根据给定的加密解密技术，在 Matrix 的帮助下解密编码的字符串

给定一个长度为N的编码（或加密）字符串S ，一个整数M 。任务是解密加密的字符串并打印出来。加密和解密技术如下：

Encryption: The original string is placed in a Matrix of M rows and N/M columns, such that the first character of the Original text or string is placed on the top-left cell to the bottom-right manner. If last row is reached, then again go to the top row, and start from the next column.

For example: If string is “geeks”, M = 2, then the string will be encrypted in below manner

Then traverse the matrix row wise and print the characters as they appear. 

Therefore the above string will be encrypted as = “ges ek”

Decryption: Traverse the matrix diagonally in the same manner as it was encrypted and find the actual string.

例子：

Input: “GSRE_ _ _E_ _K_ _ _EFGS_ _ _KOE” (Here ‘_’ means a space), 4
Output: “GEEKS FOR GEEKS”
Explanation:  See the image below for understanding the approach.

Here column number is 6. 
So, the traversing starts from (0, 0) position, then goes upto the end of the row, means from (0, 0) –> (1, 1)–>(2, 2)–>(3, 3)–>(4, 4). T
hen get to the first row, and start from the next, means from (0, 1)–>(1, 2)–>(2, 3)->(3, 4) and continues upto it reaches to the end of the given string.

Input: “GEEKSFORGEEKS”, 1
Output: “GEEKSFORGEEKS”
Explanation: There is only one row. so the decoded string will be as same as the given encoded string. 

Input: “abc de”, 2
Output: “adbec”

方法：首先，找到列数。然后，开始穿越。请按照以下步骤解决问题：

• 对于每一列，从第 i 行开始直到字符串的长度，每次遍历后将迭代值增加到column+1 。
• 因为，这里的遍历是对角进行的，这里的下一个对角字符将在列号加一之后。

For example, in the below picture, the X is a character, whose next diagonal character is XX, and the column number where X is present is 2, the next character row number is just one greater than the previous.

• 在遍历时将字符添加到空字符串中。然后检查字符串末尾是否有空格，如果有，则将其删除。

最后，打印那个字符串，这将是解码后的字符串/desired 字符串。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the desire string
string decodeString(string encodedText, int rows)
{
 
    // Stores the length of the string
    int len = encodedText.size();
 
    // Stores the number of columns
    int cols = len / rows;
 
    // Declaring an empty string
    string res;
    for (int i = 0; i < cols; ++i)
        for (int j = i; j < len; j += cols + 1)
            // Adding the characters
            // into the empty string
            res += encodedText[j];
 
    // If their any space at the end,
    // delete it
    while (res.back() == ' ') {
        res.pop_back();
    }
    return res;
}
 
// Driver Code
int main()
{
 
    string S = "GSRE   E  K   EFGS   KOE";
    int row = 4;
    cout << decodeString(S, row) << endl;
    return 0;
}

Java

// Java program for the above approach
class GFG {
 
  // Function to find the desire string
  static String decodeString(String encodedText, int rows) {
 
    // Stores the length of the string
    int len = encodedText.length();
 
    // Stores the number of columns
    int cols = len / rows;
 
    // Declaring an empty string
    String res = "";
    for (int i = 0; i < cols; ++i)
    {
      for (int j = i; j < len; j += cols + 1)
      {
        // Adding the characters
        // into the empty string
        res += encodedText.charAt(j);
      }
    }
 
    // If their any space at the end,
    // delete it
    while (res.charAt(res.length() - 1) == ' ') {
      res = res.substring(0, res.length() - 2);
    }
    return res;
  }
 
  // Driver Code
  public static void main(String args[]) {
    String S = "GSRE   E  K   EFGS   KOE";
    int row = 4;
 
    System.out.println(decodeString(S, row));
 
  }
}
 
// This code is contributed by gfgking

Python3

# Python code for the above approach
 
# Function to find the desire string
def decodeString(encodedText, rows):
 
    # Stores the length of the string
    _len = len(encodedText)
 
    # Stores the number of columns
    cols = _len // rows
 
    # Declaring an empty string
    res = "";
    for i in range(cols):
        for j in range(i, _len, cols + 1):
         
            # Adding the characters
            # into the empty string
            res += encodedText[j];
 
    # If their any space at the end,
    # delete it
    while (res[len(res) - 1] == ' '):
        res = res[0: len(res) - 1];
    return res;
 
 
# Driver Code
S = "GSRE   E  K   EFGS   KOE";
row = 4;
print(decodeString(S, row))
 
# This code is contributed by Saurabh Jaiswal

C#

// C# program for the above approach
using System;
class GFG
{
 
// Function to find the desire string
static string decodeString(string encodedText, int rows)
{
 
    // Stores the length of the string
    int len = encodedText.Length;
 
    // Stores the number of columns
    int cols = len / rows;
 
    // Declaring an empty string
    string res = "";
    for (int i = 0; i < cols; ++i) {
        for (int j = i; j < len; j += cols + 1) {
            // Adding the characters
            // into the empty string
            res += encodedText[j];
        }
    }
  
    // If their any space at the end,
    // delete it
    while (res[res.Length - 1] == ' ') {
        res = res.Remove(res.Length - 1);
    }
    return res;
}
 
// Driver Code
public static void Main()
{
    string S = "GSRE   E  K   EFGS   KOE";
    int row = 4;
 
    Console.Write(decodeString(S, row));
 
}
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

      // JavaScript code for the above approach
 
      // Function to find the desire string
      function decodeString(encodedText, rows) {
 
          // Stores the length of the string
          let len = encodedText.length;
 
          // Stores the number of columns
          let cols = Math.floor(len / rows);
 
          // Declaring an empty string
          let res = "";
          for (let i = 0; i < cols; ++i)
              for (let j = i; j < len; j += cols + 1)
               
                  // Adding the characters
                  // into the empty string
                  res += encodedText[j];
 
          // If their any space at the end,
          // delete it
          while (res[res.length - 1] == ' ') {
              res = res.slice(0, res.length - 1);
          }
          return res;
      }
 
      // Driver Code
      let S = "GSRE   E  K   EFGS   KOE";
      let row = 4;
      document.write(decodeString(S, row))
 
// This code is contributed by Potta Lokesh
  

输出

GEEKS FOR GEEKS

时间复杂度： O(M*(M/col))，接近 O(字符串长度)
辅助空间： O(N)