Given a binary string consisting only of 1’s and 0’s. Find the bit (output is either 1 or 0)whose the minimum number of contiguous sequence flips can make all bits of the string same.
Here, contiguous sequence flip means flipping a substring or 0s or 1s. For Example, in the string “00011110001110”, in one flip we can change the string to “11111110001110”. The first three continuous zeros are changed to 1.
- In one flip we can change any continuous sequence of this string.
- If both 0s and 1s are possible, print the one which comes last.
- The task is to simply print the bit 1 or 0 for which minimum sequence flips will make all bits same.
Input : str = “00011110001110”
Output : 1
Explanation: There are two contiguous sequences of 1’s and three contiguous sequences of 0’s. So flipping 1’s would lead to minimum flips.
Input: str = “010101100011”
Explanation: Since the count of groups of 0s and 1s are same and 1 comes in last.
Naive Approach: Start traversing through the string and take variables named gropupOfOnes and another groupOfZeros to count the groups of 0s and 1s. Now, compare the count of groups of one’s and zero’s. The one with the lesser count will be the answer.
If both are equal, the answer will the last character of the string.
Time complexity: O(N)
- Observe the string carefully, the character at last index of the string will have more groups in the string (If the first and the last character of the string are equal). So, the answer will be another character.
- If the first and last characters are not equal, both the characters have the same number of groups. So, the answer will be the character at last index.
Below is the implementation of the above approach:
# Python 3 program to find which bit
# sequence to be flipped
# Function to check which bit is
# to be flipped
def bitToBeFlipped( s):
# variable to store first and
# last character of string
last = s[len(s) – 1]
first = s
# Check if first and last characters
# are equal, if yes, then return
# the character which is not at last
if (last == first) :
if (last == ‘0’) :
# else return last
elif (last != first) :
# Driver Code
if __name__ == “__main__”:
s = “1101011000”
# This code is contributed by ita_c
Time Complexity: O(1)
Auxiliary Space: O(1)
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