Maximum number of 0s placed consecutively at the start and end in any rotation of a Binary String

Given a binary string S of size N, the task is to maximize the sum of the count of consecutive 0s present at the start and end of any of the rotations of the given string S.

Examples:

Input: S = “1001”
Output: 2
Explanation:
All possible rotations of the string are:
“1001”: Count of 0s at the start = 0; at the end = 0. Sum= 0 + 0 = 0.
“0011”: Count of 0s at the start = 2; at the end = 0. Sum = 2 + 0=2
“0110”: Count of 0s at the start = 1; at the end = 1. Sum= 1 + 1 = 2.
“1100”: Count of 0s at the start = 0; at the end = 2. Sum = 0 + 2 = 2
Therefore, the maximum sum possible is 2.

Input: S = “01010”
Output: 2
Explanation:
All possible rotations of the string are:
“01010”: Count of 0s at the start = 1; at the end = 1. Sum= 1+1=1
“10100”: Count of 0s at the start = 0; at the end = 2. Sum= 0+2=2
“01001”: Count of 0s at the start = 1; at the end = 0. Sum= 1+0=1
“10010”: Count of 0s at the start = 0; at the end = 1. Sum= 0+1=1
“00101”: Count of 0s at the start = 2; at the end = 0. Sum= 2+0=2
Therefore, the maximum sum possible is 2.

Naive Approach: The simplest idea is to generate all rotations of the given string and for each rotation, count the number of 0s present at the beginning and end of the string and calculate their sum. Finally, print the maximum sum obtained.

Algorithm:

1. Initialize a counter variable c0 to 0 to count the frequency of 0s in the given string.
2. Traverse the string and for each character, if it is 0, increment the value of c0 by 1.
3. If the value of c0 is equal to the length of the string n, it means that all the characters in the string are 0, so the maximum sum of consecutive 0s present at the start and end of the string will be n. Print n and return.
4. Concatenate the string with itself and store it in a new string s.
5. Initialize a variable mx to 0 to store the maximum sum of consecutive 0s present at the start and end of a string in any rotation of the given string.
6. Generate all rotations of the string s using a loop that iterates from 0 to n-1.
7. For each rotation, initialize two variables cs and ce to 0 to store the number of consecutive 0s at the start and end of the string, respectively.
8. Traverse the rotated string from the current index i to i+n-1, and for each character, if it is 0, increment the value of cs by 1, else break out of the loop.
9. Traverse the rotated string from the current index i+n-1 to i, and for each character, if it is 0, increment the value of ce by 1, else break out of the loop.
10. Calculate the sum of cs and ce and store it in a variable val.
11. Update the value of mx to the maximum of its current value and val.
12. After the loop ends, the value of mx will contain the maximum sum of consecutive 0s present at the start and end of a string in any rotation of the given string. Print mx.

 Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the maximum sum of
// consecutive 0s present at the start
// and end of a string present in any
// of the rotations of the given string
void findMaximumZeros(string str, int n)
{
    // Check if all the characters
    // in the string are 0
    int c0 = 0;

    // Iterate over characters
    // of the string
    for (int i = 0; i < n; ++i) {
        if (str[i] == '0')
            c0++;
    }

    // If the frequency of '1' is 0
    if (c0 == n) {

        // Print n as the result
        cout << n;
        return;
    }

    // Concatenate the string
    // with itself
    string s = str + str;

    // Stores the required result
    int mx = 0;

    // Generate all rotations of the string
    for (int i = 0; i < n; ++i) {

        // Store the number of consecutive 0s
        // at the start and end of the string
        int cs = 0;
        int ce = 0;

        // Count 0s present at the start
        for (int j = i; j < i + n; ++j) {
            if (s[j] == '0')
                cs++;
            else
                break;
        }

        // Count 0s present at the end
        for (int j = i + n - 1; j >= i; --j) {
            if (s[j] == '0')
                ce++;
            else
                break;
        }

        // Calculate the sum
        int val = cs + ce;

        // Update the overall
        // maximum sum
        mx = max(val, mx);
    }

    // Print the result
    cout << mx;
}

// Driver Code
int main()
{
    // Given string
    string s = "1001";

    // Store the size of the string
    int n = s.size();

    findMaximumZeros(s, n);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to find the maximum sum of
// consecutive 0s present at the start
// and end of a string present in any
// of the rotations of the given string
static void findMaximumZeros(String str, int n)
{

    // Check if all the characters
    // in the string are 0
    int c0 = 0;

    // Iterate over characters
    // of the string
    for(int i = 0; i < n; ++i)
    {
        if (str.charAt(i) == '0')
            c0++;
    }

    // If the frequency of '1' is 0
    if (c0 == n)
    {

        // Print n as the result
        System.out.print(n);
        return;
    }

    // Concatenate the string
    // with itself
    String s = str + str;

    // Stores the required result
    int mx = 0;

    // Generate all rotations of the string
    for(int i = 0; i < n; ++i)
    {

        // Store the number of consecutive 0s
        // at the start and end of the string
        int cs = 0;
        int ce = 0;

        // Count 0s present at the start
        for(int j = i; j < i + n; ++j)
        {
            if (s.charAt(j) == '0')
                cs++;
            else
                break;
        }

        // Count 0s present at the end
        for(int j = i + n - 1; j >= i; --j)
        {
            if (s.charAt(j) == '0')
                ce++;
            else
                break;
        }

        // Calculate the sum
        int val = cs + ce;

        // Update the overall
        // maximum sum
        mx = Math.max(val, mx);
    }

    // Print the result
    System.out.print(mx);
}

// Driver Code
public static void main(String[] args)
{

    // Given string
    String s = "1001";

    // Store the size of the string
    int n = s.length();

    findMaximumZeros(s, n);
}
}

// This code is contributed by susmitakundugoaldanga

Python3

# Python3 program for the above approach

# Function to find the maximum sum of
# consecutive 0s present at the start
# and end of a string present in any
# of the rotations of the given string
def findMaximumZeros(st, n):

    # Check if all the characters
    # in the string are 0
    c0 = 0

    # Iterate over characters
    # of the string
    for i in range(n):
        if (st[i] == '0'):
            c0 += 1

    # If the frequency of '1' is 0
    if (c0 == n):

        # Print n as the result
        print(n)
        return

    # Concatenate the string
    # with itself
    s = st + st

    # Stores the required result
    mx = 0

    # Generate all rotations of the string
    for i in range(n):

        # Store the number of consecutive 0s
        # at the start and end of the string
        cs = 0
        ce = 0

        # Count 0s present at the start
        for j in range(i, i + n):
            if (s[j] == '0'):
                cs += 1
            else:
                break

        # Count 0s present at the end
        for j in range(i + n - 1, i - 1, -1):
            if (s[j] == '0'):
                ce += 1
            else:
                break

        # Calculate the sum
        val = cs + ce

        # Update the overall
        # maximum sum
        mx = max(val, mx)

    # Print the result
    print(mx)

# Driver Code
if __name__ == "__main__":

    # Given string
    s = "1001"

    # Store the size of the string
    n = len(s)

    findMaximumZeros(s, n)

    # This code is contributed by ukasp.

C#

// C# program for the above approach
using System;

class GFG{

// Function to find the maximum sum of
// consecutive 0s present at the start
// and end of a string present in any
// of the rotations of the given string
static void findMaximumZeros(string str, int n)
{

    // Check if all the characters
    // in the string are 0
    int c0 = 0;

    // Iterate over characters
    // of the string
    for(int i = 0; i < n; ++i)
    {
        if (str[i] == '0')
            c0++;
    }

    // If the frequency of '1' is 0
    if (c0 == n)
    {

        // Print n as the result
        Console.Write(n);
        return;
    }

    // Concatenate the string
    // with itself
    string s = str + str;

    // Stores the required result
    int mx = 0;

    // Generate all rotations of the string
    for(int i = 0; i < n; ++i)
    {

        // Store the number of consecutive 0s
        // at the start and end of the string
        int cs = 0;
        int ce = 0;

        // Count 0s present at the start
        for(int j = i; j < i + n; ++j)
        {
            if (s[j] == '0')
                cs++;
            else
                break;
        }

        // Count 0s present at the end
        for(int j = i + n - 1; j >= i; --j)
        {
            if (s[j] == '0')
                ce++;
            else
                break;
        }

        // Calculate the sum
        int val = cs + ce;

        // Update the overall
        // maximum sum
        mx = Math.Max(val, mx);
    }

    // Print the result
    Console.Write(mx);
}

// Driver Code
public static void Main(string[] args)
{

    // Given string
    string s = "1001";

    // Store the size of the string
    int n = s.Length;

    findMaximumZeros(s, n);
}
}

// This code is contributed by AnkThon

Javascript

<script>

// Javascript program for the above approach

// Function to find the maximum sum of
// consecutive 0s present at the start
// and end of a string present in any
// of the rotations of the given string
function findMaximumZeros(str, n)
{
    // Check if all the characters
    // in the string are 0
    var c0 = 0;

    var i;
    // Iterate over characters
    // of the string
    for (i = 0; i < n; ++i) {
        if (str[i] == '0')
            c0++;
    }

    // If the frequency of '1' is 0
    if (c0 == n) {

        // Print n as the result
        document.write(n);
        return;
    }

    // Concatenate the string
    // with itself
    var s = str + str;

    // Stores the required result
    var mx = 0;
    var j;

    // Generate all rotations of the string
    for (i = 0; i < n; ++i) {

        // Store the number of consecutive 0s
        // at the start and end of the string
        var cs = 0;
        var ce = 0;

        // Count 0s present at the start
        for (j = i; j < i + n; ++j) {
            if (s[j] == '0')
                cs++;
            else
                break;
        }

        // Count 0s present at the end
        for (j = i + n - 1; j >= i; --j) {
            if (s[j] == '0')
                ce++;
            else
                break;
        }

        // Calculate the sum
        var val = cs + ce;

        // Update the overall
        // maximum sum
        mx = Math.max(val, mx);
    }

    // Print the result
    document.write(mx);
}

    // Driver Code
    // Given string
    var s = "1001";

    // Store the size of the string
    var n = s.length;

    findMaximumZeros(s, n);

</script>

Output:

2

Time Complexity: O(N²)
Auxiliary Space: O(N)

Efficient Approach: The idea is to find the maximum number of consecutive 0s in the given string. Also, find the sum of consecutive 0s at the start and the end of the string, and then print the maximum out of them. 
Follow the steps below to solve the problem:

• Check if the frequency of ‘1’ in the string, S is equal to 0 or not. If found to be true, print the value of N as the result.
• Otherwise, perform the following steps:
  • Store the maximum number of consecutive 0s in the given string in a variable, say X.
  • Initialize two variables, start as 0 and end as N-1.
  • Increment the value of cnt and start by 1 while S[start] is not equal to ‘1‘.
  • Increment the value of cnt and decrement end by 1 while S[end] is not equal to ‘1‘.
  • Print the maximum of X and cnt as a result.

Below is the implementation of the above approach: 

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the maximum sum of
// consecutive 0s present at the start
// and end of any rotation of the string str
void findMaximumZeros(string str, int n)
{
    // Stores the count of 0s
    int c0 = 0;
    for (int i = 0; i < n; ++i) {
        if (str[i] == '0')
            c0++;
    }

    // If the frequency of '1' is 0
    if (c0 == n) {

        // Print n as the result
        cout << n;
        return;
    }

    // Stores the required sum
    int mx = 0;

    // Find the maximum consecutive
    // length of 0s present in the string
    int cnt = 0;

    for (int i = 0; i < n; i++) {
        if (str[i] == '0')
            cnt++;
        else {
            mx = max(mx, cnt);
            cnt = 0;
        }
    }

    // Update the overall maximum sum
    mx = max(mx, cnt);

    // Find the number of 0s present at
    // the start and end of the string
    int start = 0, end = n - 1;
    cnt = 0;

    // Update the count of 0s at the start
    while (str[start] != '1' && start < n) {
        cnt++;
        start++;
    }

    // Update the count of 0s at the end
    while (str[end] != '1' && end >= 0) {
        cnt++;
        end--;
    }

    // Update the maximum sum
    mx = max(mx, cnt);

    // Print the result
    cout << mx;
}

// Driver Code
int main()
{
    // Given string
    string s = "1001";

    // Store the size of the string
    int n = s.size();

    findMaximumZeros(s, n);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to find the maximum sum of
// consecutive 0s present at the start
// and end of any rotation of the string str
static void findMaximumZeros(String str, int n)
{

    // Stores the count of 0s
    int c0 = 0;
    for(int i = 0; i < n; ++i)
    {
        if (str.charAt(i) == '0')
            c0++;
    }

    // If the frequency of '1' is 0
    if (c0 == n)
    {

        // Print n as the result
        System.out.print(n);
        return;
    }

    // Stores the required sum
    int mx = 0;

    // Find the maximum consecutive
    // length of 0s present in the string
    int cnt = 0;

    for(int i = 0; i < n; i++)
    {
        if (str.charAt(i) == '0')
            cnt++;
        else
        {
            mx = Math.max(mx, cnt);
            cnt = 0;
        }
    }

    // Update the overall maximum sum
    mx = Math.max(mx, cnt);

    // Find the number of 0s present at
    // the start and end of the string
    int start = 0, end = n - 1;
    cnt = 0;

    // Update the count of 0s at the start
    while (str.charAt(start) != '1' && start < n)
    {
        cnt++;
        start++;
    }

    // Update the count of 0s at the end
    while (str.charAt(end) != '1' && end >= 0)
    {
        cnt++;
        end--;
    }

    // Update the maximum sum
    mx = Math.max(mx, cnt);

    // Print the result
    System.out.println(mx);
}

// Driver Code
public static void main (String[] args)
{

    // Given string
    String s = "1001";

    // Store the size of the string
    int n = s.length();

    findMaximumZeros(s, n);
}
}

// This code is contributed by sanjoy_62

Python3

# Python3 program for the above approach

# Function to find the maximum sum of
# consecutive 0s present at the start
# and end of any rotation of the string str
def findMaximumZeros(string, n):

    # Stores the count of 0s
    c0 = 0

    for i in range(n):
        if (string[i] == '0'):
            c0 += 1

    # If the frequency of '1' is 0
    if (c0 == n):

        # Print n as the result
        print(n, end = "")
        return

    # Stores the required sum
    mx = 0

    # Find the maximum consecutive
    # length of 0s present in the string
    cnt = 0

    for i in range(n):
        if (string[i] == '0'):
            cnt += 1
        else:
            mx = max(mx, cnt)
            cnt = 0

    # Update the overall maximum sum
    mx = max(mx, cnt)

    # Find the number of 0s present at
    # the start and end of the string
    start = 0
    end = n - 1
    cnt = 0

    # Update the count of 0s at the start
    while (string[start] != '1' and start < n):
        cnt += 1
        start += 1

    # Update the count of 0s at the end
    while (string[end] != '1' and  end >= 0):
        cnt += 1
        end -= 1

    # Update the maximum sum
    mx = max(mx, cnt)

    # Print the result
    print(mx, end = "")

# Driver Code
if __name__ == "__main__":

    # Given string
    s = "1001"

    # Store the size of the string
    n = len(s)

    findMaximumZeros(s, n)

# This code is contributed by AnkThon

C#

// C# program for the above approach
using System;

class GFG{

// Function to find the maximum sum of
// consecutive 0s present at the start
// and end of any rotation of the string str
static void findMaximumZeros(string str, int n)
{

    // Stores the count of 0s
    int c0 = 0;
    for(int i = 0; i < n; ++i)
    {
        if (str[i] == '0')
            c0++;
    }

    // If the frequency of '1' is 0
    if (c0 == n)
    {

        // Print n as the result
         Console.Write(n);
        return;
    }

    // Stores the required sum
    int mx = 0;

    // Find the maximum consecutive
    // length of 0s present in the string
    int cnt = 0;

    for(int i = 0; i < n; i++)
    {
        if (str[i] == '0')
            cnt++;
        else
        {
            mx = Math.Max(mx, cnt);
            cnt = 0;
        }
    }

    // Update the overall maximum sum
    mx = Math.Max(mx, cnt);

    // Find the number of 0s present at
    // the start and end of the string
    int start = 0, end = n - 1;
    cnt = 0;

    // Update the count of 0s at the start
    while (str[start] != '1' && start < n)
    {
        cnt++;
        start++;
    }

    // Update the count of 0s at the end
    while (str[end] != '1' && end >= 0)
    {
        cnt++;
        end--;
    }

    // Update the maximum sum
    mx = Math.Max(mx, cnt);

    // Print the result
    Console.Write(mx);
}

// Driver Code
static public void Main ()
{

    // Given string
    string s = "1001";

    // Store the size of the string
    int n = s.Length;

    findMaximumZeros(s, n);
}
}

// This code is contributed by avijitmondal1998

Javascript

<script>
//Javascript program for
//the above approach

// Function to find the maximum sum of
// consecutive 0s present at the start
// and end of any rotation of the string str
function findMaximumZeros(str, n)
{
    // Stores the count of 0s
    var c0 = 0;
    for (var i = 0; i < n; ++i) {
        if (str[i] == '0')
            c0++;
    }

    // If the frequency of '1' is 0
    if (c0 == n) {

        // Print n as the result
        document.write( n);
        return;
    }

    // Stores the required sum
    var mx = 0;

    // Find the maximum consecutive
    // length of 0s present in the string
    var cnt = 0;

    for (var i = 0; i < n; i++) {
        if (str[i] == '0')
            cnt++;
        else {
            mx = Math.max(mx, cnt);
            cnt = 0;
        }
    }

    // Update the overall maximum sum
    mx = Math.max(mx, cnt);

    // Find the number of 0s present at
    // the start and end of the string
    var start = 0, end = n - 1;
    cnt = 0;

    // Update the count of 0s at the start
    while (str[start] != '1' && start < n) {
        cnt++;
        start++;
    }

    // Update the count of 0s at the end
    while (str[end] != '1' && end >= 0) {
        cnt++;
        end--;
    }

    // Update the maximum sum
    mx = Math.max(mx, cnt);

    // Print the result
    document.write( mx);
}

var s = "1001";
// Store the size of the string
var n = s.length;

findMaximumZeros(s, n);



// This code is contributed by SoumikMondal
</script>

Output:

2

Time Complexity: O(N)
Auxiliary Space: O(1)