The basic idea behind exchange arguments is given an optimal structure can you convert it to a sorted optimal structure
based on some heuristics such that the optimality still holds.
Benifit: For problem where we need to find the optimal structure we need to look for all permutations, we can look at the sorted structure and claim that each of the substructure in the sorted structure will be optimal so studying it is worth it. Hence it reduces the time to linear.
This is the heart of many greedy and dp algorithms.
Find the lexicographically smallest string obtained by concatenating strings from a dictionary, strings can be of different length.
Consider any string. You can for sure make it better by replacing a substring with another from the back if the back string is less than front string till their common length. Now the issue what do you do with strings like [a, ab, aba] etc. Here exchange argument comes to more power. Consider pair of strings, ababa or abaab which is better. So sort by this criterion! Ans is aabaab.
There are n boxes with weight wi and strength si. If one box can be placed over the other only if its weight is no greater than the strength of the bottom box. What is the maximum possible no of boxes in the tower.
Given an arrangement, under what circumstance it is always good to swap to boxes. Let F be the force(wt) that the top of both arrangement can sustain. More the F better right!
Hero has to fight some monsters. Fighting with each monster takes d[m] health and after monster dies, hero gets back a[m] health in the form of a potion. What is the min health to start the fight?
Observation:
void comparator(Monster &m1,Monster &m2){
if(m1 is good && m2 is good){
return m1.damage < m2.damage;
}
else if(m1 is good && m2 is bad){
return 1;
}
else if(m1 is bad && m2 is good){
return 0;
}
else if(m1 is bad && m2 is bad){
return m1.health > m2.health;
}
}A set of frogs have accidentally fallen to the bottom of a large pit. Their only means of escaping the pit is to jump out of it. Each frog i is described by three parameters (li,wi,hi) where li is its leap capacity, wi its weight, and hi its height. The leap capacity specifies how high that frog can jump. If a frog's leap capacity is strictly larger than the depth of the pit, the frog can directly escape the pit. However, these frogs are altruistic. Rather than selfishly saving themselves and leaving the frogs with too limited leap capacity behind, they collectively aim to save as many of them from the pit as possible.
The frogs realize that if a frog A climbs up on the back of frog B before it jumps, the first frog A stands a better chance of escaping the pit: it can escape if hB+lA is strictly larger than the depth of the pit.
Furthermore, if frog B carrying frog A on its back climbs up on the back of frog C, the situation is even better for frog A: it can now escape the pit if hC+hB+lA is strictly larger than the depth of the pit.
The frogs can build even higher piles of frogs this way, the only restriction is that no frog may carry other frogs of weight in total amounting to its own weight or heavier. Once a pile has been used to allow a frog to escape, the frogs in the pile jump back to the bottom of the pit and they can then form a new pile (possibly consisting of a different set of frogs). The question is simply how many frogs can escape the pit assuming they collaborate to maximize this number?
Quick Idea: Let us suppose a solution exist, say F1, F2, F3, ....., FN where F1 escapes before F2. We can see that W1 < W2. Once we get a sorted domain of solution we can apply other constraints through dp.
You are given a rooted tree with n (n ≤ 105) vertices. You are in the root. Moving along an edge takes 1 minute. You want to visit all vertices and get back to the root as fast as possible, what takes exactly 2·(n - 1) minutes. When we visit vertex v for the first time, the countdown is started there and after av minutes that vertex will be "ready". We want to minimize the moment when all vertices are ready. When will it happen?
Idea: Exchange two child vertex order if 2s[u]-f[u], s[u]: size of the subtree, f[u]: sol for vertex u.
https://atcoder.jp/contests/aising2020/tasks/aising2020_e
Idea:
- Observation I: call all camels red for which l >= r, and all camels blue for which r > l. Claim: all red camels should come before all blue camels. (Exch can optimize further)
- Observation II: for all red camels, you need to use job sequencing algorithm in a optimal way, similarly for blue camels. Using Priority Queue/ Max Heap, you can get O(nlogn) solution.
struct camel{
int k,l,r, diff;
};
bool operator<(const camel &ref1, const camel &ref2){
return ref1.diff < ref2.diff;
}
bool operator>(const camel &ref1, const camel &ref2){
return ref1.diff > ref2.diff;
}
const int N = 200005;
int tc,n;
void print(camel c){
cout << c.k << "," << c.l << "," << c.r << endl;
}
int main(){
std::ios::sync_with_stdio(false);
cin.tie(0);
cin >> tc;
while(tc--){
cin >> n;
vector<camel> red, blue;
for(int i=0;i<n;i++){
int k,l,r; cin >> k >> l >> r;
struct camel tmp = {k,l,r,abs(l-r)};
if(l>=r) red.eb(tmp);
else blue.eb(tmp);
}
ll ans = 0ll;
sort(all(red), [&](camel ref1, camel ref2){
return ref1.k < ref2.k;
});
priority_queue<camel> q;
const int M1 = sz(red);
int j = M1-1;
for(int i=M1;i>=1;i--){
while(j>=0 && red[j].k >= i){
q.push(red[j--]);
}
if(!q.empty()){
ans += q.top().l;
q.pop();
}
}
while(!q.empty()){
ans += q.top().r;
q.pop();
}
sort(all(blue), [&](camel ref1, camel ref2){
return ref1.k < ref2.k;
});
const int M2 = sz(blue);
j = 0;
priority_queue<camel> q2;
while(j<M2 && blue[j].k <= M1) q2.push(blue[j++]);
for(int i=M1+1;i<=n;i++){
if(!q2.empty()){
ans += q2.top().r;
q2.pop();
}
while(j<M2 && blue[j].k <= i){
q2.push(blue[j++]);
}
}
while(!q2.empty()){
ans += q2.top().l;
q2.pop();
}
cout << ans << endl;
}
return 0;
}https://codeforces.com/contest/1393/problem/C
In this problem, given a string s, example: AABCCABAB, you have to rearrange the letters in such a way that min distance between any two similar characters is as large as possible
Approach 1: O(n log n log n)
- Binary search on the distance.
- For a fixed distance. Idea is to take the character that is most frequent and has not occured in the last "distance" indices.
- Why does this logic work?
- ....solved.....|(a,b) . You have choice now to select a or b, & freq[a] > freq[b].
- Exchanging (a,b) preserve optimality. How? Take cases
- So might as well take the most frequent available character & process.
Approach 2: O(n)
- Let max frequency be m and it occures p times.
- Make m-1 bins.
- Now the smallest bin size should be maximum as possible.
- Intuition: If you need to maximize the minimum, distribute equally.
- How to ensure nothing in the bin breaks the optimality?
- abc|.....|abc|.....|abc|....|abc|....|abc|....|abc|....|abc|....|abc
- Now, find the largest frequent element. Start putting in bins one by one. If any bins left on the right, start with the next one.
- Observe that the next element will not make a complete round trip once more. so distance remains the size of the smallest bin.
https://codeforces.com/contest/1764/problem/C
https://codeforces.com/contest/1889/problem/B
There are n cities with a[i] people in each of them. You can create a road between city i and j iff connected_component[i] + connected_component[j] >= ijc. Can you connect all ?
Solution:
observation: All cities get connected to city 1 one by one in optimal setup. [Proof by construction]
Exchange Argument: In the optimal setup it is always good to swap (X + a[i] >= i*c, X + a[j] >= j*c) iff a[i] - i*c < a[j] - j*c.. Sort and verify the optimal arrangement.
Practise Problems: https://codeforces.com/contest/2023/problem/A