Encrypt given Array in single digit using inverted Pascal Triangle
Last Updated :
23 Jul, 2025
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 ith 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 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 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# 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
<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>
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 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 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))
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 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));
Time Complexity: O(N2)
Auxiliary Space: O(N)
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