Calculate sum of all numbers present in a string
Given a string containing alphanumeric characters, calculate sum of all numbers present in the string.
Examples:
Input: 1abc23 Output: 24 Input: geeks4geeks Output: 4 Input: 1abc2x30yz67 Output: 100 Input: 123abc Output: 123
Difficulty level: Rookie
The only tricky part in this question is that multiple consecutive digits are considered as one number.
The idea is very simple. We scan each character of the input string and if a number is formed by consecutive characters of the string, we increment the result by that amount.
Below is the implementation of the above idea:
C++
// C++ program to calculate sum of all numbers present // in a string containing alphanumeric characters #include <iostream> using namespace std; // Function to calculate sum of all numbers present // in a string containing alphanumeric characters int findSum(string str) { // A temporary string string temp = "" ; // holds sum of all numbers present in the string int sum = 0; // read each character in input string for ( char ch : str) { // if current character is a digit if ( isdigit (ch)) temp += ch; // if current character is an alphabet else { // increment sum by number found earlier // (if any) sum += atoi (temp.c_str()); // reset temporary string to empty temp = "" ; } } // atoi(temp.c_str()) takes care of trailing // numbers return sum + atoi (temp.c_str()); } // Driver code int main() { // input alphanumeric string string str = "12abc20yz68" ; // Function call cout << findSum(str); return 0; } |
Java
// Java program to calculate sum of all numbers present // in a string containing alphanumeric characters class GFG { // Function to calculate sum of all numbers present // in a string containing alphanumeric characters static int findSum(String str) { // A temporary string String temp = "0" ; // holds sum of all numbers present in the string int sum = 0 ; // read each character in input string for ( int i = 0 ; i < str.length(); i++) { char ch = str.charAt(i); // if current character is a digit if (Character.isDigit(ch)) temp += ch; // if current character is an alphabet else { // increment sum by number found earlier // (if any) sum += Integer.parseInt(temp); // reset temporary string to empty temp = "0" ; } } // atoi(temp.c_str()) takes care of trailing // numbers return sum + Integer.parseInt(temp); } // Driver code public static void main(String[] args) { // input alphanumeric string String str = "12abc20yz68" ; // Function call System.out.println(findSum(str)); } } // This code is contributed by AnkitRai01 |
Python3
# Python3 program to calculate sum of # all numbers present in a string # containing alphanumeric characters # Function to calculate sum of all # numbers present in a string # containing alphanumeric characters def findSum(str1): # A temporary string temp = "0" # holds sum of all numbers # present in the string Sum = 0 # read each character in input string for ch in str1: # if current character is a digit if (ch.isdigit()): temp + = ch # if current character is an alphabet else : # increment Sum by number found # earlier(if any) Sum + = int (temp) # reset temporary string to empty temp = "0" # atoi(temp.c_str1()) takes care # of trailing numbers return Sum + int (temp) # Driver code # input alphanumeric string str1 = "12abc20yz68" # Function call print (findSum(str1)) # This code is contributed # by mohit kumar |
C#
// C# program to calculate sum of // all numbers present in a string // containing alphanumeric characters using System; class GFG { // Function to calculate sum of // all numbers present in a string // containing alphanumeric characters static int findSum(String str) { // A temporary string String temp = "0" ; // holds sum of all numbers // present in the string int sum = 0; // read each character in input string for ( int i = 0; i < str.Length; i++) { char ch = str[i]; // if current character is a digit if ( char .IsDigit(ch)) temp += ch; // if current character is an alphabet else { // increment sum by number found earlier // (if any) sum += int .Parse(temp); // reset temporary string to empty temp = "0" ; } } // atoi(temp.c_str()) takes care of trailing // numbers return sum + int .Parse(temp); } // Driver code public static void Main(String[] args) { // input alphanumeric string String str = "12abc20yz68" ; // Function call Console.WriteLine(findSum(str)); } } // This code is contributed by PrinciRaj1992 |
Javascript
<script> // Javascript program to calculate // sum of all numbers present // in a string containing // alphanumeric characters // Function to calculate sum // of all numbers present // in a string containing // alphanumeric characters function findSum(str) { // A temporary string let temp = "0" ; // holds sum of all numbers // present in the string let sum = 0; // read each character in input string for (let i = 0; i < str.length; i++) { let ch = str[i]; // if current character is a digit if (!isNaN(String(ch) * 1)) temp += ch; // if current character is an alphabet else { // increment sum by number found earlier // (if any) sum += parseInt(temp); // reset temporary string to empty temp = "0" ; } } // atoi(temp.c_str()) takes care of trailing // numbers return sum + parseInt(temp); } // Driver code // input alphanumeric string let str = "12abc20yz68" ; // Function call document.write(findSum(str)); // This code is contributed by unknown2108 </script> |
100
Time complexity: O(n) where n is length of the string.
Auxiliary Space: O(n) where n is length of the string.
A Better Solution implementing Regex.
Python3
# Python3 program to calculate sum of # all numbers present in a string # containing alphanumeric characters # Function to calculate sum of all # numbers present in a string # containing alphanumeric characters import re def find_sum(str1): # Regular Expression that matches # digits in between a string return sum ( map ( int , re.findall( '\d+' , str1))) # Driver code # input alphanumeric string str1 = "12abc20yz68" # Function call print (find_sum(str1)) # This code is contributed # by Venkata Ramana B |
100
Time complexity: O(n) where n is length of the string.
Auxiliary Space: O(n) where n is length of the string.
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