Encrypt given Array in single digit using inverted Pascal Triangle

Last Updated : 30 Mar, 2023

Given an array arr[] of length N (N > 1)containing positive integers, the task is to encrypt the numbers of the array into a single digit using the inverted Pascal triangle as follows.

From the starting of the array find the sum of two adjacent elements.
Replace the sum with only the digit at the unit position of the sum.
Replace all the array elements with the values formed in this way and continue until there are only two elements left.
The last two elements are concatenated together.

Examples:

Input: arr[] = {4, 5, 6, 7}
Output: 04
Explanation:

Pascal’s encryption of [4, 5, 6, 7]

Input: arr[] = {1, 2, 3}
Output: 35
Explanation:

Pascal’s encryption of [1, 2, 3]

Input: arr[] = {14, 5}
Output: 145
Explanation: As there were two elements they are appended together

Approach: This problem can be solved using recursion based on the following idea:

Calculate the sum of all i^th with (i-1)th element and mod by 10 to get least significant digit for next operation until the whole container becomes of length 2.

Follow the steps to solve the problem:

Use a recursive function and do the following:
- Traverse numbers to calculate sum of adjacent elements and mod with 10 to get single least significant digit as numbers[i]=(numbers[i]+numbers[i+1])%10
- Delete the last element from the array, as one element will be reduced after each operation.
- Continue this procedure until only 2 elements are left.

Below is the implementation of the above approach:

C++14

// C++ code for the above approach:
 
#include <bits/stdc++.h>
using namespace std;
 
// Recursive function to find the encryption
string digitEncrypt(vector<int>& numbers)
{
    int N = numbers.size();
    string ans;
 
    // If the value of N is 2
    if (N == 2) {
        if (numbers[0] == 0 && numbers[1] == 0)
            return "00";
        else if (numbers[0] == 0)
            return "0" + to_string(numbers[1]);
        return to_string(numbers[0])
               + to_string(numbers[1]);
    }
 
    for (int i = 0; i < N - 1; i++)
        numbers[i] = (numbers[i]
                      + numbers[i + 1])
                     % 10;
 
    numbers.pop_back();
 
    return digitEncrypt(numbers);
}
 
// Drivers code
int main()
{
    vector<int> numbers = { 4, 5, 6, 7 };
 
    // Function call
    cout << digitEncrypt(numbers);
    return 0;
}

Java

// Java code for the above approach:
import java.io.*;
import java.util.*;
 
class GFG 
{
 
  // Recursive function to find the encryption
  public static String
    digitEncrypt(ArrayList<Integer> numbers)
  {
    int N = numbers.size();
 
    // If the value of N is 2
    if (N == 2) {
      if (numbers.get(0) == 0 && numbers.get(1) == 0)
        return "00";
      else if (numbers.get(0) == 0)
        return "0"
        + Integer.toString(numbers.get(1));
      else
        return Integer.toString(numbers.get(0))
        + Integer.toString(numbers.get(1));
    }
 
    for (int i = 0; i < N - 1; i++)
      numbers.set(
      i, ((numbers.get(i) + numbers.get(i + 1))
          % 10));
 
    numbers.remove(numbers.size() - 1);
 
    return digitEncrypt(numbers);
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    ArrayList<Integer> numbers = new ArrayList<Integer>(
      Arrays.asList(4, 5, 6, 7));
 
    // Function call
    System.out.print(digitEncrypt(numbers));
  }
}
 
// This code is contributed by Rohit Pradhan

Python3

# python3 code to implement the approach
 
# Recursive function to find the encryption
def digitEncrypt(numbers) :
    N = len(numbers)
     
    # If the value of N is 2
    if N == 2 :
        if numbers[0] == 0 and numbers[1] == 0 :
            return "00"
        elif numbers[0] == 0:
            return "0" + str((numbers[1]))
        return str(numbers[0])+ str(numbers[1])
     
    for i in range(0,N-1) : 
        numbers[i] = (numbers[i]+ numbers[i + 1])% 10
  
    numbers.pop()
  
    return digitEncrypt(numbers)
 
# Driver code
if __name__ == "__main__":
 
    numbers = [ 4, 5, 6, 7 ]
 
    # Function call
    print(digitEncrypt(numbers))
 
# This code is contributed by jana_sayantan.

C#

// C# code for the above approach:
 
using System;
using System.Collections.Generic;
 
class GFG {
 
    // Recursive function to find the encryption
    public static string digitEncrypt(List<int> numbers)
    {
        int N = numbers.Count;
 
        // If the value of N is 2
        if (N == 2) {
            if (numbers[0] == 0 && numbers[1] == 0)
                return "00";
            else if (numbers[0] == 0)
                return "0" + Convert.ToString(numbers[1]);
            else
                return Convert.ToString(numbers[0])
                    + Convert.ToString(numbers[1]);
        }
 
        for (int i = 0; i < N - 1; i++)
            numbers[i] = (numbers[i] + numbers[i + 1]) % 10;
 
        numbers.RemoveAt(numbers.Count - 1);
 
        return digitEncrypt(numbers);
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        List<int> numbers
            = new List<int>(new int[] { 4, 5, 6, 7 });
 
        // Function call
        Console.Write(digitEncrypt(numbers));
    }
}
 
// This code is contributed by phasing17

Javascript

<script>
// JavaScript code for the above approach:
 
// Recursive function to find the encryption
function digitEncrypt(numbers)
{
    var N = numbers.length;
    var ans;
 
    // If the value of N is 2
    if (N == 2) {
        if (numbers[0] == 0 && numbers[1] == 0)
            return "00";
        else if (numbers[0] == 0)
            return "0" + (numbers[1]);
        return to_string(numbers[0])
               + to_string(numbers[1]);
    }
 
    for (var i = 0; i < N - 1; i++)
        numbers[i] = (numbers[i]
                      + numbers[i + 1])
                     % 10;
 
    numbers.pop();
 
    return digitEncrypt(numbers);
}
 
// Drivers code
var numbers = [ 4, 5, 6, 7 ];
 
// Function call
document.write(digitEncrypt(numbers));
 
// This code is contributed by phasing17.
 
</script>

Output

Time Complexity: O(N²)
Auxiliary Space: O(N)

Approach 2: Using Queue ( just read the code you will get the approach by comments)

C++

#include <bits/stdc++.h>
using namespace std;
 
 
// iterative function to find the encryption
string digitEncrypt(vector<int> numbers){
     
    queue<int>q; 
     
    //push all elements of vector number to queue and traverse queue until we get the final result 
    for(auto x: numbers){ 
        q.push(x);
    }
     
    // when queue has only 2 elements left then stop the cycle 
    while(q.size() != 2){
        int n = q.size()-1;
        while(n--){
            int a= q.front(); // first element
            q.pop();
            int b = q.front();  // 2nd element
            int sum = (a%10) + (b%10);  // if 'a' or 'b' greater than 10 then just '%10' to get make is a single digit number
            sum = sum%10; // same with sum to make it a single digit number
            q.push(sum); // push that sum in queue for the another traversal
        }
        q.pop();
    }
     
    // now made from both the digit by converting them into string
    // we convert them into string because when a=0 we cannot do ans = a*10 + b
    // thats why we convert them into string and then add to them into our answer string
    int a= q.front();
    q.pop();
    int b = q.front();
    string ans = "";
    ans+=to_string(a);
    ans+=to_string(b);
     
    return ans;
}
 
int main()
{
    vector<int> numbers = { 4, 5, 6, 7 };
  
    cout << digitEncrypt(numbers);
    return 0;
}

Java

// Java code for the above approach:
import java.io.*;
import java.util.*;
 
class GFG {
 
    //  Iterative function to find the encryption
    public static String
    digitEncrypt(ArrayList<Integer> numbers)
    {
        Queue<Integer> q = new LinkedList<>();
 
        // push all elements of vector number to queue and
        // traverse queue until we get the final result
        for (Integer x : numbers) {
            q.add(x);
        }
 
        // when queue has only 2 elements left then stop the
        // cycle
        while (q.size() != 2) {
            int n = q.size() - 1;
            while (n > 0) {
                // first element
                int a = q.peek();
                q.remove();
                // 2nd element
                int b = q.peek();
                // if 'a' or 'b' greater than 10 then just
                // '%10' to get make is a single digit
                // number
                int sum = (a % 10) + (b % 10);
                // same with sum to make it a single digit
                // number
                sum = sum % 10;
                // push that sum in queue for the another
                // traversal
                q.add(sum);
                n--;
            }
            q.remove();
        }
 
        // now made from both the digit by converting them
        // into string we convert them into string because
        // when a=0 we cannot do ans = a*10 + b thats why we
        // convert them into string and then add to them
        // into our answer string
        int a = q.peek();
        q.remove();
        int b = q.peek();
        String ans = String.valueOf(a);
        ans += String.valueOf(b);
 
        return ans;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        ArrayList<Integer> numbers = new ArrayList<Integer>(
            Arrays.asList(4, 5, 6, 7, 8, 9));
 
        // Function call
        System.out.print(digitEncrypt(numbers));
    }
}
 
// This code is contributed by shubhamrajput6156

Python3

# Python3 code for the above approach:
 
# function to find the encryption
def digitEncrypt(numbers):
  q = []
   
  # push all elements of numbers list to queue
  for x in numbers:
      q.append(x)
 
  # traverse queue until we get the final result
  while len(q) != 2:
      n = len(q)-1
      while n > 0:
          a = q.pop(0) # first element
          b = q[0] # 2nd element
          sum = (a % 10) + (b % 10) # if 'a' or 'b' greater than 10 then just '%10' to get make is a single digit number
          sum = sum % 10 # same with sum to make it a single digit number
          q.append(sum) # push that sum in queue for the another traversal
          n -= 1
      q.pop(0)
 
  # now made from both the digit by 
  # converting them into string
  # we convert them into string because 
  # when a=0 we cannot do ans = a*10 + b
  # thats why we convert them into string 
  # and then add to them into our answer string
  a = q.pop(0)
  b = q[0]
  ans = str(a) + str(b)
 
  return ans
 
# Driver Code
if __name__ == "__main__":
   
  # Input array
  numbers = [4, 5, 6, 7]
   
  # Function Call
  print(digitEncrypt(numbers))

C#

using System;
using System.Collections.Generic;
 
public class GFG {
     
    // Iterative function to find the encryption
    public static string DigitEncrypt(List<int> numbers) {
         
        Queue<int> q = new Queue<int>();
         
        // Push all elements of the list to the queue and
        // traverse queue until we get the final result
        foreach (int x in numbers) {
            q.Enqueue(x);
        }
         
        // When queue has only 2 elements left then stop the cycle
        while (q.Count != 2) {
            int n = q.Count - 1;
            while (n > 0) {
                // First element
                int a = q.Peek();
                q.Dequeue();
                // Second element
                int b = q.Peek();
                // If 'a' or 'b' greater than 10 then just
                // '%10' to get make it a single digit number
                int sum = (a % 10) + (b % 10);
                // Same with sum to make it a single digit
                // number
                sum = sum % 10;
                // Push that sum in queue for the another
                // traversal
                q.Enqueue(sum);
                n--;
            }
            q.Dequeue();
        }
         
        // Now made from both the digit by converting them
        // into string we convert them into string because
        // when a=0 we cannot do ans = a*10 + b thats why we
        // convert them into string and then add to them
        // into our answer string
        int a1 = q.Peek();
        q.Dequeue();
        int b1 = q.Peek();
        string ans = a1.ToString() + b1.ToString();
         
        return ans;
    }
     
    // Driver Code
    public static void Main(string[] args) {
        List<int> numbers = new List<int>(new int[] {4, 5, 6, 7, 8, 9});
         
        // Function call
        Console.Write(DigitEncrypt(numbers));
    }
}

Javascript

// JavaScript code for the above approach:
function digitEncrypt(numbers) {
    const q = [];
 
    // push all elements of array 'numbers' to queue and
    // traverse queue until we get the final result
    for (let i = 0; i < numbers.length; i++) {
        q.push(numbers[i]);
    }
 
    // when queue has only 2 elements left then stop the
    // cycle
    while (q.length !== 2) {
        let n = q.length - 1;
        while (n > 0) {
            // first element
            const a = q.shift();
            // 2nd element
            const b = q[0];
            // if 'a' or 'b' greater than 10 then just
            // '%10' to get make it a single digit number
            const sum = (a % 10) + (b % 10);
            // same with sum to make it a single digit number
            const digitSum = sum % 10;
            // push that digitSum in queue for the another
            // traversal
            q.push(digitSum);
            n--;
        }
        q.shift();
    }
 
    // now made from both the digit by converting them
    // into string we convert them into string because
    // when a=0 we cannot do ans = a*10 + b thats why we
    // convert them into string and then add to them
    // into our answer string
    const a = q.shift();
    const b = q.shift();
    let ans = a.toString();
    ans += b.toString();
 
    return ans;
}
 
// Driver Code
const numbers = [4, 5, 6, 7, 8, 9];
 
// Function call
console.log(digitEncrypt(numbers));

Output

Time Complexity: O(N2)
Auxiliary Space: O(N)

Count all triplets from given Array whose bitwise XOR is equal to K

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Encrypt given Array in single digit using inverted Pascal Triangle

C++14

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Python3

C#

Javascript

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Java

Python3

C#

Javascript

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