Answer:
The time complexity of the code is O(log₇n).
Explanation:
The i is updated by 7*i.On each iteration i is multiplied by 7.So on finding the time complexity of the code given above it will come out to be log base 7.
When we divide the input by 2 the time complexity is log base 2.
So on dividing it by 7 we get the time complexity of log base 7.
Answer:
The sum of all positive even values in arr
Explanation:
We have an array named arr holding int values
Inside the method mystery:
Two variables s1 and s2 are initialized as 0
A for loop is created iterating through the arr array. Inside the loop:
num is set to the ith position of the arr (num will hold the each value in arr)
Then, we have an if statement that checks if num is greater than 0 (if it is positive number) and if num mod 2 is equal to 0 (if it is an even number). If these conditions are satisfied, num will be added to the s1 (cumulative sum). If num is less than 0 (if it is a negative number), num will be added to the s2 (cumulative sum).
When the loop is done, the value of s1 and s2 is printed.
As you can see, s1 holds the sum of positive even values in the arr
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Answer:
See explaination
Explanation:
Keep two iterators, i (for nuts array) and j (for bolts array).
while(i < n and j < n) {
if nuts[i] == bolts[j] {
We have a case where sizes match, output/return
}
else if nuts[i] < bolts[j] {
this means that size of nut is smaller than that of bolt and we should go to the next bigger nut, i.e., i+=1
}
else {
this means that size of bolt is smaller than that of nut and we should go to the next bigger bolt, i.e., j+=1
}
}
Since we go to each index in both the array only once, the algorithm take O(n) time.
