Length of the Longest Consecutive 1s in Binary Representation
Given a number N, The task is to find the length of the longest consecutive 1s series in its binary representation.
Input: N = 14
Explanation: The binary representation of 14 is 1110.
Input: N = 222
Explanation: The binary representation of 222 is 11011110.
Naive Approach: Below is the idea to solve the problem
Traverse the bits of binary representation of N and keep a track of the number of consecutive set bits, and the maximum length of consecutive 1s found so far.
Time Complexity: O(X), Here X is the length of binary representation of N.
Auxiliary Space: O(1)
Find the length of the longest consecutive 1s series using Bit Magic:
Below is the idea to solve the problem:
The idea is based on the concept that the AND of bit sequence with a left shifted by 1 version of itself effectively removes the trailing 1 from every sequence of consecutive 1s.
So the operation N = (N & (N << 1)) reduces length of every sequence of 1s by one in binary representation of N. If we keep doing this operation in a loop, we end up with N = 0. The number of iterations required to reach 0 is actually length of the longest consecutive sequence of 1s.
& 11011110 (x << 1)
11001110 (x & (x << 1))
Trailing 1 removed
Follow the below steps to implement the above approach:
- Create a variable count initialized with value 0.
- Run a while loop till N is not 0.
- In each iteration perform the operation N = (N & (N << 1))
- Increment count by one.
- Return count
Below is the Implementation of above approach:
Time Complexity: O(log X), Here X is the length of binary representation of N
Auxiliary Space: O(1)
Approach 2: Using String Conversion
Here’s a brief explanation of the algorithm:
We start by initializing max_len and cur_len to 0. Then, we iterate through the bits of the given integer n. If we encounter a 1 bit, we increment cur_len. If we encounter a 0 bit, we update max_len if cur_len is greater than max_len, and reset cur_len to 0. This is because a 0 bit breaks the current run of consecutive 1s. Finally, we return max_len.
- Initialize a variable max_len to 0 to store the length of the longest consecutive 1s found so far.
- Initialize a variable cur_len to 0 to store the length of the current consecutive 1s.
- While the integer n is not equal to 0, perform the following steps:
- If the least significant bit (LSB) of n is 1, increment cur_len.
- If the LSB of n is 0, update max_len if cur_len is greater than max_len, and reset cur_len to.
- Right shift n by 1 bit.
- If cur_len is greater than max_len, update max_len.
- Return max_len.
The length of the longest consecutive 1s in the binary representation is: 4
The time complexity of this algorithm is O(log n), where n is the given integer, since we iterate through the bits of the integer. The space complexity is O(1), since we only use constant extra space to store the variables max_len and cur_len.
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