Related Articles

# Check if a given number is one less than twice its reverse

• Last Updated : 20 Apr, 2021

Given an integer N, the task is to check if it is a solution to the equation 2 * reverse(N) – 1 = N

Examples:

Input: N = 73
Output: Yes
Explanation:
2 * reverse(N) = 2 * 37 = 74
N + 1 = 73 + 1 = 74

Input: N = 83
Output: No

Naive Approach: The simplest approach is to find the reverse of the given number and check if it satisfies the equation 2 * reverse(N) = N + 1 or not and print “Yes” or “No” accordingly.

Below is the implementation of the above approach:

## C++

 `// C++ program of the``// above approach` `#include ``using` `namespace` `std;` `// Iterative function to``// reverse digits of num``int` `rev(``int` `num)``{``    ``int` `rev_num = 0;` `    ``// Loop to extract all``    ``// digits of the number``    ``while` `(num > 0) {``        ``rev_num``            ``= rev_num * 10 + num % 10;``        ``num = num / 10;``    ``}``    ``return` `rev_num;``}` `// Function to check if N``// satisfies given equation``bool` `check(``int` `n)``{``    ``return` `2 * rev(n) == n + 1;``}` `// Driver Code``int` `main()``{``    ``int` `n = 73;``    ``if` `(check(n))``        ``cout << ``"Yes"``;``    ``else``        ``cout << ``"No"``;``    ``return` `0;``}`

## Java

 `// Java program of the``// above approach``import` `java.util.*;``class` `GFG{` `// Iterative function to``// reverse digits of num``static` `int` `rev(``int` `num)``{``  ``int` `rev_num = ``0``;` `  ``// Loop to extract all``  ``// digits of the number``  ``while` `(num > ``0``)``  ``{``    ``rev_num = rev_num * ``10` `+``              ``num % ``10``;``    ``num = num / ``10``;``  ``}``  ``return` `rev_num;``}` `// Function to check if N``// satisfies given equation``static` `boolean` `check(``int` `n)``{``  ``return` `2` `* rev(n) == n + ``1``;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``  ``int` `n = ``73``;``  ``if` `(check(n))``    ``System.out.print(``"Yes"``);``  ``else``    ``System.out.print(``"No"``);``}``}` `// This code is contributed by Princi Singh`

## Python3

 `# Python3 program of the above approach``  ` `# Iterative function to``# reverse digits of num``def` `rev(num):``     ` `    ``rev_num ``=` `0``   ` `    ``# Loop to extract all``    ``# digits of the number``    ``while` `(num > ``0``):``        ``rev_num ``=` `(rev_num ``*` `10` `+``                       ``num ``%` `10``)``        ``num ``=` `num ``/``/` `10``      ` `    ``return` `rev_num``  ` `# Function to check if N``# satisfies given equation``def` `check(n):``    ` `    ``return` `(``2` `*` `rev(n) ``=``=` `n ``+` `1``)``  ` `# Driver Code``n ``=` `73` `if` `(check(n)):``    ``print``(``"Yes"``)  ``else``:``    ``print``(``"No"``)` `# This code is contributed by code_hunt`

## C#

 `// C# program of the above approach``using` `System;` `class` `GFG{` `// Iterative function to``// reverse digits of num``static` `int` `rev(``int` `num)``{``    ``int` `rev_num = 0;``    ` `    ``// Loop to extract all``    ``// digits of the number``    ``while` `(num > 0)``    ``{``        ``rev_num = rev_num * 10 +``                      ``num % 10;``        ``num = num / 10;``    ``}``    ``return` `rev_num;``}` `// Function to check if N``// satisfies given equation``static` `bool` `check(``int` `n)``{``    ``return` `2 * rev(n) == n + 1;``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``int` `n = 73;``    ` `    ``if` `(check(n))``        ``Console.Write(``"Yes"``);``    ``else``        ``Console.Write(``"No"``);``}``}` `// This code is contributed by Amit Katiyar`

## Javascript

 ``
Output:
`Yes`

Efficient Approach: The key observation to optimize the above approach is that the numbers satisfying the given equation can be represented by:

X = 8 * 10(n-1) – 7

To check if a number X satisfies the above equation, it needs to be checked if the number (X + 7) / 8 is a power of 10 or not.

Below is the implementation of the above approach:

## C++

 `// C++ program to implement``// the above approach` `#include ``using` `namespace` `std;` `// Function to check y is a power of x``bool` `isPower(``int` `x, ``int` `y)``{``    ``// logarithm function to``    ``// calculate value``    ``int` `res1 = ``log``(y) / ``log``(x);``    ``double` `res2 = ``log``(y) / ``log``(x);` `    ``// Compare to the result1 or``    ``// result2 both are equal``    ``return` `(res1 == res2);``}` `// Function to check if N``// satisfies the equation``// 2 * reverse(n) = n + 1``bool` `check(``int` `n)``{``    ``int` `x = (n + 7) / 8;``    ``if` `((n + 7) % 8 == 0``        ``&& isPower(10, x))``        ``return` `true``;``    ``else``        ``return` `false``;``}` `// Driver Code``int` `main()``{``    ``int` `n = 73;``    ``if` `(check(n))``        ``cout << ``"Yes"``;``    ``else``        ``cout << ``"No"``;``    ``return` `0;``}`

## Java

 `// Java program to implement``// the above approach``import` `java.io.*;``import` `java.util.*;` `class` `GFG{``  ` `// Function to check y is a power of x``static` `boolean` `isPower(``int` `x, ``int` `y)``{``    ` `    ``// logarithm function to``    ``// calculate value``    ``double` `res1 = Math.log(y) / Math.log(x);``    ``double` `res2 = Math.log(y) / Math.log(x);``  ` `    ``// Compare to the result1 or``    ``// result2 both are equal``    ``return` `(res1 == res2);``}``  ` `// Function to check if N``// satisfies the equation``// 2 * reverse(n) = n + 1``static` `boolean` `check(``int` `n)``{``    ``int` `x = (n + ``7``) / ``8``;``    ` `    ``if` `((n + ``7``) % ``8` `== ``0` `&&``        ``isPower(``10``, x))``        ``return` `true``;``    ``else``        ``return` `false``;``}``  ` `// Driver Code``public` `static` `void` `main (String[] args)``{``    ``int` `n = ``73``;``    ` `    ``if` `(check(n))``        ``System.out.print(``"Yes"``);``    ``else``        ``System.out.print(``"No"``);``}``}` `// This code is contributed by code_hunt`

## Python3

 `# Python3 program to implement``# the above approach``import` `math` `# Function to check y is a power of x``def` `isPower(x, y):``    ` `    ``# logarithm function to``    ``# calculate value``    ``res1 ``=` `math.log(y) ``/``/` `math.log(x)``    ``res2 ``=` `math.log(y) ``/``/` `math.log(x)``  ` `    ``# Compare to the result1 or``    ``# result2 both are equal``    ``return` `(res1 ``=``=` `res2)`` ` `# Function to check if N``# satisfies the equation``# 2 * reverse(n) = n + 1``def` `check(n):``    ` `    ``x ``=` `(n ``+` `7``) ``/``/` `8``    ` `    ``if` `((n ``+` `7``) ``%` `8` `=``=` `0` `and``        ``isPower(``10``, x)):``        ``return` `True``    ``else``:``        ``return` `False`` ` `# Driver Code``n ``=` `73` `if` `(check(n) !``=` `0``):``    ``print``(``"Yes"``)``else``:``    ``print``(``"No"``)` `# This code is contributed by code_hunt`

## C#

 `// C# program to implement``// the above approach``using` `System;` `class` `GFG{``  ` `// Function to check y is a power of x``static` `bool` `isPower(``int` `x, ``int` `y)``{``    ` `    ``// logarithm function to``    ``// calculate value``    ``double` `res1 = Math.Log(y) / Math.Log(x);``    ``double` `res2 = Math.Log(y) / Math.Log(x);``  ` `    ``// Compare to the result1 or``    ``// result2 both are equal``    ``return` `(res1 == res2);``}``  ` `// Function to check if N``// satisfies the equation``// 2 * reverse(n) = n + 1``static` `bool` `check(``int` `n)``{``    ``int` `x = (n + 7) / 8;``    ` `    ``if` `((n + 7) % 8 == 0 &&``        ``isPower(10, x))``        ``return` `true``;``    ``else``        ``return` `false``;``}``  ` `// Driver Code``static` `public` `void` `Main ()``{``    ``int` `n = 73;``    ` `    ``if` `(check(n))``        ``Console.Write(``"Yes"``);``    ``else``        ``Console.Write(``"No"``);``}``}` `// This code is contributed by code_hunt`

## Javascript

 ``
Output:
`Yes`

Time Complexity: O(log N)
Auxiliary Space: O(1)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up