给定二进制字符串str ，任务是找到在给定二进制字符串中将所有 1 保持在一起所需的最小翻转次数，即字符串的1 之间不能有任何 0。
例子：
Input: str = “0011111100”
Output: 0
Explanation: We dont need to flip any bits because all the ones are grouped together and there is no zero between any two ones.
Input: str = “11100111000101”
Output: 4
Explanation: We can flip the 4th and 5th bit to make them 1 and flip 12th and 14th bit to make them 0. So the resulting string is “11111111000000” with 4 possible flips.
方法：为了解决上面提到的问题，我们将实现动态规划方法，其中我们将有以下状态：
- 第一个状态是 dp[i][0] ，它表示将所有零变为第 i 位所需的翻转次数。
- 第二个状态 dp[i][1]表示使当前位为 1 所需的翻转次数，从而满足问题中给出的条件。
因此，所需的答案将是使当前位为 1 的最小翻转次数 + 使当前位 0 之后的所有位为 i 的所有值的最小翻转次数。但是如果给定字符串中的所有位都是 0，那么我们就不必更改任何内容，因此我们可以检查我们的答案与使字符串全部为零所需的翻转次数之间的最小值。所以我们可以通过迭代字符串中的所有字符来计算答案，其中，
answer = min ( answer, dp[i][1] + dp[n-1][0] – dp[i][0])
where
dp[i][1] = Minimum number of flips to set current bit to 1
dp[n-1][0] – dp[i][0] = Minimum number of flips required to make all bits after i as 0
下面是上述方法的实现：
C++
//cpp implementation for Minimum number of
//flips required in a binary string such
//that all the 1’s are together
#include
using namespace std;
int minFlip(string a)
{
//Length of the binary string
int n = a.size();
vector> dp(n + 1,vector(2, 0));
//Initial state of the dp
//dp[0][0] will be 1 if the current
//bit is 1 and we have to flip it
dp[0][0] = (a[0] == '1');
//Initial state of the dp
//dp[0][1] will be 1 if the current
//bit is 0 and we have to flip it
dp[0][1] = (a[0] == '0');
for (int i = 1; i < n; i++)
{
//dp[i][0] = Flips required to
//make all previous bits zero
//+ Flip required to make current bit zero
dp[i][0] = dp[i - 1][0] + (a[i] == '1');
//dp[i][1] = mimimum flips required
//to make all previous states 0 or make
//previous states 1 satisfying the condition
dp[i][1] = min(dp[i - 1][0],
dp[i - 1][1]) + (a[i] == '0');
}
int answer = INT_MAX;
for (int i=0;i Java
// Java implementation for
// Minimum number of flips
// required in a binary string
// such that all the 1’s are together
import java.io.*;
import java.util.*;
class GFG{
static int minFlip(String a)
{
// Length of the binary string
int n = a.length();
int dp[][] = new int[n + 1][2];
// Initial state of the dp
// dp[0][0] will be 1 if
// the current bit is 1
// and we have to flip it
if(a.charAt(0) == '1')
{
dp[0][0] = 1 ;
}
else dp[0][0] = 0;
// Initial state of the dp
// dp[0][1] will be 1 if
// the current bit is 0
// and we have to flip it
if(a.charAt(0) == '0')
dp[0][1] = 1;
else dp[0][1] = 0;
for (int i = 1; i < n; i++)
{
// dp[i][0] = Flips required to
// make all previous bits zero
// Flip required to make current
// bit zero
if(a.charAt(i) == '1')
{
dp[i][0] = dp[i - 1][0] + 1;
}
else dp[i][0] = dp[i - 1][0];
// dp[i][1] = mimimum flips
// required to make all previous
// states 0 or make previous states
// 1 satisfying the condition
if(a.charAt(i) == '0')
{
dp[i][1] = Math.min(dp[i - 1][0],
dp[i - 1][1]) + 1;
}
else dp[i][1] = Math.min(dp[i - 1][0],
dp[i - 1][1]);
}
int answer = Integer.MAX_VALUE;
for (int i = 0; i < n; i++)
{
answer = Math.min(answer, dp[i][1] +
dp[n - 1][0] -
dp[i][0]);
}
//Minimum of answer and flips
//required to make all bits 0
return Math.min(answer,
dp[n - 1][0]);
}
// Driver code
public static void main (String[] args)
{
String s = "1100111000101";
System.out.println(minFlip(s));
}
}
// This code is contributed by satvikshrivas26Python3
# Python implementation for Minimum number of
# flips required in a binary string such
# that all the 1’s are together
def minFlip(a):
# Length of the binary string
n = len(a)
dp =[[0, 0] for i in range(n)]
# Initial state of the dp
# dp[0][0] will be 1 if the current
# bit is 1 and we have to flip it
dp[0][0]= int(a[0]=='1')
# Initial state of the dp
# dp[0][1] will be 1 if the current
# bit is 0 and we have to flip it
dp[0][1]= int(a[0]=='0')
for i in range(1, n):
# dp[i][0] = Flips required to
# make all previous bits zero
# + Flip required to make current bit zero
dp[i][0]= dp[i-1][0]+int(a[i]=='1')
# dp[i][1] = mimimum flips required
# to make all previous states 0 or make
# previous states 1 satisfying the condition
dp[i][1]= min(dp[i-1])+int(a[i]=='0')
answer = 10**18
for i in range(n):
answer = min(answer,
dp[i][1]+dp[n-1][0]-dp[i][0])
# Minimum of answer and flips
# required to make all bits 0
return min(answer, dp[n-1][0])
# Driver code
s = "1100111000101"
print(minFlip(s))C#
// C# implementation for
// Minimum number of
// flips required in
// a binary string such
// that all the 1’s are together
using System;
class GFG{
static int minFlip(string a)
{
//Length of the binary string
int n = a.Length;
int [,]dp=new int[n + 1, 2];
//Initial state of the dp
//dp[0][0] will be 1 if the current
//bit is 1 and we have to flip it
dp[0, 0] = (a[0] == '1' ? 1 : 0);
//Initial state of the dp
//dp[0][1] will be 1 if the current
//bit is 0 and we have to flip it
dp[0, 1] = (a[0] == '0' ? 1 : 0);
for (int i = 1; i < n; i++)
{
//dp[i][0] = Flips required to
//make all previous bits zero
//+ Flip required to make current bit zero
dp[i, 0] = dp[i - 1, 0] +
(a[i] == '1' ? 1 : 0);
//dp[i][1] = mimimum flips required
//to make all previous states 0 or make
//previous states 1 satisfying the condition
dp[i, 1] = Math.Min(dp[i - 1, 0],
dp[i - 1, 1]) +
(a[i] == '0' ? 1 : 0);
}
int answer = int.MaxValue;
for (int i = 0; i < n; i++)
{
answer = Math.Min(answer, dp[i, 1] +
dp[n - 1, 0] - dp[i, 0]);
}
//Minimum of answer and flips
//required to make all bits 0
return Math.Min(answer, dp[n - 1, 0]);
}
// Driver code
public static void Main(string[] args)
{
string s = "1100111000101";
Console.Write(minFlip(s));
}
}
// This code is contributed by Rutvik_56Javascript
输出：
4如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。