# C++ Program to Find 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 **0**s present at the start and end of any of the rotations of the given string **S**.

**Examples:**

Input:S = “1001”Output:2Explanation:

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:2Explanation:

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.

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;` `}` |

**Output:**

2

**Time Complexity: **O(N^{2})**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 **0**s 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.

- Store the maximum number of consecutive 0s in the given string in a variable, say

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;` `}` |

**Output:**

2

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

Please refer complete article on Maximum number of 0s placed consecutively at the start and end in any rotation of a Binary String for more details!