How to memoize a recursive problem to avoid re-calculating subproblems? - c++

Question

I am trying to solve this problem:
https://www.geeksforgeeks.org/count-possible-decodings-given-digit-sequence/
Example:
Input:
Input str = "121"
Total decoding:: 3 :: ABA AU LA
I am able to code this problem through recursion. But the code fails to process a bigger input sequence (for e.g., i/p str = 11111111111111111111111111111111111111111)
This is happening because I am calculating sub-problems again-and-again.
Can anyone help me by letting me know how to memoize below sample code?
PS - I know there are other ways to solve this problem. But I don't want to do that. I want to memoize this solution only. It will help me to build my concept. Please help.
Here is the code:
#include "iostream"
#include <iostream>
#include <vector>
#include <string>
using namespace std;
namespace solution3
{
void solve(string str, string& out, vector<string>& v)
{
if (str.size() == 0)
{
v.push_back(out);
return;
}
//we have 2 choices:
//ch#1: take 1st char of str
//ch#2: take 1st and 2nd chars of str
if (str.size() >= 1)//ch#1: take 1st char of str
{
string out1 = out;
string str1 = str;
int num1 = stoi(str.substr(0, 1)); // converting string at index 0 to integer
if (num1) // we will not consider if the string at index 0 is zero.
{
out1.push_back(('#' + num1)); //<-- It will conevrt 1 into A; 2 into B; and so on.
str1 = str1.erase(0, 1);//erase the index 0 from str1.
solve(str1, out1, v);
}
}
if (str.size() >= 2)//ch#2: take 1st and 2nd chars of str
{
string out2 = out;
string str2 = str;
int num2 = stoi(str.substr(0, 2)); // converting string at index 0 and 1 to integer
// checking if num2 is a valid number for decoding.
// num2 should be - NON-ZERO, 1st char is not ZERO, is within the range of 1 and 26.
if (num2 && str[0] != '0' && num2 > 0 && num2 <= 26)
{
out2.push_back(('#' + num2));
//Erase 1st two chars from str
str2 = str2.erase(0, 1);//erase the index 0 from str1.
str2 = str2.erase(0, 1);//erase the index 0 from str1.
solve(str2, out2, v);
}
}
}
void alphacode(string str)
{
string out;
vector<string> v; //<-- To store all the Decodings
solve(str, out, v);
cout << "Total decoding:: " << v.size() << ":: ";
for (int i = 0; i < v.size(); i++)
cout << v[i] << " ";
cout << endl;
}
}
int main()
{
string str = "25114";
cout << "IpStr:: " << str << endl;
solution3::alphacode(str);
cout << "----------------" << endl;
str = "1111111111";
cout << "IpStr:: " << str << endl;
solution3::alphacode(str);
cout << "----------------" << endl;
str = "3333333333";
cout << "IpStr:: " << str << endl;
solution3::alphacode(str);
cout << "----------------" << endl;
str = "202";
cout << "IpStr:: " << str << endl;
solution3::alphacode(str);
cout << "----------------" << endl;
str = "2010";
cout << "IpStr:: " << str << endl;
solution3::alphacode(str);
cout << "----------------" << endl;
str = "1111111111111111111111111111111"; //<-- takes too much time! How to solve this?
cout << "IpStr:: " << str << endl;
solution3::alphacode(str);
return 0;
}

Answer 1

You can memoize each substring that you are currently working with, which you're forming after deleting one or two characters, depending on the case. Something like this:
#include "iostream"
#include <iostream>
#include <vector>
#include <string>
#include <map>
using namespace std;
map<string, vector<string>> dp;
namespace solution3
{
void solve(string str, string& out, vector<string>& v)
{
if (str.size() == 0)
{
v.push_back(out);
return;
}
//we have 2 choices:
//ch#1: take 1st char of str
//ch#2: take 1st and 2nd chars of str
if(dp.find(str) != dp.end()) {
vector<string> current = dp[str];
for(string s: current) {
v.push_back(s);
}
return;
}
if (str.size() >= 1)//ch#1: take 1st char of str
{
string out1 = out;
string str1 = str;
int num1 = stoi(str.substr(0, 1)); // converting string at index 0 to integer
if (num1) // we will not consider if the string at index 0 is zero.
{
out1.push_back(('#' + num1)); //<-- It will conevrt 1 into A; 2 into B; and so on.
str1 = str1.erase(0, 1);//erase the index 0 from str1.
solve(str1, out1, v);
}
}
if (str.size() >= 2)//ch#2: take 1st and 2nd chars of str
{
string out2 = out;
string str2 = str;
int num2 = stoi(str.substr(0, 2)); // converting string at index 0 and 1 to integer
// checking if num2 is a valid number for decoding.
// num2 should be - NON-ZERO, 1st char is not ZERO, is within the range of 1 and 26.
if (num2 && str[0] != '0' && num2 > 0 && num2 <= 26)
{
out2.push_back(('#' + num2));
//Erase 1st two chars from str
str2 = str2.erase(0, 1);//erase the index 0 from str1.
str2 = str2.erase(0, 1);//erase the index 0 from str1.
solve(str2, out2, v);
}
}
dp[str] = v;
}
void alphacode(string str)
{
string out;
vector<string> v; //<-- To store all the Decodings
solve(str, out, v);
cout << "Total decoding:: " << v.size() << ":: ";
// for (int i = 0; i < v.size(); i++)
// cout << v[i] << " ";
cout << endl;
}
}
int main()
{
string str = "25114";
cout << "IpStr:: " << str << endl;
solution3::alphacode(str);
cout << "----------------" << endl;
str = "1111111111";
cout << "IpStr:: " << str << endl;
solution3::alphacode(str);
cout << "----------------" << endl;
str = "3333333333";
cout << "IpStr:: " << str << endl;
solution3::alphacode(str);
cout << "----------------" << endl;
str = "202";
cout << "IpStr:: " << str << endl;
solution3::alphacode(str);
cout << "----------------" << endl;
str = "2010";
cout << "IpStr:: " << str << endl;
solution3::alphacode(str);
cout << "----------------" << endl;
str = "1111111111111111111111111111111"; //<-- takes too much time! How to solve this?
cout << "IpStr:: " << str << endl;
solution3::alphacode(str);
return 0;
}