给定一个由0和1 组成的二进制字符串S 。任务是使用以下操作使给定的字符串成为替代字符序列：
- 从开头删除一些前缀并将其附加到结尾。
- 翻转给定字符串中的部分或每一位。
打印要翻转以使给定字符串交替的最小位数。
例子：
Input: S = “001”
Output: 0
Explanation:
No need to flip any element we can get alternating sequence by using left rotation: 010.
Input: S = “000001100”
Output: 3
Explanation:
Following steps to find minimum flips to get alternating string:
1. After rotating string 6 times towards left we will get: 100000001
2. Now we can apply flip operation as following: 101000001 -> 101010001 -> 101010101
Thus, minimum flips to make string alternating is 3.
朴素的方法：朴素的方法是采用所有N 种可能的组合并计算要在每个字符串翻转的最小位数。打印所有此类组合中的最小计数。
时间复杂度： O(N 2 )，其中 N 是字符串的长度。
辅助空间： O(N)
有效方法：这可以通过观察最终字符串将是“101010…”或“010101…”类型中的任一种来解决，这样所有的1要么位于奇数位置，要么位于偶数位置。请按照以下步骤解决问题：
- 创建一个前缀和数组，其中pref[i]表示索引 i 之前所需的更改次数。
- 为上述两种模式创建前缀数组。
- 检查每个i ，如果substring[0, i]附加在最后需要翻转多少个字符。
- 打印上述步骤中所有子串之间的最小翻转次数。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function that finds the minimum
// number of flips to make the
// binary string alternating if
// left circular rotation is allowed
int MinimumFlips(string s, int n)
{
int a[n];
for(int i = 0; i < n; i++)
{
a[i] = (s[i] == '1' ? 1 : 0);
}
// Initialize prefix arrays to store
// number of changes required to put
// 1s at either even or odd position
int oddone[n + 1];
int evenone[n + 1];
oddone[0] = 0;
evenone[0] = 0;
for(int i = 0; i < n; i++)
{
// If i is odd
if (i % 2 != 0)
{
// Update the oddone
// and evenone count
oddone[i + 1] = oddone[i] +
(a[i] == 1 ? 1 : 0);
evenone[i + 1] = evenone[i] +
(a[i] == 0 ? 1 : 0);
}
// Else i is even
else
{
// Update the oddone
// and evenone count
oddone[i + 1] = oddone[i] +
(a[i] == 0 ? 1 : 0);
evenone[i + 1] = evenone[i] +
(a[i] == 1 ? 1 : 0);
}
}
// Initialize minimum flips
int minimum = min(oddone[n], evenone[n]);
// Check if substring[0, i] is
// appended at end how many
// changes will be required
for(int i = 0; i < n; i++)
{
if (n % 2 != 0)
{
minimum = min(minimum,
oddone[n] -
oddone[i + 1] +
evenone[i + 1]);
minimum = min(minimum,
evenone[n] -
evenone[i + 1] +
oddone[i + 1]);
}
}
// Return minimum flips
return minimum;
}
// Driver Code
int main()
{
// Given String
string S = "000001100";
// Length of given string
int n = S.length();
// Function call
cout << (MinimumFlips(S, n));
}
// This code is contributed by chitranayal Java
// Java program for the above approach
import java.util.*;
class GFG {
// Function that finds the minimum
// number of flips to make the
// binary string alternating if
// left circular rotation is allowed
static int MinimumFlips(String s, int n)
{
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = (s.charAt(i) == '1' ? 1 : 0);
}
// Initialize prefix arrays to store
// number of changes required to put
// 1s at either even or odd position
int[] oddone = new int[n + 1];
int[] evenone = new int[n + 1];
oddone[0] = 0;
evenone[0] = 0;
for (int i = 0; i < n; i++) {
// If i is odd
if (i % 2 != 0) {
// Update the oddone
// and evenone count
oddone[i + 1]
= oddone[i]
+ (a[i] == 1 ? 1 : 0);
evenone[i + 1]
= evenone[i]
+ (a[i] == 0 ? 1 : 0);
}
// Else i is even
else {
// Update the oddone
// and evenone count
oddone[i + 1]
= oddone[i]
+ (a[i] == 0 ? 1 : 0);
evenone[i + 1]
= evenone[i]
+ (a[i] == 1 ? 1 : 0);
}
}
// Initialize minimum flips
int minimum = Math.min(oddone[n],
evenone[n]);
// Check if substring[0, i] is
// appended at end how many
// changes will be required
for (int i = 0; i < n; i++) {
if (n % 2 != 0) {
minimum = Math.min(minimum,
oddone[n]
- oddone[i + 1]
+ evenone[i + 1]);
minimum = Math.min(minimum,
evenone[n]
- evenone[i + 1]
+ oddone[i + 1]);
}
}
// Return minimum flips
return minimum;
}
// Driver Code
public static void main(String[] args)
{
// Given String
String S = "000001100";
// Length of given string
int n = S.length();
// Function call
System.out.print(MinimumFlips(S, n));
}
}Python3
# Python3 program for the above approach
# Function that finds the minimum
# number of flips to make the
# binary string alternating if
# left circular rotation is allowed
def MinimumFlips(s, n):
a = [0] * n
for i in range(n):
a[i] = 1 if s[i] == '1' else 0
# Initialize prefix arrays to store
# number of changes required to put
# 1s at either even or odd position
oddone = [0] * (n + 1)
evenone = [0] * (n + 1)
for i in range(n):
# If i is odd
if(i % 2 != 0):
# Update the oddone
# and evenone count
if(a[i] == 1):
oddone[i + 1] = oddone[i] + 1
else:
oddone[i + 1] = oddone[i] + 0
if(a[i] == 0):
evenone[i + 1] = evenone[i] + 1
else:
evenone[i + 1] = evenone[i] + 0
# Else i is even
else:
# Update the oddone
# and evenone count
if (a[i] == 0):
oddone[i + 1] = oddone[i] + 1
else:
oddone[i + 1] = oddone[i] + 0
if (a[i] == 1):
evenone[i + 1] = evenone[i] + 1
else:
evenone[i + 1] = evenone[i] + 0
# Initialize minimum flips
minimum = min(oddone[n], evenone[n])
# Check if substring[0, i] is
# appended at end how many
# changes will be required
for i in range(n):
if(n % 2 != 0):
minimum = min(minimum,
oddone[n] -
oddone[i + 1] +
evenone[i + 1])
minimum = min(minimum,
evenone[n] -
evenone[i + 1] +
oddone[i + 1])
# Return minimum flips
return minimum
# Driver Code
# Given String
S = "000001100"
# Length of given string
n = len(S)
# Function call
print(MinimumFlips(S, n))
# This code is contributed by Shivam SinghC#
// C# program for the above approach
using System;
class GFG{
// Function that finds the minimum
// number of flips to make the
// binary string alternating if
// left circular rotation is allowed
static int MinimumFlips(String s, int n)
{
int[] a = new int[n];
for (int i = 0; i < n; i++)
{
a[i] = (s[i] == '1' ? 1 : 0);
}
// Initialize prefix arrays to store
// number of changes required to put
// 1s at either even or odd position
int[] oddone = new int[n + 1];
int[] evenone = new int[n + 1];
oddone[0] = 0;
evenone[0] = 0;
for (int i = 0; i < n; i++)
{
// If i is odd
if (i % 2 != 0)
{
// Update the oddone
// and evenone count
oddone[i + 1] = oddone[i] +
(a[i] == 1 ? 1 : 0);
evenone[i + 1] = evenone[i] +
(a[i] == 0 ? 1 : 0);
}
// Else i is even
else
{
// Update the oddone
// and evenone count
oddone[i + 1] = oddone[i] +
(a[i] == 0 ? 1 : 0);
evenone[i + 1] = evenone[i] +
(a[i] == 1 ? 1 : 0);
}
}
// Initialize minimum flips
int minimum = Math.Min(oddone[n],
evenone[n]);
// Check if substring[0, i] is
// appended at end how many
// changes will be required
for (int i = 0; i < n; i++)
{
if (n % 2 != 0)
{
minimum = Math.Min(minimum, oddone[n] -
oddone[i + 1] +
evenone[i + 1]);
minimum = Math.Min(minimum, evenone[n] -
evenone[i + 1] +
oddone[i + 1]);
}
}
// Return minimum flips
return minimum;
}
// Driver Code
public static void Main(String[] args)
{
// Given String
String S = "000001100";
// Length of given string
int n = S.Length;
// Function call
Console.Write(MinimumFlips(S, n));
}
}
// This code is contributed by Rajput-JiJavascript
输出：
3时间复杂度： O(N)，其中 N 是给定字符串的长度。
辅助空间： O(N)
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。