Q:

# Number following the pattern

Given a pattern containing only (I)s and (D)s. (I) stands for increasing and (D) for decreasing. Devise an algorithm to print the minimum number following that pattern. Digits are from 1-9 and digits can not repeat

The pattern & number to be generated

1. Length of minimum number = string length+1
Hence, maximum string length possible is 8, since we can construct only with different digits (1-9)
2. 'I' means the next digit will be 1 greater than the current & 'D' means the next digit will be 1 less than the current digit
"II" → 123
"DD" → 321

The problem can be used with help of stack. The concept is to create stack with consecutive number same as depth of a local contiguous sequence of 'D'.

Prerequisite:

1. Input pattern, string s
2. Stack st to store the digits
```    Function findMinFromPattern(string s)
1.  Declare a stack st;
2.  FOR i=0 : pattern length
EnQueue (st, i+1); //push i+1 at first, i+1 becuase i is 0-indexed
IF (entire pattern is processed || s[i] =='I')
While(stack is not empty){
Pop and print
End While
END IF
END FOR
END FUNCTION
```

C++ Implementation

``````#include <bits/stdc++.h>
using namespace std;

void findMinFromPattern(string s){
stack<int> st; //stack declared using STL
for(int i=0;i<=s.length();i++){
//push i+1 at first, i+1 becuase i is 0-indexed
st.push(i+1);
//when string length is processed or pattern in I
if(s.length()==i || s[i]=='I' ){
//pop and print until stack is empty
while(!st.empty()){
cout<<st.top();
st.pop();
}
}
}
cout<<endl;
}

int main(){
cout<<"enter pattern\n";
string s;
cin>>s;

if(s.length()>8){
cout<<"Not possible,length>8\n";
}
else{
cout<<"The minimum number generated is:\n";
findMinFromPattern(s);//function to print
}

return 0;
}
``````

Output

```First run:
enter pattern
IIDDIDD
The minimum number generated is:
12543876

Second run:
enter pattern
IIIIIIIIDDDDIII
Not possible,length>8```
` `

Example with explanation:

```Pattern string:
IIDDIDD

0 th iteration
i=0
EnQueue(st,i+1)
Stack status:
1
S[i]=='I'
So pop and print until stack becomes empty
Thus printing:
1
Output up to now:
1
Stack status;
Empty stack
-------------------------------------------------------------

1st iteration
i=1
EnQueue(st,i+1)
Stack status:
2
S[i]=='I'
So pop and print until stack becomes empty
Thus printing:
2
Output up to now:
12
Stack status;
Empty stack
-------------------------------------------------------------

2nd iteration
i=2
EnQueue(st,i+1)
Stack status:
3
S[i]=='D'
Nothing to do
Output up to now:
12
Stack status;
3
-------------------------------------------------------------

3rd iteration
i=3
EnQueue(st,i+1)
Stack status:
3
4(top)
S[i]=='D'
Nothing to do
Output up to now:
12
Stack status;
3
4(top)
-------------------------------------------------------------

4th iteration
i=4
EnQueue(st,i+1)
Stack status:
3
4
5(top)
S[i]=='I'
So pop and print until stack becomes empty
Thus printing:
5, then 4,then 3
Output up to now:
12543
Stack status;
Empty stack
-------------------------------------------------------------

5th iteration
i=5
EnQueue(st,i+1)
Stack status:
6
S[i]=='D'
Nothing to do
Output up to now:
12543
Stack status;
6
-------------------------------------------------------------

6th iteration
i=6
EnQueue(st,i+1)
Stack status:
6
7(top)
S[i]=='D'
Nothing to do
Output up to now:
12543
-------------------------------------------------------------

7th iteration
i=7
EnQueue(st,i+1)
Stack status:
6
7
8(top)
Entire string is processed
Pop and print until stack becomes empty
Print:
8, then 7, then 6
Output up to now:
12543876
Exit:
Minimum no is:
12543876```