 # SALARY - Editorial

Author: Ivan Zdomsky
Tester: Anton Lunyov
Editorialist: Anton Lunyov

CAKEWALK

None

### PROBLEM:

You are given the array W, W, …, W[N] of salaries of the workers. At each operation you can increase by 1 salaries of any N − 1 workers. Your goal is to make salaries of all the workers to be equal. What is the minimum number of operations needed for this?

### QUICK EXPLANATION:

The answer is just sumW − N * minW, where sumW is the sum of all W[i] and minW is the minimum over all W[i]. See author’s solution and tester’s first solution as a reference.

### EXPLANATION:

The operation can be treated as follows: we at first decrease salary of some worker by 1 and then increase salaries of all workers by 1. But why do we need to do the second part? Since we want all salaries to be equal the second part of the operation could be simply ignored. So we may assume that at each operation we decrease salary of some worker by 1.

Now if we have some salary greater the minimum salary then without applying operation to it we can’t achieve the goal in any way - the minimum could only decreases during operations so this salary will be always not equal to the minimum one. Hence we need to apply at least W[i] − minW operations for the i-th worker. The summation of this over all i is exactly sumW − N * minW. But, clearly, applying exactly W[i] − minW operations to the i-th worker (for all i) makes all salaries to be equal to minW, which is our goal. Therefore, this number of operations is also sufficient. Hence it is the answer as stated above.

### ALTERNATIVE SOLUTION:

The constraints were quite moderate. So alternatively one could model the process of applying all the operations. But the following naive implementation will get TLE: at each step we at first check whether all salaries are equal, if no, then choose the worker with the maximal salary and increase by 1 salaries of all other works. Such solution has complexity O(maxW * N * N) in the worst case that have the form W = 0, W = maxW, W = maxW, …, W[N] = maxW, where maxW = 10000. Since we also have like 100 tests in a file, such solution could consume more than 8 seconds on one test file. And we indeed have test files having all 100 tests of this or similar form.

In order to get AC with modeling one should at least figure it out the first step of the solution explained above: instead of increasing salaries of N − 1 workers one should decrease salary of just one worker at each operation. But even this will get TLE if it would be implemented as above. The simplest way to make such solution fast enough is to decrease by one all maximum salaries at one step (so we perform several operations at once). Then the solution will have complexity O(maxW * N) and passes the TL easily. Namely now at each step we at first check whether all salaries are equal and if no we find the maximal salary and then decrease by 1 all salaries equal to this maximum. See tester’s second solutions as a reference.

### AUTHOR’S AND TESTER’S SOLUTIONS:

Author’s solution can be found here.
Tester’s first solution can be found here.
Tester’s second solution can be found here.

### RELATED PROBLEMS:

18 Likes

I think any Cakewalk problem can be used here as related problem Especially if it also has almost no pre-requisites or very simple ones… For example, LEBOMBS and also maybe, SAD.

3 Likes

I tried doing this in the opposite way!
I took max of the array and
for each element of the array a[i]
sum+=max-a[i]
but I dunno why I was getting wrong answer!
anybody help?

2 Likes

though we are getting answer from tester’s second approach but can someone plz elaborate what is the reason behind decreasing maximum element by 1…is there any logic or it is just an observation?

@herman look this thing in this way… increasing salary of (n-1) workers is like all the (n-1) workers get 1 point each and the 1 worker left gets a 0 …but we are concerned with the number of moves we need to perform to make all salaries equal which will be equal to the scenario where we just decrease the salary of a worker by 1… i.e now instead of giving 1 point we give 0 point each to (n-1) workers and -1 point to the unlucky worker and this unlucky one is the one having salary higher than the minimum salary…

I hope u will get my point…

23 Likes

I used the same logic but getting wrong ans during submission … Any reason ???

1 Like

I didn’t use this approach. Here is my approach.

1. I determine the MAX of the array and the MIN. The difference of MAX and MIN gives rise to DELTA.

2. Then, I consider the first element with value MAX and increase the salaries of all the other elements by delta.

3. The above can give rise to a new set of MAX and MIN and thus DELTA. I re-compute DELTA and iterate the above operation of selecting the element with a value of MAX and increase salaries of other elements by DELTA.

4. This operation is repeated until DELTA becomes equal to 0. Yet I am getting a wrong answer.

Can you please see point out my mistake. This problem is very simple yet 2 Likes

I did what the question said by increasing 1 to get all the salaries equal in an O(N) solution for each case.

1- Sort the list of salaries in increasing order.

2- In each step and starting from the last term I need to make seq[i - 1] = seq[i]. Each step needs
(N - i) * (seq[i] - seq[i - 1]).

The final answer becomes (seq[N - 1] - seq[N - 2]) + (2 * (seq[N - 2] - seq[N - 3]) + … + (N - 1) * (seq - seq).

I think this is a simple explanation. In case u need an example I’m ready to give one 7 Likes

Can someone explain this to me in a little bit easier way?

Can anyone pls tell me how to optimise this code : https://www.codechef.com/viewsolution/16553743

I am using Merge Sort AFAIK its the best sorting algorithm for time complexity but still Code Chef compiler says time limit exceeded. Kindly help me out in optimising it !

in the given test case

3

1 2 3

can i can do it in 2 steps?
by stopping 3 for the first

``````2 3 3
and now stopping 2 and 3 we get
3 3 3``````
2 Likes

@Vartult: You can’t stop 2 numbers at a time according to the question,“choose some worker and increase by 1 salary of each worker, except the salary of the chosen worker”.Here it is given as worker and not as workers. So 2 3 3 >>> 3 4 3 >>> 4 4 4

what can I do to optimise this code?
https://www.codechef.com/viewsolution/20272381

#include
#include<stdio.h>
#include<conio.h>
using namespace std;
int main()
{
int t,n,a,small=10001,sum=0,tot_sum=0;
cin>>t;
for(int i=0;i<t;i++)
{
cin>>n;
for(int j=0;j<n;j++)
{
cin>>a[j];
if(small>a[j])
{
small=a[j];
}
}
for(int k=0;k<n;k++)
{
sum=a[k]-small;
tot_sum+=sum;
}
cout<<tot_sum;

``````}
``````

}

Its output is correct …why it is not getting accepted

here is my approach using binary_search!

``````def main():
#codechef question SALARY
t = input()
t = int(t)
while t > 0:
n = input()
n = int(n)
val = list(map(int, input().split(" ")))
initial_sum = sum(val)
min_value = min(val)
left = 0
right = 10000000000
while left <= right:
mid = (left + right) // 2
may_be = initial_sum + (mid * (n-1))
mean = may_be / n
diff = mean - min_value
if diff == mid:
break
elif diff < mid:
right = mid - 1
else:
left = mid + 1
print(mid)
t -= 1
``````

if name == ‘main’:
main()

Can anyone please tell me the derivation of that formula?

3 Likes

I am getting correct ans for my test case. But codechef compiler is giving time sigtstp error.

``````

int T,N,moves,minW;
long sum = 0;
cin >> T;

for(int t=0; t<T; t++)
{
cin >> N;
vector<int> W(N);

for(int i=0; i<N; i++)
{
cin >> W[i];
sum += W[i];
}

minW = *min_element(W.begin(),W.end());

moves= sum -N*minW;
cout <<moves << endl;
}

return 0;
}

``````

Actually you get TLE and it is expected for your solution.
The test case where your program works too slow is described in the editorial. But I repeat here:
T = 100
N = 100 for each test
salaries are {0, 10000, 10000, …, 10000} for each test

1 Like

Consider test case:
3
0 2 2