Max distance between first and last indexed element in Binary String

Given a binary string S of length N along with Y[] of size K. Then you have to perform K number of queries for each i (1 <= i <= K) where you have to output the maximum distance between the first and last indexed character of S among all the queries, by doing the following operation for each query:

• Reverse the value of S_i, If it is 0 then 1 or vice-versa.
• The distance between two adjacent characters is 2 if the elements are equal otherwise 1.

Note: After each query String S resets to its initial value. 

Examples:

Input: N = 4, K = 2, S = 0110, Y[] = {3, 1}   
Output: 5
Explanation:

• First Query: After changing S₃ from 1 to 0_, S = 0100, Then characters at the first and second index are not equal so the distance between them = 1. The second and third characters are also not equal, Then the distance between them is 1. The third and fourth characters are the same, So the distance between them is equal to 2. Therefore, the total distance between the first and last indexed character is 1+1+2 = 4. Then S resets to its initial value.  
• Second Query: After changing S₁ from 0 to 1, S = 1110, Then characters at the first and second index are equal so the distance between them = 2. The second and third characters are also equal, Then the distance between them is 2. The third and fourth characters are not the same, So the distance between them is equal to 1. Therefore, the total distance between the first and last indexed character is 2+2+1 = 5. 

The maximum distance between the first and last characters among all the queries is 5. Therefore, the output is 5. 

Input: N = 6, K = 6, S = 010110,  Y[] = { 2, 1, 3, 3, 4, 5 }
Output: 9
Explanation: It can be verified that the maximum distance between the first and last indexed character will be 5. Therefore, the output is 5.

Approach: Implement the idea below to solve the problem

The problem is a mathematical concept based and can be solved by using some mathematics.

Steps were taken to solve the problem:

• Create a character array let's say Arr[] and initialize it with characters of S.
• Create two variables sum=0 and max=0.
• Run a loop for j=0 to less than N-1 and follow the below-mentioned steps under the scope of the loop:
  • if(Arr[j] == Arr[j + 1]) then sum+=2 else sum+=1
• Run a loop for j =0 to less than K and follow the below-mentioned steps under the scope of the loop:
  •  if (a > 0), Where a = Y[i] -1
    • if (Arr[a] == Arr[a - 1]) then sum-=1 else sum+=1 
  •  (a < N - 1), Where a = Y[i] -1
    •  (Arr[a] == Arr[a +1]) then sum-=1 else sum+=1 
  • Update max, formally max=sum>max?sum: max
• Output the value of max.

Below is the code to implement the approach:

// C++ code to implement the approach

#include <bits/stdc++.h>
using namespace std;

// Method for calculating maximum distance
void Max_Distance(int N, int K, string S, int Y[])
{

  // Variable for holding
  // sum of distances
  int sum = 0;

  int max = 0;

  // Loop for calculating sum
  for (int j = 0; j < N - 1; j++) {
    if (S[j] == S[j + 1])
      sum = sum + 2;
    else
      sum = sum + 1;
  }

  // Loop running number of
  // times query asked
  for (int j = 0; j < K; j++) {
    int a = Y[j] - 1;
    if (a > 0) {
      if (S[a] == S[a - 1])
        sum = sum - 1;
      else
        sum = sum + 1;
    }
    if (a < N - 1)
      if (S[a] == S[a + 1])
        sum = sum - 1;
    else
      sum = sum + 1;

    // Updating maximum length
    max = sum > max ? sum : max;
    if (S[a] == '1')
      S[a] = '0';
    else
      S[a] = '1';
  }

  // Printing maximum distance
  cout<< max << endl;
}

int main()
{

  // Inputs according to statement
  int N = 6;
  int K = 6;
  string S = "010110";
  int Y[] = { 2, 1, 3, 3, 4, 5 };

  // Function call
  Max_Distance(N, K, S, Y);
}

// Java code to implement the approach

import java.io.*;
import java.lang.*;
import java.util.*;

class GFG {

    // Driver Function
    public static void main(String[] args)
        throws java.lang.Exception
    {
        // Inputs according to statement
        int N = 6;
        int K = 6;
        String S = "010110";
        int Y[] = { 2, 1, 3, 3, 4, 5 };

        // Function call
        Max_Distance(N, K, S, Y);
    }

    // Method for calculating maximum distance
    static void Max_Distance(int N, int K, String S,
                             int[] Y)
    {

        // Array of characters for
        // storing characters of S
        char arr[] = S.toCharArray();

        // Variable for holding
        // sum of distances
        int sum = 0;

        int max = 0;

        // Loop for calculating sum
        for (int j = 0; j < N - 1; j++) {
            if (arr[j] == arr[j + 1])
                sum = sum + 2;
            else
                sum = sum + 1;
        }

        // Loop running number of
        // times query asked
        for (int j = 0; j < K; j++) {
            int a = Y[j] - 1;
            if (a > 0) {
                if (arr[a] == arr[a - 1])
                    sum = sum - 1;
                else
                    sum = sum + 1;
            }
            if (a < N - 1)
                if (arr[a] == arr[a + 1])
                    sum = sum - 1;
                else
                    sum = sum + 1;

            // Updating maximum length
            max = sum > max ? sum : max;
            if (arr[a] == '1')
                arr[a] = '0';
            else
                arr[a] = '1';
        }

        // Printing maximum distance
        System.out.println(max);
    }
}

def Max_Distance(N, K, S, Y):
    # Variable for holding sum of distances
    sum = 0

    # Variable for holding maximum distance
    max_dist = 0

    # Loop for calculating sum
    for j in range(N - 1):
        if S[j] == S[j + 1]:
            sum += 2
        else:
            sum += 1

    # Loop running number of times query asked
    for j in range(K):
        a = Y[j] - 1
        if a > 0:
            if S[a] == S[a - 1]:
                sum -= 1
            else:
                sum += 1
        if a < N - 1:
            if S[a] == S[a + 1]:
                sum -= 1
            else:
                sum += 1

        # Updating maximum length
        max_dist = max(sum, max_dist)
        if S[a] == '1':
            S = S[:a] + '0' + S[a+1:]
        else:
            S = S[:a] + '1' + S[a+1:]

    # Printing maximum distance
    print(max_dist)


# Inputs according to statement
N = 6
K = 6
S = "010110"
Y = [2, 1, 3, 3, 4, 5]

# Function call
Max_Distance(N, K, S, Y)

using System;

public class Program
{
    // Method for calculating maximum distance
    public static void Max_Distance(int N, int K, string S, int[] Y)
    {
        // Variable for holding sum of distances
        int sum = 0;
        int max = 0;

        // Loop for calculating sum
        for (int j = 0; j < N - 1; j++)
        {
            if (S[j] == S[j + 1])
            {
                sum += 2;
            }
            else
            {
                sum += 1;
            }
        }

        // Loop running number of times query asked
        for (int j = 0; j < K; j++)
        {
            int a = Y[j] - 1;

            if (a > 0)
            {
                if (S[a] == S[a - 1])
                {
                    sum -= 1;
                }
                else
                {
                    sum += 1;
                }
            }

            if (a < N - 1)
            {
                if (S[a] == S[a + 1])
                {
                    sum -= 1;
                }
                else
                {
                    sum += 1;
                }
            }
            else
            {
                sum += 1;
            }

            // Updating maximum length
            max = sum > max ? sum : max;

            if (S[a] == '1')
            {
                S = S.Remove(a, 1).Insert(a, "0");
            }
            else
            {
                S = S.Remove(a, 1).Insert(a, "1");
            }
        }

        // Printing maximum distance
        Console.WriteLine(max);
    }

    public static void Main()
    {
        // Inputs according to statement
        int N = 6;
        int K = 6;
        string S = "010110";
        int[] Y = { 2, 1, 3, 3, 4, 5 };

        // Function call
        Max_Distance(N, K, S, Y);
    }
}

// javascript code to implement the approach

// Method for calculating maximum distance
function Max_Distance(N, K, s, Y)
{

  // Variable for holding
  // sum of distances
  let sum = 0;
  let S = s.split('');
  let max = 0;

  // Loop for calculating sum
  for (let j = 0; j < N - 1; j++) {
    if (S[j] == S[j + 1])
      sum = sum + 2;
    else
      sum = sum + 1;
  }

  // Loop running number of
  // times query asked
  for (let j = 0; j < K; j++) {
    let a = Y[j] - 1;
    if (a > 0) {
      if (S[a] == S[a - 1])
        sum = sum - 1;
      else
        sum = sum + 1;
    }
    if (a < N - 1)
      if (S[a] == S[a + 1])
        sum = sum - 1;
    else
      sum = sum + 1;

    // Updating maximum length
    max = sum > max ? sum : max;
    if (S[a] == '1')
      S[a] = '0';
    else
      S[a] = '1';
  }

  // Printing maximum distance
  console.log(max);
}


// Inputs according to statement
let N = 6;
let K = 6;
let S = "010110";
let Y = [2, 1, 3, 3, 4, 5];

// Function call
Max_Distance(N, K, S, Y);

// The code is contributed by Arushi Jindal. 
    

9

Time Complexity: O(N), As String S is traversed to calculate distance. 
Auxiliary Space: O(N), As an array for storing characters, is used because the string is immutable in Java.

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Article Tags :

• Strings
• DSA
• Arrays

Practice Tags :

• Arrays
• Strings