Thursday, July 27th

 Keep all computers powered

Given a number of computers, and batteries of various sizes, where batteries can be freely swapped between computers, how long can we keep all computers simultaneously powered?


There are a couple of solutions to this problem.  Here we will use binary search, building off its use these past 2 days.

Similar to yesterday, binary search can be used while testing a target solution t.  In this case, given a collection of batteries, can we keep all the computers running for t minutes?  

To determine so, we iterate over all the batteries and sum up the available power e:

  • if a given battery length is greater than t, then we add only t to the available power, because the battery won't be available after t
  • Otherwise, if a battery is smaller than t, we add all its capacity to available power

    long e=0;
for(long b : batteries) {
    e+=min(t,b);
}

This capacity can be tested to see if it can power the batteries for the target time by multiplying the target time  t by n the number of computers.  If so, we set the left hand side of the binary search window to t, otherwise set.t-1 to the right side

    if( e>= (long)n*t) {
    l=t;
} else {
r=t-1;
}

Putting this all together, we start with original indices with left equal to 1, and right equal to the sum of all power divided by the number of computers, representing a perfect distribution

    long s = 0;
for(auto b : batteries) {
s += b;
}
long l=1, r=s/n;

Full implementation below:

    long long maxRunTime(int n, vector<int>& batteries) {
long s = 0;
for(auto b : batteries) {
s += b;
}
long l=1, r=s/n;
long t, e;
while(l<r) {
t= r-(r-l)/2;
long e=0;
for(long b : batteries) {
e+=min(t,b);
}
if( e>= (long)n*t) {
l=t;
} else {
r=t-1;
}
}
return l;
}

The solution here demonstrates some tricky details in binary search implementations.  Here is a blog entry about the details of binary search, and a paper off of arXiv discussing the same issues.

Shout out to the Editorial for helpful intuition

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