Given a number of the form 11..1 such that number of digits in it is smaller than 10, find square of the number.
Examples:
Input : 111111 Output : 12345654321 Input : 1111 Output : 1234321
Squares of 11…1 (where length is smaller than 10) are called Demlo numbers. A demlo number is number which consist of 1, 2, 3….n, n-1, n-2, …..1.
Demlo numbers are:
1 = 1 11 * 11 = 121 111 * 111 = 12321 1111 * 1111 = 1234321
To find demlo number, first we find append n numbers which are increasing and then append numbers which are decreasing.
C++
// CPP program to print DemloNumber #include <bits/stdc++.h> using namespace std; // To return demlo number. This function assumes // that the length of str is smaller than 10. string printDemlo(string str) { int len = str.length(); string res = ""; // Add numbers to res upto size // of str and then add number // reverse to it for (int i = 1; i <= len; i++) res += char(i + '0'); for (int i = len - 1; i >= 1; i--) res += char(i + '0'); return res; } // Driver program to test printDemlo() int main() { string str = "111111"; cout << printDemlo(str); return 0; } |
Java
// Java program to print DemloNumber public class Main { // To return demlo number. This function assumes // that the length of str is smaller than 10. static String printDemlo(String str) { int len = str.length(); String res = ""; // Add numbers to res upto size // of str and then add number // reverse to it for (int i = 1; i <= len; i++) res += Integer.toString(i); for (int i = len - 1; i >= 1; i--) res += Integer.toString(i); return res; } // Driver program to test printDemlo() public static void main(String[] args) { String str = "111111"; System.out.println(printDemlo(str)); } } |
Python3
# Python program to print Demlo Number # To return demlo number # Length of s is smaller than 10 def printDemlo(s): l = len(s) res = "" # Add numbers to res upto size # of s then add in reverse for i in range(1,l+1): res = res + str(i) for i in range(l-1,0,-1): res = res + str(i) return res # Driver Code s = "111111" print (printDemlo(s)) # Contributed by Harshit Agrawal |
C#
// C# program to print DemloNumber using System; public class GFG { // To return demlo number. This function assumes // that the length of str is smaller than 10. static String printDemlo(String str) { int len = str.Length; String res = ""; // Add numbers to res upto size // of str and then add number // reverse to it for (int i = 1; i <= len; i++) res += i.ToString(); for (int i = len - 1; i >= 1; i--) res += i.ToString(); return res; } // Driver program to test printDemlo() public static void Main() { String str = "111111"; Console.WriteLine(printDemlo(str)); } } // This code is contributed by mits |
PHP
<?php // PHP program to print DemloNumber // To return demlo number. This function // assumes that the length of str is // smaller than 10. function printDemlo($str) { $len = strlen($str); $res = ""; // Add numbers to res upto size // of str and then add number // reverse to it for ($i = 1; $i <= $len; $i++) $res .= chr($i + 48); for ($i = $len - 1; $i >= 1; $i--) $res .= chr($i + 48); return $res; } // Driver Code $str = "111111"; echo printDemlo($str); // This code is contributed by mits ?> |
Javascript
<script> // Javascript program to print DemloNumber // To return demlo number. This function assumes // that the length of str is smaller than 10. function printDemlo(str) { let len = str.length; let res = ""; // Add numbers to res upto size // of str and then add number // reverse to it for (let i = 1; i <= len; i++) res += i.toString(); for (let i = len - 1; i >= 1; i--) res += i.toString(); return res; } // driver program let str = "111111"; document.write(printDemlo(str)); // This code is contributed by susmitakundugoaldanga. </script> |
Output:
12345654321
Time complexity: O(log N) where N is the length of string “str” of number n
Auxiliary space: O(1) because it is using constant space
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