Convert all substrings of length ‘k’ from base ‘b’ to decimal

A string defining a valid number is given. Output all the base conversions of substrings of length ‘k’ from base ‘b’ to base 10.

Examples:

Input :  str = "12212",
k = 3, b = 3.
Output : 17 25 23
Explanation :
All the substrings of length 'k' are : 122, 221, 212.
Base conversion can be computed using the formula.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Method 1 (Simple)

A simple approach is to use simple base conversion technique. For a base b number str, its decimal equivalent is str*b0 + str*b1 + str*b2 + … + str[n-1]*bn-1

C++

 // Simple C++ program to convert all substrings from // decimal to given base. #include using namespace std;    int substringConversions(string str, int k, int b) {     for (int i=0; i + k <= str.size(); i++)     {         // Saving substring in sub         string sub = str.substr(i, k);                            // Evaluating decimal for current substring         // and printing it.         int sum = 0, counter = 0;         for (int i = sub.size() - 1; i >= 0; i--)         {             sum = sum + ((sub.at(i) - '0') * pow(b, counter));             counter++;         }                 cout << sum << " ";     } }    // Driver code int main() {     string str = "12212";     int b = 3, k = 3;     substringConversions(str, b, k);         return 0; }

Java

 // Simple Java program to convert all substrings from // decimal to given base.    class GFG {    static void substringConversions(String str, int k, int b) {     for (int i=0; i + k <= str.length(); i++)     {         // Saving substring in sub         String sub = str.substring(i, i+k);                         // Evaluating decimal for current substring         // and printing it.         int sum = 0, counter = 0;         for (int j = sub.length() - 1; j >= 0; j--)         {             sum = (int) (sum + ((sub.charAt(j) - '0') *                                     Math.pow(b, counter)));             counter++;         }              System.out.print(sum + " ");     } }    // Driver code public static void main(String[] args) {     String str = "12212";     int b = 3, k = 3;     substringConversions(str, b, k);  } }    // This code is contributed by 29AjayKumar

Python3

 # Simple Python3 program to convert  # all substrings from decimal to given base. import math    def substringConversions(s, k, b):            l = len(s);     for i in range(l):                    if((i + k) < l + 1):                            # Saving substring in sub             sub = s[i : i + k];                                 # Evaluating decimal for current              # substring and printing it.             sum, counter = 0, 0;             for i in range(len(sub) - 1, -1, -1):                        sum = sum + ((ord(sub[i]) - ord('0')) *                                    pow(b, counter));                 counter += 1;                            print(sum , end = " ");    # Driver code s = "12212"; b, k = 3, 3; substringConversions(s, b, k);     # This code is contributed  # by Princi Singh

C#

 // Simple C# program to convert all substrings from // decimal to given base. using System;     class GFG {    static void substringConversions(String str, int k, int b) {     for (int i = 0; i + k <= str.Length; i++)     {         // Saving substring in sub         String sub = str.Substring(i, k);                         // Evaluating decimal for current substring         // and printing it.         int sum = 0, counter = 0;         for (int j = sub.Length - 1; j >= 0; j--)         {             sum = (int) (sum + ((sub[j] - '0') *                                     Math.Pow(b, counter)));             counter++;         }              Console.Write(sum + " ");     } }    // Driver code public static void Main(String[] args) {     String str = "12212";     int b = 3, k = 3;     substringConversions(str, b, k);  } }    /* This code is contributed by PrinciRaj1992 */

Output:

17 25 23

Time Complexity : O(n*k)

Method 2 (Using sliding window)

We can use Sliding Window technique to solve it in linear time. Every time we slide the window, we will subtract the weight of first element i.e. (element * pow(b, k-1) ). Now multiplying the previous sum with ‘b’ will increase weight of every element 3 times which is required. Also we will simply add the new element in window because its weight will be element * pow(b, 0).

Below is the implementation :

C++

 // Efficient C++ program to convert all substrings from // decimal to given base. #include using namespace std;    int substringConversions(string str, int k, int b) {    int i = 0, sum = 0, counter = k-1;        // Computing the decimal of first window     for (i; i < k; i++)     {         sum = sum + ((str.at(i) - '0') * pow(b, counter));         counter--;     }     cout << sum << " ";            // prev stores the pervious decimal     int prev = sum;                   // Computing decimal equivalents of all other windows     sum = 0, counter = 0;     for (i; i < str.size(); i++)     {         // Subtracting weight of the element pushed out of window         sum = prev - ((str.at(i - k) - '0') * pow(b, k-1));                    // Multiplying the decimal by base to formulate other window          sum = sum * b;                    // Adding the new element of window to sum         sum = sum + (str.at(i) - '0');                    // Decimal of current window         cout << sum << " ";                    // Updating prev         prev = sum;                    counter++;     } }    // Driver code int main() {     string str = "12212";     int b = 3, k = 3;     substringConversions(str, b, k);         return 0; }

Java

 // Efficient Java program to convert  // all substrings from decimal to given base. import java.util.*;    class GFG  { static void substringConversions(String str,                                   int k, int b) {     int i = 0, sum = 0, counter = k-1;        // Computing the decimal of first window     for (i = 0; i < k; i++)     {         sum = (int) (sum + ((str.charAt(i) - '0') *                               Math.pow(b, counter)));         counter--;     }     System.out.print(sum + " ");            // prev stores the pervious decimal     int prev = sum;                    // Computing decimal equivalents of all other windows     sum = 0; counter = 0;     for (; i < str.length(); i++)     {         // Subtracting weight of the element          // pushed out of window         sum = (int) (prev - ((str.charAt(i - k) - '0') *                               Math.pow(b, k - 1)));                    // Multiplying the decimal by base         // to formulate other window          sum = sum * b;                    // Adding the new element of window to sum         sum = sum + (str.charAt(i) - '0');                    // Decimal of current window         System.out.print(sum + " ");                    // Updating prev         prev = sum;                    counter++;     } }    // Driver code public static void main(String[] args) {     String str = "12212";     int b = 3, k = 3;     substringConversions(str, b, k); } }    // This code is contributed by Rajput-Ji

Python3

 # Simple Python3 program to convert all # substrings from decimal to given base. import math as mt    def substringConversions(str1, k, b):        for i in range(0, len(str1) - k + 1):                # Saving subin sub         sub = str1[i:k + i]                    # Evaluating decimal for current          # substring and printing it.         Sum = 0         counter = 0         for i in range(len(sub) - 1, -1, -1):             Sum = (Sum + ((ord(sub[i]) - ord('0')) *                             pow(b, counter)))             counter += 1                            print(Sum, end = " ")        # Driver code str1 = "12212" b = 3 k = 3 substringConversions(str1, b, k)     # This code is contributed by  # Mohit Kumar 29

C#

 // Efficient C# program to convert  // all substrings from decimal to given base. using System;    class GFG  { static void substringConversions(String str,                                   int k, int b) {     int i = 0, sum = 0, counter = k-1;        // Computing the decimal of first window     for (i = 0; i < k; i++)     {         sum = (int) (sum + ((str[i] - '0') *                               Math.Pow(b, counter)));         counter--;     }     Console.Write(sum + " ");            // prev stores the pervious decimal     int prev = sum;                    // Computing decimal equivalents      // of all other windows     sum = 0; counter = 0;     for (; i < str.Length; i++)     {         // Subtracting weight of the element          // pushed out of window         sum = (int) (prev - ((str[i - k] - '0') *                                Math.Pow(b, k - 1)));                    // Multiplying the decimal by base         // to formulate other window          sum = sum * b;                    // Adding the new element of window to sum         sum = sum + (str[i] - '0');                    // Decimal of current window         Console.Write(sum + " ");                    // Updating prev         prev = sum;                    counter++;     } }    // Driver code public static void Main(String[] args) {     String str = "12212";     int b = 3, k = 3;     substringConversions(str, b, k); } }    // This code is contributed by Princi Singh

Output:

17 25 23

Time Complexity: O(n)

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