PROBLEM LINK:

Practice
Contest

Author:Shiplu
Tester:Mahbubul Hasan
Editorialist:Jingbo Shang

DIFFICULTY:

Simple

PREREQUISITES:

Sort

PROBLEM:

Given a random sequence of N numbers, a[0..N-1], sort them in order and report the difference between the sum of numbers in odd positions and the sum of numbers in even position.

EXPLANATION:

If a quick sort is directly applied, the time complexity is O(N logN). But the range of numbers is small, within 1,000,000. Therefore, we can use the counting sort, that is,

count[0 .. 999999] = 0
for i = 0 to N-1 do
    count[a[i]] ++;
end

Go through the count array again, we can get the order. With this algorithm, we can solve this problem in O(X) time, where X is the range of all numbers.

AUTHOR'S AND TESTER'S SOLUTIONS:

Author's solution can be found here.
Tester's solution can be found here.