Given a binary string S, the task is to find the minimum number of Powers of 2 required to express a S.
Examples:
Input: S = “111”
Output: 2
Explanation:
Two possible representations of “111” (= 7) using powers of 2 are:
20 + 21 + 22 = 1 + 2 + 4 = 7
23 – 20 = 8 – 1 = 7
Therefore, the minimum powers of 2 required to represent S is 2.
Input: S = “1101101”
Output: 4
Explanation:
1101101 can be represented as 27 – 24 – 22 + 20.
Approach:
The key observation to solve this problem is that we can replace any consecutive sequence of 1 by using only two powers of 2.
Illustration:
S = “1001110”
The sequence of 3 consecutive 1’s, “1110” can be expressed as 24 – 21
Therefore, the given str
Follow the steps below to solve the problem:
- Reverse the string S.
- Iterate over the string S.
- Replace every substring of 1’s lying within indices [L, R] by placing 1 at R+1 and -1 at L.
- Once the entire string is traversed, count the number of non-zero values in the string as the required answer.
Below is the implementation of the above approach:
C++
// C++ Program to implement the// above approach#include <bits/stdc++.h>using namespace std;// Function to return the minimum// distinct powers of 2 required// to express svoid findMinimum(string s){ int n = s.size(); int x[n + 1] = { 0 }; // Reverse the string to // start from lower powers reverse(s.begin(), s.end()); for (int i = 0; i < n; i++) { // Check if the character is 1 if (s[i] == '1') { if (x[i] == 1) { x[i + 1] = 1; x[i] = 0; } // Add in range provided range else if (i and x[i - 1] == 1) { x[i + 1] = 1; x[i - 1] = -1; } else x[i] = 1; } } // Initialize the counter int c = 0; for (int i = 0; i <= n; i++) { // Check if the character // is not 0 if (x[i] != 0) // Increment the counter c++; } // Print the result cout << c << endl;}// Driver Codeint main(){ string str = "111"; findMinimum(str); return 0;} |
Java
// Java program to implement the// above approachimport java.util.*;class GFG{// Function to return the minimum// distinct powers of 2 required// to express sstatic void findMinimum(String s){ int n = s.length(); int[] x = new int[n + 1]; StringBuilder s2 = new StringBuilder(s); // Reverse the string to // start from lower powers s2.reverse(); String s3 = s2.toString(); for(int i = 0; i < n; i++) { // Check if the character is 1 if (s3.charAt(i) == '1') { if (x[i] == 1) { x[i + 1] = 1; x[i] = 0; } // Add in range provided range else if (1 <= i && (i & x[i - 1]) == 1) { x[i + 1] = 1; x[i - 1] = -1; } else x[i] = 1; } } // Initialize the counter int c = 0; for(int i = 0; i <= n; i++) { // Check if the character // is not 0 if (x[i] != 0) // Increment the counter c++; } // Print the result System.out.println(c);}// Driver codepublic static void main(String[] args){ String str = "111"; findMinimum(str);}}// This code is contributed by offbeat |
Python3
# Python3 program to implement the # above approach # Function to return the minimum# distinct powers of 2 required# to express sdef findMinimum(s): n = len(s) x = [0] * (n + 1) # Reverse the string to # start from lower powers s = s[::-1] for i in range(n): # Check if the character is 1 if(s[i] == '1'): if(x[i] == 1): x[i + 1] = 1 x[i] = 0 # Add in range provided range elif(i and x[i - 1] == 1): x[i + 1] = 1 x[i - 1] = -1 else: x[i] = 1 # Initialize the counter c = 0 for i in range(n + 1): # Check if the character # is not 0 if(x[i] != 0): # Increment the counter c += 1 # Print the result print(c)# Driver Codeif __name__ == '__main__': str = "111" # Function Call findMinimum(str)# This code is contributed by Shivam Singh |
C#
// C# program to implement the// above approachusing System;using System.Text;class GFG{// Function to return the minimum// distinct powers of 2 required// to express sstatic void findMinimum(String s){ int n = s.Length; int[] x = new int[n + 1]; StringBuilder s2 = new StringBuilder(s); // Reverse the string to // start from lower powers s2 = reverse(s2.ToString()); String s3 = s2.ToString(); for(int i = 0; i < n; i++) { // Check if the character is 1 if (s3[i] == '1') { if (x[i] == 1) { x[i + 1] = 1; x[i] = 0; } // Add in range provided range else if (1 <= i && (i & x[i - 1]) == 1) { x[i + 1] = 1; x[i - 1] = -1; } else x[i] = 1; } } // Initialize the counter int c = 0; for(int i = 0; i <= n; i++) { // Check if the character // is not 0 if (x[i] != 0) // Increment the counter c++; } // Print the result Console.WriteLine(c);}static StringBuilder 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 new StringBuilder(String.Join("", a));}// Driver codepublic static void Main(String[] args){ String str = "111"; findMinimum(str);}}// This code is contributed by Rohit_ranjan |
Javascript
<script> // Javascript Program to implement the// above approach// Function to return the minimum// distinct powers of 2 required// to express sfunction findMinimum(s){ var n = s.length; var x = Array(n+1).fill(0); // Reverse the string to // start from lower powers s.reverse(); for (var i = 0; i < n; i++) { // Check if the character is 1 if (s[i] == '1') { if (x[i] == 1) { x[i + 1] = 1; x[i] = 0; } // Add in range provided range else if (i && x[i - 1] == 1) { x[i + 1] = 1; x[i - 1] = -1; } else x[i] = 1; } } // Initialize the counter var c = 0; for (var i = 0; i <= n; i++) { // Check if the character // is not 0 if (x[i] != 0) // Increment the counter c++; } // Print the result document.write( c );}// Driver Codevar str = "111".split('');findMinimum(str);</script> |
Output:
2
Time Complexity: O(N)
Auxiliary Space: O(N)
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