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.
Below is the implementation of the above approach:
Java
// Java program for the above approachimport 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 stringstatic 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 Codepublic 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 |
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:
Java
// Java program for the above approachimport 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 strstatic 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 Codepublic 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 |
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!
