# Generate a sequence with the given operations

• Difficulty Level : Easy
• Last Updated : 14 May, 2021

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:

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

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<``int``> 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<``int``> ans = ``new` `List<``int``>();``    ``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`

## Javascript

 ``
Output:
`[0, 4, 1, 3, 2]`

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