Calculate sum of all numbers present in a string using recursion

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

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

C#

// C# program to calculate sum of all numbers
// present in a string containing alphanumeric
// characters
 
using System;
using System.Linq;
using System.Collections.Generic;
 
class GFG 
{
    static bool isdigit(char c)
    {
        if(c>='0' && c<='9')
            return true;
        return false;
    }
    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 (isdigit(str[i])) {
                return str[i];
            }
            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 Int32.Parse(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
    static public void Main()
    {
        // input alphanumeric string
        string str = "12abc20yz68";
     
        // Function call
        Console.Write(findSum(str));
     
    }
}

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

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

Tags:

#DSA #Strings #Strings

Approach: