYesterday){
salesTrend = 1;
}
Explanation:
The if/else statement is more explicit. The first if condition check if soldYesterday is greater than soldToday, if true, then -1 is assigned to salesTrend.
Else if soldToday is greater than soldYesterday, if true, then 1 is assigned to salesTrend.
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:
Adding more links to the page.
Explanation:
User experience design, UED, is a concept or a process of software development application life cycle that graphically presents a clients project to reflect the needs, want and interactivity of the users.
There are certain principles that governs the design of user interactive web and mobile interfaces. A user interface has to be simple, easy to interact with, give a good feel, signing into email, phone, laptop, mobile app etc, should also be made easy.
Covering the screen with link, makes it difficult to navigate without triggering a link.
Answer:
in computer science, an instruction is a single operation of a processor defined by the instruction set
Explanation:
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