Generate a sequence with the given operations

Given a string which contains only (increase) and (decrease). The task is to return any permutation of integers [0, 1, …, N] where N ≤ Length of S such that for all i = 0, …, N-1:

1. If S[i] == “D”, then A[i] > A[i+1]
2. If S[i] == “I”, then A[i] < A[i+1].

Note that output must contain distinct elements.

Examples:

Input: S = “DDI”
Output: [3, 2, 0, 1]

Input: S = “IDID”
Output: [0, 4, 1, 3, 2]

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: If S == “I”, then choose as the first element. Similarly, if S == “D”, then choose as the first element. Now for every operation, choose the next maximum element which hasn’t been chosen before from the range [0, N] and for the operation, choose the next minimum.

Below is the implementation of the above approach:

C++

 //C++ Implementation of above approach  #include using namespace std;     // function to find minimum required permutation      void  StringMatch(string s)      {     int lo=0, hi = s.length(), len=s.length();      vector ans;     for (int x=0;x

Java

 // Java Implementation of above approach  import java.util.*;    class GFG  {    // function to find minimum required permutation  static void StringMatch(String s)  {     int lo=0, hi = s.length(), len=s.length();      Vector ans = new Vector<>();     for (int x = 0; x < len; x++)     {         if (s.charAt(x) == 'I')          {             ans.add(lo) ;             lo += 1;         }         else         {             ans.add(hi) ;             hi -= 1;         }     }             ans.add(lo) ;     System.out.print("[");     for(int i = 0; i < ans.size(); i++)     {         System.out.print(ans.get(i));         if(i != ans.size()-1)             System.out.print(",");     }     System.out.print("]"); }    // Driver code public static void main(String[] args) {     String S = "IDID";     StringMatch(S);  } }    // This code is contributed by Rajput-Ji

Python

 # Python Implementation of above approach    # function to find minimum required permutation def StringMatch(S):     lo, hi = 0, len(S)     ans = []     for x in S:         if x == 'I':             ans.append(lo)             lo += 1         else:             ans.append(hi)             hi -= 1        return ans + [lo]    # Driver code S = "IDID" print(StringMatch(S))

C#

 // C# Implementation of above approach  using System; using System.Collections.Generic;    class GFG  {    // function to find minimum required permutation  static void StringMatch(String s)  {     int lo=0, hi = s.Length, len=s.Length;      List ans = new List();     for (int x = 0; x < len; x++)     {         if (s[x] == 'I')          {             ans.Add(lo) ;             lo += 1;         }         else         {             ans.Add(hi) ;             hi -= 1;         }     }             ans.Add(lo) ;     Console.Write("[");     for(int i = 0; i < ans.Count; i++)     {         Console.Write(ans[i]);         if(i != ans.Count-1)             Console.Write(",");     }     Console.Write("]"); }    // Driver code public static void Main(String[] args) {     String S = "IDID";     StringMatch(S);  } }    // This code is contributed by 29AjayKumar

Output:

[0, 4, 1, 3, 2]

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