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

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++ 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 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 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# 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`

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++ 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 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 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# 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`

Output:
```Yes
```

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

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