# Word Pattern II in C++

Suppose we have a pattern and a string called str, we have to check whether str follows the same pattern or not. Here pattern follow means a full match, such that there is a bijection between a letter in pattern and a non-empty substring in str.

So, if the input is like pattern is "abaa", str is "orangegreenorangeorange", then the output will be true

To solve this, we will follow these steps −

• Define a function solve(), this will take i, j, ptr, s, a map m, one set called used,

• if i >= size of s and j >= size of ptr, then −

• return true

• if i >= size of s or j >= size of ptr, then −

• return false

• if ptr[j] is in m, then −

• req := m[ptr[j]]

• len := size of req

• if len > size of s, then −

• return false

• if substring of s from index (i to len-1) is same as req and solve(i + len, j + 1, ptr, s, m, used), then −

• return true

• return false

• Otherwise

• x := ptr[j]

• for initialize k := i, when k < size of s, update (increase k by 1), do −

• temp := substring of s from index (i to k - i)

• if temp is in used, then −

• m[x] := temp

• insert temp into used

• if solve(k + 1, j + 1, ptr, s, m, used), then −

• return true

• delete x from m

• delete temp from used

• return false

• From the main method do the following −

• Define one map m

• Define one set used

• return solve(0, 0, ptr, s, m, used)

## Example

Let us see the following implementation to get a better understanding −

Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
bool solve(int i, int j, string ptr, string s, map <char, string>& m, set<string>& used){
if (i >= s.size() && j >= ptr.size()) {
return true;
}
if (i >= s.size() || j >= ptr.size())
return false;
if (m.count(ptr[j])) {
string req = m[ptr[j]];
int len = req.size();
if (len > s.size() - i)
return false;
if ((s.substr(i, len) == req) && solve(i + len, j + 1, ptr, s, m, used))
return true;
return false;
}
else {
char x = ptr[j];
for (int k = i; k < s.size(); k++) {
string temp = s.substr(i, k - i + 1);
;
if (used.count(temp))
continue;
m[x] = temp;
used.insert(temp);
if (solve(k + 1, j + 1, ptr, s, m, used))
return true;
m.erase(x);
used.erase(temp);
}
}
return false;
}
bool wordPatternMatch(string ptr, string s) {
map<char, string> m;
set<string> used;
return solve(0, 0, ptr, s, m, used);
}
};
main(){
Solution ob;
cout << (ob.wordPatternMatch("abaa", "orangegreenorangeorange"));
}

## Input

"abaa" "orangegreenorangeorange"

## Output

1