Remove consecutive duplicates in a string to make the smallest string
A tree would definitely get you the shortest string(s).
The tree solution:
- Define a
State
(node) for each currentstring Input
and all its removable sub-strings'int[] Indexes
. - Create the tree: For each
int index
create anotherState
and add it to the parent stateState[] Children
. - A
State
with no possible removable sub-strings has no childrenChildren = null
. - Get all Descendants
State[]
of your rootState
. Order them by their shorteststring Input
. And that is/are your answer(s).
Test cases:
string result = FindShortest("AAAABBBAC"); // AC
string result2 = FindShortest("AABBAAAC"); // AABBC
string result3 = FindShortest("BAABCCCBBA"); // B
The Code:
Note: Of-course everyone is welcome to enhance the following code in terms of performance and/or fixing any bug.
class Program
{
static void Main(string[] args)
{
string result = FindShortest("AAAABBBAC"); // AC
string result2 = FindShortest("AABBAAAC"); // AABBC
string result3 = FindShortest("BAABCCCBBA"); // B
}
// finds the FIRST shortest string for a given input
private static string FindShortest(string input)
{
// all possible removable strings' indexes
// for this given input
int[] indexes = RemovableIndexes(input);
// each input string and its possible removables are a state
var state = new State { Input = input, Indexes = indexes };
// create the tree
GetChildren(state);
// get the FIRST shortest
// i.e. there would be more than one answer sometimes
// this could be easily changed to get all possible results
var result =
Descendants(state)
.Where(d => d.Children == null || d.Children.Length == 0)
.OrderBy(d => d.Input.Length)
.FirstOrDefault().Input;
return result;
}
// simple get all descendants of a node/state in a tree
private static IEnumerable<State> Descendants(State root)
{
var states = new Stack<State>(new[] { root });
while (states.Any())
{
State node = states.Pop();
yield return node;
if (node.Children != null)
foreach (var n in node.Children) states.Push(n);
}
}
// creates the tree
private static void GetChildren(State state)
{
// for each an index there is a child
state.Children = state.Indexes.Select(
i =>
{
var input = RemoveAllAt(state.Input, i);
return input.Length < state.Input.Length && input.Length > 0
? new State
{
Input = input,
Indexes = RemovableIndexes(input)
}
: null;
}).ToArray();
foreach (var c in state.Children)
GetChildren(c);
}
// find all possible removable strings' indexes
private static int[] RemovableIndexes(string input)
{
var indexes = new List<int>();
char d = input[0];
int count = 1;
for (int i = 1; i < input.Length; i++)
{
if (d == input[i])
count++;
else
{
if (count >= 3)
indexes.Add(i - count);
// reset
d = input[i];
count = 1;
}
}
if (count >= 3)
indexes.Add(input.Length - count);
return indexes.ToArray();
}
// remove all duplicate chars starting from an index
private static string RemoveAllAt(string input, int startIndex)
{
string part1, part2;
int endIndex = startIndex + 1;
int i = endIndex;
for (; i < input.Length; i++)
if (input[i] != input[startIndex])
{
endIndex = i;
break;
}
if (i == input.Length && input[i - 1] == input[startIndex])
endIndex = input.Length;
part1 = startIndex > 0 ? input.Substring(0, startIndex) : string.Empty;
part2 = endIndex <= (input.Length - 1) ? input.Substring(endIndex) : string.Empty;
return part1 + part2;
}
// our node, which is
// an input string &
// all possible removable strings' indexes
// & its children
public class State
{
public string Input;
public int[] Indexes;
public State[] Children;
}
}
I propose O(n^2) solution with dynamic programming.
Let's introduce notation. Prefix and suffix of length l of string A denoted by P[l] and S[l]. And we call our procedure Rcd.
- Rcd(A) = Rcd(Rcd(P[n-1])+S[1])
- Rcd(A) = Rcd(P[1]+Rcd(S[n-1]))
Note that outer Rcd in the RHS is trivial. So, that's our optimal substructure. Based on this i came up with the following implementation:
#include <iostream>
#include <string>
#include <vector>
#include <cassert>
using namespace std;
string remdupright(string s, bool allowEmpty) {
if (s.size() >= 3) {
auto pos = s.find_last_not_of(s.back());
if (pos == string::npos && allowEmpty) s = "";
else if (pos != string::npos && s.size() - pos > 3) s = s.substr(0, pos + 1);
}
return s;
}
string remdupleft(string s, bool allowEmpty) {
if (s.size() >= 3) {
auto pos = s.find_first_not_of(s.front());
if (pos == string::npos && allowEmpty) s = "";
else if (pos != string::npos && pos >= 3) s = s.substr(pos);
}
return s;
}
string remdup(string s, bool allowEmpty) {
return remdupleft(remdupright(s, allowEmpty), allowEmpty);
}
string run(const string in) {
vector<vector<string>> table(in.size());
for (int i = 0; i < (int)table.size(); ++i) {
table[i].resize(in.size() - i);
}
for (int i = 0; i < (int)table[0].size(); ++i) {
table[0][i] = in.substr(i,1);
}
for (int len = 2; len <= (int)table.size(); ++len) {
for (int pos = 0; pos < (int)in.size() - len + 1; ++pos) {
string base(table[len - 2][pos]);
const char suffix = in[pos + len - 1];
if (base.size() && suffix != base.back()) {
base = remdupright(base, false);
}
const string opt1 = base + suffix;
base = table[len - 2][pos+1];
const char prefix = in[pos];
if (base.size() && prefix != base.front()) {
base = remdupleft(base, false);
}
const string opt2 = prefix + base;
const string nodupopt1 = remdup(opt1, true);
const string nodupopt2 = remdup(opt2, true);
table[len - 1][pos] = nodupopt1.size() > nodupopt2.size() ? opt2 : opt1;
assert(nodupopt1.size() != nodupopt2.size() || nodupopt1 == nodupopt2);
}
}
string& res = table[in.size() - 1][0];
return remdup(res, true);
}
void testRcd(string s, string expected) {
cout << s << " : " << run(s) << ", expected: " << expected << endl;
}
int main()
{
testRcd("BAABCCCBBA", "B");
testRcd("AABBAAAC", "AABBC");
testRcd("AAAA", "");
testRcd("AAAABBBAC", "C");
}
You can check default and run your tests here.