给定一个二进制字符串str和一个二维数组Q [] []，表示形式为{L，R}的查询。在每个查询中，切换索引[L，R]中存在的二进制字符串的所有字符。任务是通过执行所有查询来打印二进制字符串。
例子：
Input: str = “101010”, Q[][] = { {0, 1}, {2, 5}, {2, 3}, {1, 4}, {0, 5} }
Output: 111000
Explanation:
Query 1: Toggling all the characters of str in the indices [0, 1]. Therefore, str becomes “011010”.
Query 2: Toggling all the characters of str in the indices [2, 5]. Therefore, str becomes “010101”.
Query 3: Toggling all the characters of str in the indices [2, 3]. Therefore, str becomes “011001”.
Query 4: Toggling all the characters of str in the indices [1, 4]. Therefore, str becomes “000111”.
Query 5: Toggling all the characters of str in the indices [0, 5]. Therefore, str becomes “111000”.
Therefore, the required binary string is “111000”.
Input: str = “00101”, Q[][]={ {0, 2}, {2, 3}, {1, 4} }
Output: 10000
天真的方法：解决问题的最简单方法是遍历所有查询数组并切换索引[L，R]中的所有字符。最后，打印二进制字符串。
时间复杂度： O(M * N)，其中N表示二进制字符串的长度。
辅助空间： O(1)
高效方法：为了优化上述方法，其思想是使用Prefix Sum数组技术。请按照以下步骤解决问题：
- 初始化一个数组，例如prefixCnt []，以存储二进制字符串的第i个元素被切换的次数的值。
- 遍历查询Q [] []数组，并为每个查询更新prefixCnt [Q [i] [0]] + = 1和prefixCnt [Q [i] [1] + 1]-= 1的值。
- 遍历prefixCnt []数组，并获取prefixCnt []数组的前缀总和。
- 最后，遍历二进制字符串并检查prefixCnt [i]％2 == 1的值。如果发现是正确的，则切换二进制字符串的第i个元素。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to find the binary string by
// performing all the given queries
string toggleQuery(string str, int Q[][2],
int M)
{
// Stores length of the string
int N = str.length();
// prefixCnt[i]: Stores number
// of times str[i] toggled by
// performing all the queries
int prefixCnt[N + 1] = { 0 };
for (int i = 0; i < M; i++) {
// Update prefixCnt[Q[i][0]]
prefixCnt[Q[i][0]] += 1;
// Update prefixCnt[Q[i][1] + 1]
prefixCnt[Q[i][1] + 1] -= 1;
}
// Calculate prefix sum of prefixCnt[i]
for (int i = 1; i <= N; i++) {
prefixCnt[i] += prefixCnt[i - 1];
}
// Traverse prefixCnt[] array
for (int i = 0; i < N; i++) {
// If ith element toggled
// odd number of times
if (prefixCnt[i] % 2) {
// Toggled i-th element
// of binary string
str[i] = '1' - str[i] + '0';
}
}
return str;
}
// Driver Code
int main()
{
string str = "101010";
int Q[][2] = { { 0, 1 }, { 2, 5 },
{ 2, 3 }, { 1, 4 },
{ 0, 5 } };
int M = sizeof(Q) / sizeof(Q[0]);
cout << toggleQuery(str, Q, M);
} Java
// Java program to implement
// the above approach
import java.util.*;
class GFG{
// Function to find the binary
// String by performing all the
// given queries
static String toggleQuery(char[] str,
int Q[][], int M)
{
// Stores length of the String
int N = str.length;
// prefixCnt[i]: Stores number
// of times str[i] toggled by
// performing all the queries
int prefixCnt[] = new int[N + 1];
for (int i = 0; i < M; i++)
{
// Update prefixCnt[Q[i][0]]
prefixCnt[Q[i][0]] += 1;
// Update prefixCnt[Q[i][1] + 1]
prefixCnt[Q[i][1] + 1] -= 1;
}
// Calculate prefix sum of
// prefixCnt[i]
for (int i = 1; i <= N; i++)
{
prefixCnt[i] += prefixCnt[i - 1];
}
// Traverse prefixCnt[] array
for (int i = 0; i < N; i++)
{
// If ith element toggled
// odd number of times
if (prefixCnt[i] % 2 == 1)
{
// Toggled i-th element
// of binary String
str[i] = (char)('1' -
str[i] + '0');
}
}
return String.valueOf(str);
}
// Driver Code
public static void main(String[] args)
{
String str = "101010";
int Q[][] = {{0, 1}, {2, 5},
{2, 3}, {1, 4},
{0, 5}};
int M = Q.length;
System.out.print(
toggleQuery(str.toCharArray(),
Q, M));
}
}
// This code is contributed by shikhasingrajputPython3
# Python3 program to implement
# the above approach
# Function to find the binary by
# performing all the given queries
def toggleQuery(strr, Q, M):
strr = [i for i in strr]
# Stores length of the strring
N = len(strr)
# prefixCnt[i]: Stores number
# of times strr[i] toggled by
# performing all the queries
prefixCnt = [0] * (N + 1)
for i in range(M):
# Update prefixCnt[Q[i][0]]
prefixCnt[Q[i][0]] += 1
# Update prefixCnt[Q[i][1] + 1]
prefixCnt[Q[i][1] + 1] -= 1
# Calculate prefix sum of prefixCnt[i]
for i in range(1, N + 1):
prefixCnt[i] += prefixCnt[i - 1]
# Traverse prefixCnt[] array
for i in range(N):
# If ith element toggled
# odd number of times
if (prefixCnt[i] % 2):
# Toggled i-th element
# of binary strring
strr[i] = (chr(ord('1') -
ord(strr[i]) +
ord('0')))
return "".join(strr)
# Driver Code
if __name__ == '__main__':
strr = "101010";
Q = [ [ 0, 1 ],[ 2, 5 ],
[ 2, 3 ],[ 1, 4 ],
[ 0, 5 ] ]
M = len(Q)
print(toggleQuery(strr, Q, M))
# This code is contributed by mohit kumar 29C#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to find the binary
// String by performing all the
// given queries
static String toggleQuery(char[] str,
int [,]Q ,
int M)
{
// Stores length of the String
int N = str.Length;
// prefixCnt[i]: Stores number
// of times str[i] toggled by
// performing all the queries
int []prefixCnt = new int[N + 1];
for (int i = 0; i < M; i++)
{
// Update prefixCnt[Q[i,0]]
prefixCnt[Q[i, 0]] += 1;
// Update prefixCnt[Q[i,1] + 1]
prefixCnt[Q[i, 1] + 1] -= 1;
}
// Calculate prefix sum of
// prefixCnt[i]
for (int i = 1; i <= N; i++)
{
prefixCnt[i] += prefixCnt[i - 1];
}
// Traverse prefixCnt[] array
for (int i = 0; i < N; i++)
{
// If ith element toggled
// odd number of times
if (prefixCnt[i] % 2 == 1)
{
// Toggled i-th element
// of binary String
str[i] = (char)('1' -
str[i] + '0');
}
}
return String.Join("", str);
}
// Driver Code
public static void Main(String[] args)
{
String str = "101010";
int [,]Q = {{0, 1}, {2, 5},
{2, 3}, {1, 4},
{0, 5}};
int M = Q.GetLength(0);
Console.Write(toggleQuery(str.ToCharArray(),
Q, M));
}
}
// This code is contributed by shikhasingrajput输出：
111000
时间复杂度： O(N + | Q |)，其中N是二进制字符串的长度。
辅助空间： O(N)