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:
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. |
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:
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 |
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!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
