Sum of all prefixes of given numeric string

Last Updated : 08 Mar, 2022

Given string str having N characters representing an integer, the task is to calculate the sum of all possible prefixes of the given string.

Example:

Input: str = "1225"
Output: 1360
Explanation: The prefixes of the given string are 1, 12, 122, and 1225 and their sum will be 1 + 12 + 122 + 1225 = 1360.

Input: str = "20"
Output: 22

Approach: The given problem is an implementation-based problem and can be solved by iterating over all the prefixes of the string and maintaining their sum in a string. The sum of two integers represented as strings can be done using the approach discussed here.

Below is the implementation of the above approach:

C++

// C++ program of the above approach #include <bits/stdc++.h> using namespace std;  // Function for finding sum of larger numbers string findSum(string str1, string str2) {     // Before proceeding further, make     // sure length of str2 is larger     if (str1.length() > str2.length())         swap(str1, str2);      // Stores resulting sum     string str = "";      // Calculate length of both string     int n1 = str1.length(), n2 = str2.length();      // Reverse both of strings     reverse(str1.begin(), str1.end());     reverse(str2.begin(), str2.end());      int carry = 0;     for (int i = 0; i < n1; i++) {         // Compute sum of current         // digits and carry         int sum             = ((str1[i] - '0')                + (str2[i] - '0')                + carry);         str.push_back(sum % 10 + '0');          // Carry for next step         carry = sum / 10;     }      // Add remaining digits     for (int i = n1; i < n2; i++) {         int sum = ((str2[i] - '0') + carry);         str.push_back(sum % 10 + '0');         carry = sum / 10;     }      // Add remaining carry     if (carry)         str.push_back(carry + '0');      // Reverse string     reverse(str.begin(), str.end());      // Return Answer     return str; }  // Function to find sum of all prefixes // of a string representing a number string sumPrefix(string str) {     // Stores the desired sum     string sum = "0";      // Stores the current prefix     string curPre = "";      // Loop to iterate str     for (int i = 0; i < str.length(); i++) {         // Update current prefix         curPre += str[i];          // Update Sum         sum = findSum(curPre, sum);     }      // Return Answer     return sum; }  // Driver Code int main() {     string str = "1225";     cout << sumPrefix(str);      return 0; }

Java

// Java program of the above approach import java.util.*; class GFG{  // Function for finding sum of larger numbers static String findSum(String str1, String str2) {        // Before proceeding further, make     // sure length of str2 is larger     if (str1.length() > str2.length()) {         String s = str1;         str1=str2;         str2=s;     }      // Stores resulting sum     String str = "";      // Calculate length of both String     int n1 = str1.length(), n2 = str2.length();      // Reverse both of Strings     str1 = reverse(str1);     str2 = reverse(str2);      int carry = 0;     for (int i = 0; i < n1; i++) {         // Compute sum of current         // digits and carry         int sum             = ((str1.charAt(i) - '0')                + (str2.charAt(i) - '0')                + carry);         str+=(char)((sum % 10 + '0'));          // Carry for next step         carry = sum / 10;     }      // Add remaining digits     for (int i = n1; i < n2; i++) {         int sum = ((str2.charAt(i) - '0') + carry);         str+=(char)(sum % 10 + '0');         carry = sum / 10;     }      // Add remaining carry     if (carry > 0)         str += (carry + '0');      // Reverse String     str = reverse(str);      // Return Answer     return str; }  // Function to find sum of all prefixes // of a String representing a number static String sumPrefix(String str) {        // Stores the desired sum     String sum = "0";      // Stores the current prefix     String curPre = "";      // Loop to iterate str     for (int i = 0; i < str.length(); i++)     {                // Update current prefix         curPre += (char)(str.charAt(i));          // Update Sum         sum = findSum(curPre, sum);     }      // Return Answer     return sum; } static String reverse(String input) {     char[] a = input.toCharArray();     int l, r = a.length - 1;     for (l = 0; l < r; l++, r--) {         char temp = a[l];         a[l] = a[r];         a[r] = temp;     }     return String.valueOf(a); }    // Driver Code public static void main(String[] args) {     String str = "1225";     System.out.print(sumPrefix(str)); } }  // This code is contributed by shikhasingrajput

Python3

# Python code for the above approach  # Function for finding sum of larger numbers def findSum(str1, str2):      # Before proceeding further, make     # sure length of str2 is larger     if (len(str1) > len(str2)):         temp = str1;         str1 = str2;         str2 = temp;      # Stores resulting sum     str = [];      # Calculate length of both string     n1 = len(str1)     n2 = len(str2)      # Reverse both of strings     str1 = list(str1)     str2 = list(str2)     str1.reverse();     str2.reverse();      carry = 0;     for i in range(n1):                # Compute sum of current         # digits and carry         sum = ((ord(str1[i]) - ord('0')) + (ord(str2[i]) - ord('0')) + carry);         str.append(chr(sum % 10 + ord('0')));          # Carry for next step         carry = sum // 10      # Add remaining digits     for i in range(n1, n2):         sum = ((ord(str2[i]) - ord('0')) + carry);         str.append(chr(sum % 10 + ord('0')));         carry = sum // 10      # Add remaining carry     if (carry):         str.append(chr(carry + ord('0')));      # Reverse string     str.reverse();      # Return Answer     return ''.join(str)  # Function to find sum of all prefixes # of a string representing a number def sumPrefix(str):      # Stores the desired sum     sum = "0";      # Stores the current prefix     curPre = "";      # Loop to iterate str     for i in range(len(str)):         # Update current prefix         curPre += str[i];          # Update Sum         sum = findSum(curPre, sum);          # Return Answer     return sum;  # Driver Code str = "1225"; print(sumPrefix(str));  # This code is contributed by Saurabh Jaiswal

// C# program of the above approach using System;  public class GFG{  // Function for finding sum of larger numbers static String findSum(String str1, String str2) {        // Before proceeding further, make     // sure length of str2 is larger     if (str1.Length > str2.Length) {         String s = str1;         str1=str2;         str2=s;     }      // Stores resulting sum     String str = "";      // Calculate length of both String     int n1 = str1.Length, n2 = str2.Length;      // Reverse both of Strings     str1 = reverse(str1);     str2 = reverse(str2);      int carry = 0;     for (int i = 0; i < n1; i++)      {                // Compute sum of current         // digits and carry         int sum             = ((str1[i] - '0')                + (str2[i] - '0')                + carry);         str+=(char)((sum % 10 + '0'));          // Carry for next step         carry = sum / 10;     }      // Add remaining digits     for (int i = n1; i < n2; i++) {         int sum = ((str2[i] - '0') + carry);         str+=(char)(sum % 10 + '0');         carry = sum / 10;     }      // Add remaining carry     if (carry > 0)         str += (carry + '0');      // Reverse String     str = reverse(str);      // Return Answer     return str; }  // Function to find sum of all prefixes // of a String representing a number static String sumPrefix(String str) {        // Stores the desired sum     String sum = "0";      // Stores the current prefix     String curPre = "";      // Loop to iterate str     for (int i = 0; i < str.Length; i++)     {                // Update current prefix         curPre += (char)(str[i]);          // Update Sum         sum = findSum(curPre, sum);     }      // Return Answer     return sum; } static String reverse(String input) {     char[] a = input.ToCharArray();     int l, r = a.Length - 1;     for (l = 0; l < r; l++, r--) {         char temp = a[l];         a[l] = a[r];         a[r] = temp;     }     return String.Join("",a); }    // Driver Code public static void Main(String[] args) {     String str = "1225";     Console.Write(sumPrefix(str)); } }  // This code is contributed by 29AjayKumar

JavaScript

<script>         // JavaScript code for the above approach          // Function for finding sum of larger numbers         function findSum(str1, str2)         {                      // Before proceeding further, make             // sure length of str2 is larger             if (str1.length > str2.length) {                 let temp = str1;                 str1 = str2;                 str2 = temp;             }              // Stores resulting sum             let str = [];              // Calculate length of both string             let n1 = str1.length, n2 = str2.length;              // Reverse both of strings             str1 = str1.split('')             str2 = str2.split('')             str1.reverse();             str2.reverse();              let carry = 0;             for (let i = 0; i < n1; i++) {                 // Compute sum of current                 // digits and carry                 let sum                     = ((str1[i].charCodeAt(0) - '0'.charCodeAt(0))                         + (str2[i].charCodeAt(0) - '0'.charCodeAt(0))                         + carry);                 str.push(String.fromCharCode(sum % 10 + '0'.charCodeAt(0)));                  // Carry for next step                 carry = Math.floor(sum / 10);             }              // Add remaining digits             for (let i = n1; i < n2; i++) {                 let sum = ((str2[i].charCodeAt(0) - '0'.charCodeAt(0)) + carry);                 str.push(String.fromCharCode(sum % 10 + '0'.charCodeAt(0)));                 carry = Math.floor(sum / 10);             }              // Add remaining carry             if (carry)                 str.push(String.fromCharCode(carry + '0'.charCodeAt(0)));              // Reverse string             str.reverse();              // Return Answer             return str.join('');         }          // Function to find sum of all prefixes         // of a string representing a number         function sumPrefix(str)          {                      // Stores the desired sum             let sum = "0";              // Stores the current prefix             let curPre = "";              // Loop to iterate str             for (let i = 0; i < str.length; i++) {                 // Update current prefix                 curPre += str[i];                  // Update Sum                 sum = findSum(curPre, sum);             }              // Return Answer             return sum;         }          // Driver Code         let str = "1225";         document.write(sumPrefix(str));         // This code is contributed by Potta Lokesh     </script>

Output

Time Complexity: O(N²)
Auxiliary Space: O(N)

Sum of all substrings of a string representing a number | Set 1

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Sum of all prefixes of given numeric string

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