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.

Method 1 (Simple)



A simple approach is to use simple base conversion technique. For a base b number str, its decimal equivalent is str[0]*b0 + str[1]*b1 + str[2]*b2 + … + str[n-1]*bn-1

C++

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// Simple C++ program to convert all substrings from
// decimal to given base.
#include <bits/stdc++.h>
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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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 */

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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++

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// Efficient C++ program to convert all substrings from
// decimal to given base.
#include <bits/stdc++.h>
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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

17 25 23

Time Complexity: O(n)

This article is contributed by Rohit Thapliyal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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