表示给定二进制数所需的2的幂次方的最小数目

给定二进制字符串S ，任务是找到表示S所需的2的最小幂。
例子：

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.

方法：
解决此问题的关键发现是，我们可以仅使用2的2的幂来替换任何连续的1序列。

Illustration: 
S = “1001110” 
The sequence of 3 consecutive 1’s, “1110” can be expressed as 2⁴ – 2¹ 
Therefore, the given str

请按照以下步骤解决问题：

• 反转字符串S。
• 遍历字符串S。
• 通过将在1 R + 1和-1为L取代1米的卧指数为[L，R]内的每个子串。
• 一旦整个字符串被遍历，算上字符串所需答案非零值的数量。

下面是上述方法的实现：

C++

// C++ Program to implement the
// above approach
#include 
using namespace std;
  
// Function to return the minimum
// distinct powers of 2 required
// to express s
void 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 Code
int main()
{
    string str = "111";
  
    findMinimum(str);
  
    return 0;
}

Java

// Java program to implement the
// above approach
import java.util.*;
  
class GFG{
  
// Function to return the minimum
// distinct powers of 2 required
// to express s
static 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 code
public 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 s
def 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 Code
if __name__ == '__main__':
  
    str = "111"
      
    # Function Call
    findMinimum(str)
  
# This code is contributed by Shivam Singh

C#

// C# program to implement the
// above approach
using System;
using System.Text;
  
class GFG{
  
// Function to return the minimum
// distinct powers of 2 required
// to express s
static 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 code
public static void Main(String[] args)
{
    String str = "111";
  
    findMinimum(str);
}
}
  
// This code is contributed by Rohit_ranjan

输出：

2

时间复杂度： O(N)
辅助空间： O(N)

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