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:
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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++
// 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 codeint 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 codepublic 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 mathdef 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 codes = "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 codepublic static void Main(String[] args){ String str = "12212"; int b = 3, k = 3; substringConversions(str, b, k);}}/* This code is contributed by PrinciRaj1992 */ |
Javascript
<script>// Simple Javascript program to convert// all substrings from decimal to given base.function substringConversions(str, k, b){ for(let i = 0; i + k <= str.length; i++) { // Saving substring in sub let sub = str.substring(i, i+k); // Evaluating decimal for current substring // and printing it. let sum = 0, counter = 0; for(let j = sub.length - 1; j >= 0; j--) { sum = (sum + ((sub[j].charCodeAt(0) - '0'.charCodeAt(0)) * Math.pow(b, counter))); counter++; } document.write(sum + " "); }}// Driver codelet str = "12212";let b = 3, k = 3;substringConversions(str, b, k); // This code is contributed by patel2127</script> |
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 <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 previous 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 codeint 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 previous 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 codepublic 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 mtdef 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 codestr1 = "12212"b = 3k = 3substringConversions(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 previous 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 codepublic static void Main(String[] args){ String str = "12212"; int b = 3, k = 3; substringConversions(str, b, k);}}// This code is contributed by Princi Singh |
Javascript
<script>// Efficient Javascript program to convert// all substrings from decimal to given base.function substringConversions(str, k, b){ let i = 0, sum = 0, counter = k-1; // Computing the decimal of first window for(i = 0; i < k; i++) { sum = (sum + ((str[i].charCodeAt(0) - '0'.charCodeAt(0)) * Math.pow(b, counter))); counter--; } document.write(sum + " "); // prev stores the previous decimal let 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 = (prev - ((str[i - k].charCodeAt(0) - '0'.charCodeAt(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].charCodeAt(0) - '0'.charCodeAt(0)); // Decimal of current window document.write(sum + " "); // Updating prev prev = sum; counter++; }}// Driver codelet str = "12212";let b = 3, k = 3;substringConversions(str, b, k);// This code is contributed by unknown2108</script> |
Output:
17 25 23
Time Complexity: O(n)
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