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Number following the pattern

In this article, we are going to see how to generate a sequence following a specific pattern? This problem has been featured in coding round of Amazon.
Submitted by Radib Kar, on February 10, 2019 [Last updated : March 20, 2023]

Problem Description

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't repeat.

Example

    Input:
    IIDDIDD  

    Output:
    12543876

Solution of Number following the pattern

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 for Number following the pattern

#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