Form minimum number from given sequence

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

Examples:

Input: D        Output: 21
Input: I        Output: 12
Input: DD       Output: 321
Input: II       Output: 123
Input: DIDI     Output: 21435
Input: IIDDD    Output: 126543
Input: DDIDDIID Output: 321654798

Source: Amazon Interview Question

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Below are some important observations

Since digits can’t repeat, there can be at most 9 digits in output.

Also number of digits in output is one more than number of characters in input. Note that the first character of input corresponds to two digits in output.

Idea is to iterate over input array and keep track of last printed digit and maximum digit printed so far. Below is the implementation of above idea.

C++

 // C++ program to print minimum number that can be formed // from a given sequence of Is and Ds #include using namespace std;    // Prints the minimum number that can be formed from // input sequence of I's and D's void PrintMinNumberForPattern(string arr) {     // Initialize current_max (to make sure that     // we don't use repeated character     int curr_max = 0;        // Initialize last_entry (Keeps track for     // last printed digit)     int last_entry = 0;        int j;        // Iterate over input array     for (int i=0; i

PHP



Output:

1 3 2 5 4
1 2
3 2 1
1 2 3
2 1 4 3 5
1 2 6 5 4 3
3 2 1 6 5 4 7 9 8

This solution is suggested by Swapnil Trambake.

Alternate Solution:
Let’s observe a few facts in case of minimum number:

• The digits can’t repeat hence there can be 9 digits at most in output.
• To form a minimum number , at every index of the output, we are interested in the minimum number which can be placed at that index.

The idea is to iterate over the entire input array , keeping track of the minimum number (1-9) which can be placed at that position of the output.

The tricky part of course occurs when ‘D’ is encountered at index other than 0. In such a case we have to track the nearest ‘I’ to the left of ‘D’ and increment each number in the output vector by 1 in between ‘I’ and ‘D’.

We cover the base case as follows:

• If the first character of input is ‘I’ then we append 1 and 2 in the output vector and the minimum available number is set to 3 .The index of most recent ‘I’ is set to 1.
• If the first character of input is ‘D’ then we append 2 and 1 in the output vector and the minimum available number is set to 3, and the index of most recent ‘I’ is set to 0.

Now we iterate the input string from index 1 till its end and:

• If the character scanned is ‘I’ ,minimum value which has not been used yet is appended to the output vector .We increment the value of minimum no. available and index of most recent ‘I’ is also updated.
• If the character scanned is ‘D’ at index i of input array, we append the ith element from output vector in the output and track the nearest ‘I’ to the left of ‘D’ and increment each number in the output vector by 1 in between ‘I’ and ‘D’.

Following is the program for the same:

C++

 // C++ program to print minimum number that can be formed // from a given sequence of Is and Ds #include using namespace std;    void printLeast(string arr) {     // min_avail represents the minimum number which is     // still available for inserting in the output vector.     // pos_of_I keeps track of the most recent index     // where 'I' was encountered w.r.t the output vector     int min_avail = 1, pos_of_I = 0;        //vector to store the output     vectorv;        // cover the base cases     if (arr=='I')     {         v.push_back(1);         v.push_back(2);         min_avail = 3;         pos_of_I = 1;     }     else     {         v.push_back(2);         v.push_back(1);         min_avail = 3;         pos_of_I = 0;     }        // Traverse rest of the input     for (int i=1; i

Java

 // Java program to print minimum number that can be formed  // from a given sequence of Is and Ds  import java.io.*; import java.util.*; public class GFG {           static void printLeast(String arr)        {               // min_avail represents the minimum number which is                // still available for inserting in the output vector.                // pos_of_I keeps track of the most recent index                // where 'I' was encountered w.r.t the output vector                int min_avail = 1, pos_of_I = 0;                   //vector to store the output               ArrayList al = new ArrayList<>();                                // cover the base cases               if (arr.charAt(0) == 'I')                {                    al.add(1);                    al.add(2);                    min_avail = 3;                    pos_of_I = 1;                }                   else               {                   al.add(2);                   al.add(1);                   min_avail = 3;                    pos_of_I = 0;                }                  // Traverse rest of the input               for (int i = 1; i < arr.length(); i++)               {                    if (arr.charAt(i) == 'I')                    {                        al.add(min_avail);                        min_avail++;                        pos_of_I = i + 1;                    }                    else                    {                        al.add(al.get(i));                        for (int j = pos_of_I; j <= i; j++)                             al.set(j, al.get(j) + 1);                           min_avail++;                    }               }                  // print the number               for (int i = 0; i < al.size(); i++)                    System.out.print(al.get(i) + " ");               System.out.println();        }              // Driver code        public static void main(String args[])        {               printLeast("IDID");                printLeast("I");                printLeast("DD");                printLeast("II");                printLeast("DIDI");                printLeast("IIDDD");                printLeast("DDIDDIID");         } } // This code is contributed by rachana soma

Output:

1 3 2 5 4
1 2
3 2 1
1 2 3
2 1 4 3 5
1 2 6 5 4 3
3 2 1 6 5 4 7 9 8

This solution is suggested by Ashutosh Kumar.

Method 3
We can that when we encounter I, we got numbers in increasing order but if we encounter ‘D’, we want to have numbers in decreasing order. Length of the output string is always one more than the input string. So loop is from 0 till the length of the sting. We have to take numbers from 1-9 so we always push (i+1) to our stack. Then we check what is the resulting character at the specified index.So,there will be two cases which are as follows:-
Case 1: If we have encountered I or we are at the last character of input string,then pop from the stack and add it to the end of the output string until the stack gets empty.
Case 2: If we have encountered D, then we want the numbers in decreasing order.so we just push (i+1) to our stack.

C++

 // C++ program to print minimum number that can be formed // from a given sequence of Is and Ds #include using namespace std;    // Function to decode the given sequence to construct // minimum number without repeated digits void PrintMinNumberForPattern(string seq) {     // result store output string     string result;        // create an empty stack of integers     stack stk;        // run n+1 times where n is length of input sequence     for (int i = 0; i <= seq.length(); i++)     {         // push number i+1 into the stack         stk.push(i + 1);            // if all characters of the input sequence are         // processed or current character is 'I'         // (increasing)         if (i == seq.length() || seq[i] == 'I')         {             // run till stack is empty             while (!stk.empty())             {                 // remove top element from the stack and                 // add it to solution                 result += to_string(stk.top());                 result += " ";                 stk.pop();             }         }     }        cout << result << endl; }    // main function int main() {     PrintMinNumberForPattern("IDID");     PrintMinNumberForPattern("I");     PrintMinNumberForPattern("DD");     PrintMinNumberForPattern("II");     PrintMinNumberForPattern("DIDI");     PrintMinNumberForPattern("IIDDD");     PrintMinNumberForPattern("DDIDDIID");     return 0; }

Java

 import java.util.Stack;    // Java program to print minimum number that can be formed // from a given sequence of Is and Ds class GFG {    // Function to decode the given sequence to construct // minimum number without repeated digits     static void PrintMinNumberForPattern(String seq) {         // result store output string         String result = "";            // create an empty stack of integers         Stack stk = new Stack();            // run n+1 times where n is length of input sequence         for (int i = 0; i <= seq.length(); i++) {             // push number i+1 into the stack             stk.push(i + 1);                // if all characters of the input sequence are             // processed or current character is 'I'             // (increasing)             if (i == seq.length() || seq.charAt(i) == 'I') {                 // run till stack is empty                 while (!stk.empty()) {                     // remove top element from the stack and                     // add it to solution                     result += String.valueOf(stk.peek());                     result += " ";                     stk.pop();                 }             }         }            System.out.println(result);     }    // main function     public static void main(String[] args) {         PrintMinNumberForPattern("IDID");         PrintMinNumberForPattern("I");         PrintMinNumberForPattern("DD");         PrintMinNumberForPattern("II");         PrintMinNumberForPattern("DIDI");         PrintMinNumberForPattern("IIDDD");         PrintMinNumberForPattern("DDIDDIID");     } } // This code is contributed by PrinciRaj1992

C#

 // C# program to print minimum number that can be formed // from a given sequence of Is and Ds using System; using System.Collections; public class GFG {     // Function to decode the given sequence to construct // minimum number without repeated digits     static void PrintMinNumberForPattern(String seq) {         // result store output string         String result = "";             // create an empty stack of integers         Stack stk = new Stack();             // run n+1 times where n is length of input sequence         for (int i = 0; i <= seq.Length; i++) {             // push number i+1 into the stack             stk.Push(i + 1);                 // if all characters of the input sequence are             // processed or current character is 'I'             // (increasing)             if (i == seq.Length || seq[i] == 'I') {                 // run till stack is empty                 while (stk.Count!=0) {                     // remove top element from the stack and                     // add it to solution                     result += String.Join("",stk.Peek());                     result += " ";                     stk.Pop();                 }             }         }             Console.WriteLine(result);     }     // main function     public static void Main() {         PrintMinNumberForPattern("IDID");         PrintMinNumberForPattern("I");         PrintMinNumberForPattern("DD");         PrintMinNumberForPattern("II");         PrintMinNumberForPattern("DIDI");         PrintMinNumberForPattern("IIDDD");         PrintMinNumberForPattern("DDIDDIID");     } } // This code is contributed by 29AjayKumar

Output:

1 3 2 5 4
1 2
3 2 1
1 2 3
2 1 4 3 5
1 2 6 5 4 3
3 2 1 6 5 4 7 9 8

Time Complexity : O(n)
Auxiliary Space : O(n)

This method is contributed by Roshni Agarwal.

Method 4 (Using two pointers)
Observation

1. Since we have to find minimum number without repeating digits, maximum length of output can be 9 (using each 1-9 digits once)
2. Length of the output will be exactly one greater than input length.
3. The idea is to iterate over the string and do the following if current character is ‘I’ or string is ended.
1. Assign count in increasing order to each element from current-1 to the next left index of ‘I’ (or starting index is reached).
2. Increase the count by 1.
Input  :  IDID
Output : 13254

Input  :  I
Output : 12

Input  :  DD
Output : 321

Input  :  II
Output : 123

Input  :  DIDI
Output : 21435

Input  :  IIDDD
Output : 126543

Input  :  DDIDDIID
Output : 321654798

Below is the implementation of above approach:

C++

 // C++ program of above approach #include   using namespace std;       // Returns minimum number made from given sequence without repeating digits  string getMinNumberForPattern(string seq)  {      int n = seq.length();        if (n >= 9)         return "-1";        string result(n+1, ' ');         int count = 1;          // The loop runs for each input character as well as      // one additional time for assigning rank to remaining characters     for (int i = 0; i <= n; i++)      {          if (i == n || seq[i] == 'I')         {             for (int j = i - 1 ; j >= -1 ; j--)             {                 result[j + 1] = '0' + count++;                 if(j >= 0 && seq[j] == 'I')                     break;             }         }     }     return result; }      // main function  int main()  {     string inputs[] = {"IDID", "I", "DD", "II", "DIDI", "IIDDD", "DDIDDIID"};        for (string input : inputs)     {         cout << getMinNumberForPattern(input) << " ";     }     return 0;  }

Java

 // Java program of above approach import java.io.IOException;    public class Test {     // Returns minimum number made from given sequence without repeating digits     static String getMinNumberForPattern(String seq)     {         int n = seq.length();            if (n >= 9)             return "-1";            char result[] = new char[n + 1];            int count = 1;            // The loop runs for each input character as well as          // one additional time for assigning rank to each remaining characters         for (int i = 0; i <= n; i++)         {             if (i == n || seq.charAt(i) == 'I')             {                 for (int j = i - 1; j >= -1; j--)                 {                     result[j + 1] = (char) ((int) '0' + count++);                     if (j >= 0 && seq.charAt(j) == 'I')                         break;                 }             }         }         return new String(result);     }            public static void main(String[] args) throws IOException     {         String inputs[] = { "IDID", "I", "DD", "II", "DIDI", "IIDDD", "DDIDDIID" };            for(String input : inputs)         {             System.out.println(getMinNumberForPattern(input));         }     } }

Python3

 # Python3 program of above approach        # Returns minimum number made from  # given sequence without repeating digits  def getMinNumberForPattern(seq):     n = len(seq)         if (n >= 9):         return "-1"        result = [None] * (n + 1)         count = 1        # The loop runs for each input character      # as well as one additional time for      # assigning rank to remaining characters     for i in range(n + 1):         if (i == n or seq[i] == 'I'):             for j in range(i - 1, -2, -1):                 result[j + 1] = int('0' + str(count))                 count += 1                 if(j >= 0 and seq[j] == 'I'):                      break     return result        # Driver Code  if __name__ == '__main__':     inputs = ["IDID", "I", "DD", "II",                "DIDI", "IIDDD", "DDIDDIID"]     for Input in inputs:         print(*(getMinNumberForPattern(Input)))    # This code is contributed by PranchalK

C#

 // C# program of above approach  using System;  class GFG {         // Returns minimum number made from given // sequence without repeating digits  static String getMinNumberForPattern(String seq)  {      int n = seq.Length;         if (n >= 9)          return "-1";         char []result = new char[n + 1];         int count = 1;         // The loop runs for each input character     // as well as one additional time for     // assigning rank to each remaining characters      for (int i = 0; i <= n; i++)      {          if (i == n || seq[i] == 'I')          {              for (int j = i - 1; j >= -1; j--)              {                  result[j + 1] = (char) ((int) '0' + count++);                  if (j >= 0 && seq[j] == 'I')                      break;              }          }      }      return new String(result);  }     // Driver Code public static void Main() {      String []inputs = { "IDID", "I", "DD", "II",                          "DIDI", "IIDDD", "DDIDDIID" };         foreach(String input in inputs)      {          Console.WriteLine(getMinNumberForPattern(input));      }  }  }     // This code is contributed by Rajput-Ji

Output:

13254
12
321
123
21435
126543
321654798

Time Complexity : O(N)
Auxiliary Space : O(1)

This solution is suggested by Brij Desai.