Calculate sum of all numbers present in a string

Difficulty Level : Easy
Last Updated : 27 Jan, 2023

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Given a string S containing alphanumeric characters, The task is to calculate the sum of all numbers present in the string.

Examples:

Input: 1abc23
Output: 24
Explanation: 1 + 23 = 24

Input: geeks4geeks
Output: 4

Input: 1abc2x30yz67
Output: 100

Approach:

Scan each character of the input string and if a number is formed by consecutive characters of the string, then increment the result by that amount. The only tricky part of this question is that multiple consecutive digits are considered one number.

Follow the below steps to implement the idea:

Create an empty string temp and an integer sum.
Iterate over all characters of the string.
- If the character is a numeric digit add it to temp.
- Else convert temp string to number and add it to sum, empty temp.
Return sum + number obtained from temp.

Below is the implementation of the above approach:

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
import java.io.*;
 
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>

Output

Time complexity: O(N) where n is length of the string.
Auxiliary Space: O(N) where n is length of the string.

Calculate sum of all numbers present in a string using recursion

The idea is to recursively traverse over the string and find out the numbers then add these numbers to the result, at last return the result.

Follow the below steps to implement the idea:

Create an empty string temp and an integer sum.
Recursively traverse the characters for every index i from 0 to length – 1.
- If i = N-1 then check if current character is a digit return str[i] – ‘0’.
- Else return 0.
- If str[i] is a digit.
  - Run a for loop with counter j from i to N – 1.
    - If the character is a numeric digit add it to temp.
    - Else break.
  - Return sum of numeric value of temp + recur for index j.

Below is the implementation of the above approach:

C++

// C++ program to calculate sum of all numbers
// present in a string containing alphanumeric
// characters
#include <iostream>
using namespace std;
 
int solve(string& str, int i, int n)
{
    // if string is empty
    if (i >= n)
        return 0;
 
    // if on the last index
    if (i == n - 1) {
 
        // if last digit is numeric
        if (isdigit(str[i])) {
            return str[i] - '0';
        }
        else {
            return 0;
        }
    }
 
    // if current char is digit
    // then sum the consecutive digits
    if (isdigit(str[i])) {
 
        // declared an empty string
        string temp = "";
        int j;
 
        // start from that index
        // sum all the consecutive digits
        for (j = i; j < n; j++) {
 
            // if current char is digit
            // add it to the temp string
            if (isdigit(str[j]))
                temp += str[j];
 
            // if it is not a digit
            // break instantly
            else
                break;
        }
 
        // add the number associated to temp
        // with the answer recursion will bring
        return stoi(temp) + solve(str, j, n);
    }
 
    // else call from the next index
    else {
        solve(str, i + 1, n);
    }
}
 
int findSum(string str)
{
    // recursiven function
    return solve(str, 0, str.size());
}
 
// Driver code
int main()
{
    // input alphanumeric string
    string str = "12abc20yz68";
 
    // Function call
    cout << findSum(str);
 
    return 0;
}

Java

import java.util.Scanner;
 
class Main {
  static int solve(String str, int i, int n) {
    // if string is empty
    if (i >= n)
      return 0;
 
    // if on the last index
    if (i == n - 1) {
 
      // if last digit is numeric
      if (Character.isDigit(str.charAt(i))) {
        return str.charAt(i) - '0';
      }
      else {
        return 0;
      }
    }
 
    // if current char is digit
    // then sum the consecutive digits
    if (Character.isDigit(str.charAt(i))) {
 
      // declared an empty string
      String temp = "";
      int j;
 
      // start from that index
      // sum all the consecutive digits
      for (j = i; j < n; j++) {
 
        // if current char is digit
        // add it to the temp string
        if (Character.isDigit(str.charAt(j)))
          temp += str.charAt(j);
 
        // if it is not a digit
        // break instantly
        else
          break;
      }
 
      // add the number associated to temp
      // with the answer recursion will bring
      return Integer.parseInt(temp) + solve(str, j, n);
    }
 
    // else call from the next index
    else {
      return solve(str, i + 1, n);
    }
  }
 
  static int findSum(String str) {
    // recursiven function
    return solve(str, 0, str.length());
  }
 
  // Driver code
  public static void main(String[] args) {
    // input alphanumeric string
    String str = "12abc20yz68";
 
    // Function call
    System.out.println(findSum(str));
  }
}
// This code contributed by Ajax

Python3

def findSum(str):
    # variable to store sum
    result = 0
    temp = ""
     
    for i in range(len(str)):
        if str[i].isnumeric():
            temp += str[i]
            if i == len(str) - 1:
                result += int(temp)
        else:
            if temp != "":
                result += int(temp)
                temp = ""
 
    return result
 
# driver code
if __name__ == "__main__":
    # input alphanumeric string
    str = "12abc20yz68"
    print(findSum(str))
 
#This code contributed by Shivam Tiwari

Javascript

function findSum(str) {
    // variable to store sum
    let result = 0;
    let temp = "";
     
    for (let i = 0; i < str.length; i++) {
        if (!isNaN(str[i])) {
            temp += str[i];
            if (i === str.length - 1) {
                result += parseInt(temp);
            }
        } else {
            if (temp !== "") {
                result += parseInt(temp);
                temp = "";
            }
        }
    }
 
    return result;
}
 
// driver code
console.log(findSum("12abc20yz68"));
 
// This code is contributed by Shivam Tiwari

Output

Time Complexity: O(N), where N is the size of the given string.
Auxiliary Space: O(N), in worst case it can cost O(N) recursive calls

Calculate sum of all numbers present in a string using Regex in Python:

The idea is to use inbuilt function Python RegEx.

Below is the Implementation of above approach:

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

Output

Time complexity: O(n) where n is length of the string.
Auxiliary Space: O(n) where n is length of the string.

This article is contributed by Aditya Goel. 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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