Given a binary string **S** of length **N** (*1 ≤ N ≤ 10 ^{5}*), the task is to print a permutation

**A**consisting of integers

**0**to

**N**satisfying the following conditions:

- If
**S[i] = 1**, then**A[i] < A[i + 1]**. - If
**S[i] = 0**, then**A[i] > A[i + 1]**.

**Examples:**

Input:S = “001”Output:[3, 2, 0, 1]Explanation:

S[0] = 0 and A[0] (= 3) > A[1] (= 2)

S[1] = 0 and A[1] (= 2) > A[2] (= 0)

S[2] = 1 and A[2] (= 0) > A[3] (= 1)

Input:S = “1010”Output:[0, 4, 1, 3, 2]

**Approach:** Follow the steps below to solve the problem:

- Assign the lowest number possible if the character encountered in the string is
**‘1’**. - Assign the highest number possible if the character encountered in the string is
**‘0’**. - Set two pointers
**lo (= 0)**, storing the current lowest possible, and**hi (= N)**, storing the current highest possible. - Traverse the string and append the values to the resultant array, say
**re**s, from the range using the two pointers accordingly. - After complete traversal of the string, only one value is remaining to be appended. Append
**lo**to the last index. - Print the array
**res**as the required result.

Below is the implementation of the above approach:

## C++

`// C++ program to implement` `// the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to print the` `// required sequence` `void` `StringPattern(string S)` `{` ` ` `// Length of the string` ` ` `int` `n = (` `int` `)S.size();` ` ` `// Pointers to store lowest and` ` ` `// highest possible values` ` ` `int` `l = 0, r = n;` ` ` `// Stores the required answer` ` ` `vector<` `int` `> res(n + 1, 0);` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` `switch` `(S[i]) {` ` ` `// If current character is '1'` ` ` `case` `'1'` `:` ` ` `// Assign smallest` ` ` `// possible value` ` ` `res[i] = l++;` ` ` `break` `;` ` ` `// If current character is '0'` ` ` `case` `'0'` `:` ` ` `// Assign largest` ` ` `// possible value` ` ` `res[i] = r--;` ` ` `break` `;` ` ` `}` ` ` `}` ` ` `// At this point, l == r ，` ` ` `// only one value is not assigned yet.` ` ` `res[n] = l;` ` ` `for` `(` `int` `el : res) {` ` ` `cout << el << ` `" "` `;` ` ` `}` `}` `// Driver Code` `int` `main()` `{` ` ` `string s = ` `"001"` `;` ` ` `StringPattern(s);` ` ` `return` `0;` `}` |

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## Java

`// Java Implementation of above approach ` `import` `java.util.*; ` `class` `GFG ` `{ ` `// function to find minimum required permutation ` `static` `void` `StringPattern(String s) ` `{ ` ` ` `int` `n = s.length();` ` ` `int` `l = ` `0` `, r = n; ` ` ` `int` `[] res = ` `new` `int` `[n + ` `1` `]; ` ` ` `for` `(` `int` `x = ` `0` `; x < n; x++) ` ` ` `{ ` ` ` `if` `(s.charAt(x) == ` `'1'` `) ` ` ` `{ ` ` ` `res[x] = l++;` ` ` `} ` ` ` `else` `if` `(s.charAt(x) == ` `'0'` `)` ` ` `{ ` ` ` `res[x] = r--;` ` ` `} ` ` ` `} ` ` ` `res[n] = l;` ` ` `for` `(` `int` `i = ` `0` `; i < n+` `1` `; i++) ` ` ` `{ ` ` ` `System.out.print(res[i]+` `" "` `); ` ` ` `}` `} ` `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` ` ` `String s = ` `"001"` `;` ` ` `StringPattern(s); ` `} ` `} ` `// This code is contributed by mohit kumar 29.` |

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## Python3

`# Python program to implement` `# the above approach` `# Function to print the` `# required sequence` `def` `StringPattern(S):` ` ` `# Stores the lowest, highest` ` ` `# possible values that can be` ` ` `# appended, size of the string` ` ` `lo, hi ` `=` `0` `, ` `len` `(S)` ` ` `# Stores the required result` ` ` `ans ` `=` `[]` ` ` `# Traverse the string` ` ` `for` `x ` `in` `S:` ` ` `# If current character is '1'` ` ` `if` `x ` `=` `=` `'1'` `:` ` ` `# Append lowest` ` ` `# possible value` ` ` `ans.append(lo)` ` ` `lo ` `+` `=` `1` ` ` `# If current character is '0'` ` ` `else` `:` ` ` `# Append highest` ` ` `# possible value` ` ` `ans.append(hi)` ` ` `hi ` `-` `=` `1` ` ` `return` `ans ` `+` `[lo]` `# Driver Code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `# Input` ` ` `s ` `=` `'001'` ` ` `print` `(StringPattern(s))` |

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## C#

`// C# Implementation of above approach ` `using` `System;` `class` `GFG ` `{ ` `// function to find minimum required permutation ` `static` `void` `StringPattern(String s) ` `{ ` ` ` `int` `n = s.Length;` ` ` `int` `l = 0, r = n; ` ` ` `int` `[] res = ` `new` `int` `[n + 1]; ` ` ` `for` `(` `int` `x = 0; x < n; x++) ` ` ` `{ ` ` ` `if` `(s[x] == ` `'1'` `) ` ` ` `{ ` ` ` `res[x] = l++;` ` ` `} ` ` ` `else` `if` `(s[x] == ` `'0'` `)` ` ` `{ ` ` ` `res[x] = r--;` ` ` `} ` ` ` `} ` ` ` `res[n] = l;` ` ` `for` `(` `int` `i = 0; i < n + 1; i++) ` ` ` `{ ` ` ` `Console.Write(res[i] + ` `" "` `); ` ` ` `}` `} ` `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` ` ` `String s = ` `"001"` `;` ` ` `StringPattern(s); ` `} ` `} ` `// This code is contributed by 29AjayKumar` |

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**Output:**

3 2 0 1

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

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