Queries to flip characters of a binary string in given range
Last Updated :
21 Jun, 2021
Given a binary string, str and a 2D array Q[][] representing queries of the form {L, R}. In each query, toggle all the characters of the binary strings present in the indices [L, R]. The task is to print the binary string by performing all the queries.
Examples:
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
Naive Approach: The simplest approach to solve the problem is to traverse all the queries arrays and toggle all the characters in the indices [L, R]. Finally, print the binary string.
Time Complexity: O(M * N), Where N denotes the length of the binary string.
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach the idea is to use the Prefix Sum array technique. Follow the steps below to solve the problem:
Below is the implementation of the above approach:
C++
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
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 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 shikhasingrajput
# 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 string
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 string
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 29
// 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
<script>
// Javascript program to implement
// the above approach
// Function to find the binary
// String by performing all the
// given queries
function toggleQuery(str, Q, M)
{
// Stores length of the String
let N = str.length;
// prefixCnt[i]: Stores number
// of times str[i] toggled by
// performing all the queries
let prefixCnt = new Array(N + 1);
for(let i = 0; i < N + 1; i++)
{
prefixCnt[i] = 0;
}
for (let 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 (let i = 1; i <= N; i++)
{
prefixCnt[i] += prefixCnt[i - 1];
}
// Traverse prefixCnt[] array
for (let 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] = String.fromCharCode('1'.charCodeAt(0) -
str[i].charCodeAt(0) + '0'.charCodeAt(0));
}
}
return (str).join("");
}
// Driver Code
let str = "101010";
let Q = [[0, 1], [2, 5],
[2, 3], [1, 4],
[0, 5]];
let M = Q.length;
document.write(
toggleQuery(str.split(""),
Q, M));
// This code is contributed by patel2127
</script>
Time Complexity: O(N + |Q|), Where N is the length of binary string.
Auxiliary Space: O(N)
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