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Explain time complexity of the code given in the picture
So basically you have two loops, the top one is quite simple, but the second one is a tad bit more tricky.
for (i = n/2; i < n; i) // ~n/2 so O(n)
for(j = 0; j < n; j = 2 * j) // How many times does this run for n
So j doubles after every iteration until we reach n, so if we double n, j will only do one extra iteration! Alternatively, you could say that log_2 n is how many times we can double j to reach n.
So the time complexity is O(n log n).
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How to solve a problem with Kotlin when I want to generate random numbers in time period (eg. 1 sec) and than than that numbers collect somehow and make Sum of that numbers.
Thanks
The routines of what you wanted are very simple. First, run a loop of how many random numbers that you want. Run a for loop that many times. Each time you generate a random number, add the value of that random number to a variable to keep track of the sum. Add that random number to a list. Then let the program go to sleep for a specific amount of time in milliseconds. The program below is a short example.
Note that, if the random number generated is a negative number, it will be flip back to its positive value. That will not work for integer.min value. However, this is just an example.
import java.util.Random
import kotlinx.coroutines.*
fun main() {
val howManyNumber = 5
val delayTime = 1000L // ms
val randomNumbers = mutableListOf<Int>()
var sum = 0
for (i in 1..howManyNumber){
var randN = Random().nextInt()
randN = if (randN > 0) randN else randN * -1
sum += randN
randomNumbers.add(randN)
Thread.sleep(delayTime)
}
println(sum)
println(randomNumbers)
}
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This may have been asked before but none of the answers I saw worked for me. I tried Lucas Theorem,Fermat's theorem but none of them worked. Is there an efficient way to find the value of:
nCr mod 10^9+7 where n<=10^9 and r<=1000
Any help will be very useful
n is large while r is small, you are better off compute nCr by n(n-1)...(n-r+1)/(1*2*...*r)
You may need to find multiplicate inverse of 1, 2, ... r mod 10^9+7
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*warning: iteration 5u invokes undefined behavior [-Waggressive-loop-optimizations]
if([user1 isEqualToString:account1[i].name])
^
main.m:33:2: note: containing loop
for(i=0;i<=6;i++)
^
please somebody correct this code
Almost certainly you have:
Thing account1[6] = { ... };
for (i = 0; i <= 6; i++)
{
if ([user1 isEqualToString:account1[i].name])
and the compiler knows that <= 6 will go beyond the bounds of the array (last index is 5 not 6).
To correct:
for (i = 0; i < 6; i++)
^
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Why can't you specify a val or var type in for loop's in Kotlin. For example, I would like to be able to do
for (var i in 0...data.size - 1) {
for (j in 0..bytes.size - 1) {
bytes[j] = data[i++]//cant do i++ in current kotlin because "i" is val
}
//do stuff
}
But instead I have to do this
var i = 0
while (i < data.size) {
for (j in 0..bytes.size - 1) {
bytes[j] = data[i++]
}
//do stuff
}
Your example is slightly different from Java's typical for(int i=0;i<data.size;i++) example.
In the Kotlin version 'i' is actually an element of the range in which case i++ doesn't make sense. It just so happens that the range you have is a list of indexes.
The way you are using the Kotlin for loop is much closer to Java's foreach loop for(i : indexes).
I think that because Kotlin is a language that tries to make easy to respect most of the functional programming concepts, it prefers to prohibit that sort of behaviour. Also, a possible problem that your initial code could encount is an OutOfBoundException in the situation where the bytes array has more elements than the data array.
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An electric current, I, in amps, is given by
I=cos(wt)+√(8)sin(wt),
where w≠0 is a constant. What are the maximum and minimum values of I?
I have tried finding the derivative, but after that, I do not know how to solve for 0 because of the constant w.
Well David,you can convert this function into one trigonometric function by multiplying and dividing it by
√(1^2 + 8) i.e, 3. So your function becomes like this
I = 3*(1/3 cos(wt) + √8/3 sin(wt))
= 3* sin(wt + atan(1/√8))
Now, you can easily say its maximum value is
I = 3 amp
and minimum value is
I = 0 amp.