PROBLEM LINK:
Author: Puneet Gupta
Editorialist: Puneet Gupta
DIFFICULTY:
EASY
PREREQUISITES:
Greedy
PROBLEM:
Given the number of R, G and B ballons, find the maximum number N of tables, that can be decorated so that three balloons attached to some table shouldn’t have the same color.
EXPLANATION:
The order of the balloons isn’t important, so instead or r, g, b, we’ll call them a[0], a[1], a[2]
and sort them in ascending order. We’ll now have a[0] <= a[1] <= a[2]
.
There are two case:

2*(a[0]+a[1]) <= a[2]
. In this case, we can takea[0]
sets of (1, 0, 2) anda[1]
sets of (0, 1, 2), so the answer isa[0]+a[1]
. 
2*(a[0]+a[1]) > a[2]
. In this case, we can continuously take a set of two balloon froma[2]
and a balloon frommax(a[0], a[1])
until a point thata[2] <= max(a[0], a[1])
. At this point,max(a[0], a[1])a[2] <= 1
, and sincemax(a[0], a[1])  min(a[0], a[1]) <= 1
too,max(a[0], a[1], a[2])  min(a[0], a[1], a[2]) <= 1
. All we have to do is take all possible (1, 1, 1) group left. Since we only take the balloons in group of 3,(a[0]+a[1]+a[2])%3
doesn’t change, so there will be at most(a[0]+a[1]+a[2])% 3
balloons wasted. We go back to the beginning now. The answer is(a[0]+a[1]+a[2])/3
.
AUTHOR’S SOLUTION:
Author’s solution can be found here.