Calculate sum of all numbers present in a string (original) (raw)

Last Updated : 20 Mar, 2023

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

Try It Yourself

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 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

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 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

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:

C++14 `

#include #include

// Function to calculate sum of all // numbers present in a string // containing alphanumeric characters int findSum(std::string str) { // Regular Expression that matches // digits in between a string std::regex pattern("\d+"); std::smatch match; int sum = 0; while (std::regex_search(str, match, pattern)) { sum += stoi(match[0].str()); str = match.suffix().str(); } return sum; }

// Driver code int main() { // input alphanumeric string std::string str = "12abc20yz68";

// Function call
std::cout << findSum(str) << std::endl;
return 0;

} // This code is contributed by Shivam Tiwari

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

JavaScript

// 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 find_sum(str1) { // Regular Expression that matches // digits in between a string return str1.match(/\d+/g).reduce((acc, val) => acc + parseInt(val), 0); }

// Driver code // input alphanumeric string const str1 = "12abc20yz68";

// Function call console.log(find_sum(str1));

Java

import java.util.regex.*;

public class Main {

// Function to calculate sum of all
// numbers present in a string
// containing alphanumeric characters
public static int findSum(String str)
{
    // Regular Expression that matches
    // digits in between a string
    Pattern pattern = Pattern.compile("\\d+");
    Matcher matcher = pattern.matcher(str);
    int sum = 0;
    while (matcher.find()) {
        sum += Integer.parseInt(matcher.group());
        str = matcher.replaceFirst("");
        matcher = pattern.matcher(str);
    }
    return sum;
}

// Driver code
public static void main(String[] args)
{
    // input alphanumeric string
    String str = "12abc20yz68";

    // Function call
    System.out.println(findSum(str));
}

}

C#

using System; using System.Text.RegularExpressions;

public class GFG { // Function to calculate sum of all // numbers present in a string // containing alphanumeric characters public static int FindSum(string str) { // Regular Expression that matches // digits in between a string Regex pattern = new Regex(@"\d+"); Match matcher = pattern.Match(str); int sum = 0; while (matcher.Success) { sum += Int32.Parse(matcher.Value); str = pattern.Replace(str, "", 1, matcher.Index); matcher = pattern.Match(str); } return sum; }

// Main method
static public void Main()
{
    // input alphanumeric string
    string str = "12abc20yz68";

    // Function call
    Console.WriteLine(FindSum(str));
}

}

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