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# Check length of a string is equal to the number appended at its last

Given a string that (may) be appended with a number at last. You need to find whether the length of string excluding that number is equal to that number. For example for “helloworld10”, answer is True as helloworld consist of 10 letters. Length of String is less than 10, 000.

Examples :

```Input:  str = "geeks5"
Output:  Yes
Explanation : As geeks is of 5 length and at
last number is also 5.

Input:  str = "geeksforgeeks15"
Output:  No
Explanation: As geeksforgeeks is of 13 length and
at last number is 15 i.e. not equal```

A Naive approach is to traverse from starting and retrieve the number from string and check if length of string – digits in the number = number or Not

An efficient method is to do following steps

1. Traverse string from end and keep storing the number till it is smaller than the length of the overall string.
2. If the number is equal to length of string except that number’s digits then return true.
3. Else return false.

Implementation:

## C++

 `// C++ program to check if size of string is appended``// at the end or not.``#include ``using` `namespace` `std;` `// Function to find if given number is equal to``// length or not``bool` `isequal(string str)``{``    ``int` `n = str.length();` `    ``// Traverse string from end and find the number``    ``// stored at the end.``    ``// x is used to store power of 10.``    ``int` `num = 0, x = 1, i = n - 1;``    ``for` `(i = n - 1; i >= 0; i--) {``        ``if` `(``'0'` `<= str[i] && str[i] <= ``'9'``) {``            ``num = (str[i] - ``'0'``) * x + num;``            ``x = x * 10;``            ``if``(num>=n)``                ``return` `false``;``        ``}``        ``else``            ``break``;``    ``}` `    ``// Check if number is equal to string length except``    ``// that number's digits``    ``return` `num == i + 1;``}` `// Drivers code``int` `main()``{``    ``string str = ``"geeksforgeeks13"``;``    ``isequal(str) ? cout << ``"Yes"` `: cout << ``"No"``;``    ``return` `0;``}`

## Java

 `// Java program to check if size of``// string is appended at the end or not.``import` `java.io.*;` `class` `GFG {` `    ``// Function to find if given number is``    ``// equal to length or not``    ``static` `boolean` `isequal(String str)``    ``{``        ``int` `n = str.length();` `        ``// Traverse string from end and find the number``        ``// stored at the end.``        ``// x is used to store power of 10.``        ``int` `num = ``0``, x = ``1``, i = n - ``1``;``        ``for` `(i = n - ``1``; i >= ``0``; i--)``        ``{``            ``if` `(``'0'` `<= str.charAt(i) &&``                ``str.charAt(i) <= ``'9'``)``            ``{``                ``num = (str.charAt(i) - ``'0'``) * x + num;``                ``x = x * ``10``;``                ``if``(num>=n)``                    ``return` `false``;``            ``}``            ``else``                ``break``;``        ``}` `        ``// Check if number is equal to string``        ``// length except that number's digits``        ``return` `num == i + ``1``;``    ``}` `    ``// Drivers code``    ``static` `public` `void` `main(String[] args)``    ``{``        ``String str = ``"geeksforgeeks13"``;``        ``if` `(isequal(str))``            ``System.out.println(``"Yes"``);``        ``else``            ``System.out.println(``"No"``);``    ``}``}` `// This Code is contributed by vt_m.`

## Python3

 `# Python 3 program to check if size of``# string is appended at the end or not.` `# Function to find if given number``# is equal to length or not``def` `isequal(``str``):` `    ``n ``=` `len``(``str``)` `    ``# Traverse string from end and``    ``# find the number stored at the end.``    ``# x is used to store power of 10.``    ``num ``=` `0``    ``x ``=` `1``    ``i ``=` `n ``-` `1``    ``for` `i ``in` `range``(n ``-` `1``, ``-``1``,``-``1``) :``        ``if` `(``'0'` `<``=` `str``[i] ``and` `str``[i] <``=` `'9'``) :``            ``num ``=` `(``ord``(``str``[i]) ``-` `ord``(``'0'``)) ``*` `x ``+` `num``            ``x ``=` `x ``*` `10``            ``if` `(num>``=``n):``                ``return` `false``    ` `        ``else``:``            ``break` `    ``# Check if number is equal to string``    ``# length except that number's digits``    ``return` `num ``=``=` `i ``+` `1` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``    ` `    ``str` `=` `"geeksforgeeks13"``    ``print``(``"Yes"``) ``if` `isequal(``str``) ``else` `print``(``"No"``)` `# This code is contributed by ChitraNayal`

## C#

 `// C# program to check if size of``// string is appended at the end or not.``using` `System;` `class` `GFG {` `    ``// Function to find if given number``    ``// is equal to length or not``    ``static` `bool` `isequal(``string` `str)``    ``{``        ``int` `n = str.Length;` `        ``// Traverse string from end and find the number``        ``// stored at the end.``        ``// x is used to store power of 10.``        ``int` `num = 0, x = 1, i = n - 1;``        ``for` `(i = n - 1; i >= 0; i--)``        ``{``            ``if` `(``'0'` `<= str[i] && str[i] <= ``'9'``) {``                ``num = (str[i] - ``'0'``) * x + num;``                ``x = x * 10;``                ``if``(num>=n)``                    ``return` `false``;``            ``}``            ``else``                ``break``;``        ``}` `        ``// Check if number is equal to string``        ``// length except that number's digits``        ``return` `num == i + 1;``    ``}` `    ``// Drivers code``    ``static` `public` `void` `Main()``    ``{``        ``string` `str = ``"geeksforgeeks13"``;``        ``if` `(isequal(str))``            ``Console.WriteLine(``"Yes"``);``        ``else``            ``Console.WriteLine(``"No"``);``    ``}``}` `// This Code is contributed by vt_m.`

## PHP

 `= 0; ``\$i``--)``    ``{``        ``if` `(``'0'` `<= ``\$str``[``\$i``] &&``                   ``\$str``[``\$i``] <= ``'9'``)``        ``{``            ``\$num` `= (``\$str``[``\$i``] - ``'0'``) *``                           ``\$x` `+ ``\$num``;``            ``\$x` `= ``\$x` `* 10;``            ``if``(``\$num``>=``\$n``)``                ``return` `false;``        ``}``        ``else``            ``break``;``    ``}` `    ``// Check if number is equal``    ``// to string length except``    ``// that number's digits``    ``return` `\$num` `== ``\$i` `+ 1;``}` `// Driver code``\$str` `= ``"geeksforgeeks13"``;``if``(isequal(``\$str``))``echo` `"Yes"` `;``else``echo` `"No"``;``return` `0;` `// This code is contributed by nitin mittal.``?>`

## Javascript

 ``

Output

`Yes`

Time complexity: O(n) where n is length of the string
Auxiliary space: O(1)

#### Approach#2: Using is.digit()

First, the length of the input string is calculated, and 1 is subtracted from it to exclude the last character, which is expected to be a number. Then, the isdigit() method is used to check if the last character of the string is a digit. If the last character is a digit, the code converts it into an integer and compares it with the length of the string. If they are equal, it means the length of the string is equal to the number appended at the end of the string.

#### Algorithm

1. Initialize a variable str with the input string.
2. Calculate the length of the string and store it in the variable length.
3. Subtract 1 from length to exclude the last character from the length calculation.
4. Check if the last character of the string is a digit using the isdigit() method.
5. If the last character is a digit, convert it into an integer using the int() function.
6. Compare the integer with length.
7. If they are equal, print “Yes”; otherwise, print “No”.

## Python3

 `str` `=` `"geeksforgeeks15"``length ``=` `len``(``str``) ``-` `1`  `# subtract 1 to exclude the last character, which should be a number``if` `str``[length:].isdigit() ``and` `int``(``str``[length:]) ``=``=` `length:``    ``print``(``"Yes"``)``else``:``    ``print``(``"No"``)`

Output

`No`

Time Complexity: O(1), The algorithm has constant time complexity because it performs a fixed number of operations, regardless of the length of the input string.
Space Complexity: O(1), The algorithm uses a fixed amount of memory to store the input string, length, and a few variables, so the space complexity is constant.

This article is contributed by Sahil Chhabra (akku). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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