Decrypt the encoded string with help of Matrix as per given encryption decryption technique - GeeksforGeeks

14 Mar, 2022
Given an encoded (or encrypted) string S of length N, an integer M. The task is to decrypt the encrypted string and print it. The encryption and decryption techniques are given as:
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.
Examples:
Input: “GSRE_ _ _E_ _K_ _ _EFGS_ _ _KOE” (Here ‘_’ means a space), 4Output: “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). Then 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”, 1Output: “GEEKSFORGEEKS”Explanation: There is only one row. so the decoded string will be as same as the given encoded string. 
Input: “abc de”, 2Output: “adbec”
Approach: First, find the number of columns. Then, start traversing. Follow the below steps to solve the problem:
For each of the columns, go up to the length of the string starting from the i’th row, and after each traversal increase the iterating value to column+1.
Because, here traversal is done diagonally, and here the next diagonal character will be after column number plus one.
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.
At the time of traversing add the characters into an empty string.  Then check, if there is any space at the end of the string or not, if it is, then just delete it.
At last, print that string, and this will be the decoded string/desired string.
Below is the implementation of the above approach:
C++
Java
Python3
C#
Javascript
// C++ program for the above approach#include <bits/stdc++.h>using namespace std; // Function to find the desire stringstring 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 Codeint main(){     string S = "GSRE   E  K   EFGS   KOE";    int row = 4;    cout << decodeString(S, row) << endl;    return 0;}
// Java program for the above approachclass 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
# Python code for the above approach # Function to find the desire stringdef 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 CodeS = "GSRE   E  K   EFGS   KOE";row = 4;print(decodeString(S, row)) # This code is contributed by Saurabh Jaiswal
// C# program for the above approachusing System;class GFG{ // Function to find the desire stringstatic 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 Codepublic 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.
<script>      // 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  </script>
GEEKS FOR GEEKS
Time Complexity: O(M*(M/col)) which is near about O(length of the string)Auxiliary Space: O(N)
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