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 = 24Input: neveropen4neveropen
Output: 4Input: 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 charactersint 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 codeint 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 charactersimport 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 charactersdef 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 stringstr1 = "12abc20yz68"# Function callprint(findSum(str1))# This code is contributed# by mohit kumar |
C#
// C# program to calculate sum of// all numbers present in a string// containing alphanumeric charactersusing 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
100
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.
- Run a for loop with counter j from i to N – 1.
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 codeint 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 codeif __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// charactersusing 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 codeconsole.log(findSum("12abc20yz68"));// This code is contributed by Shivam Tiwari |
Output
100
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 <iostream>#include <regex>// Function to calculate sum of all// numbers present in a string// containing alphanumeric charactersint 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 codeint 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 charactersimport redef find_sum(str1): # Regular Expression that matches # digits in between a string return sum(map(int, re.findall('\d+', str1)))# Driver code# input alphanumeric stringstr1 = "12abc20yz68"# Function callprint(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 charactersfunction 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 stringconst str1 = "12abc20yz68";// Function callconsole.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)); }} |
Output
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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