给定一个二进制字符串S ,任务是找到表达 S 所需的 2 的最小幂数。
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:
解决这个问题的关键观察是我们可以只使用 2 的两个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 。
- 迭代字符串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
Javascript
输出:
2
时间复杂度: O(N)
辅助空间: O(N)
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