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Give an integer n. We can flip exactly one bit. Write code to find the length of the longest sequence of 1 s you could create.
Examples:
Input : 1775
Output : 8
Binary representation of 1775 is 11011101111.
After flipping the highlighted bit, we get
consecutive 8 bits. 11011111111.
Input : 12
Output : 3
Input : 15
Output : 5
Input : 71
Output: 4
Binary representation of 71 is 1000111.
After flipping the highlighted bit, we get
consecutive 4 bits. 1001111.
A simple solution is to store the binary representation of a given number in a binary array. Once we have elements in a binary array, we can apply the methods discussed here.
An efficient solution is to walk through the bits in the binary representation of the given number. We keep track of the current 1's sequence length and the previous 1's sequence length. When we see a zero, update the previous Length:
We update max length by comparing the following two:
Below is the implementation of the above idea :
4 8 5
Time Complexity: O(log2n)
Auxiliary Space: O(1)
Approach 2: Sliding Window
In this approach, we can use a sliding window to count the length of the longest consecutive 1's. We can use two pointers to define a window and keep track of the number of flips we have made. When we encounter a 0, we can flip it and move the right pointer to the right. If the number of flips we have made is greater than 1, we can move the left pointer to the right until we have made only one flip. We can then update the maximum length obtained so far.
Here's the code:
4 8 5
Time Complexity: O(n), where n is the number of bits in the binary representation of the given number.
Space Complexity: O(1)
This article is contributed by Mr. Somesh Awasthi.