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# Find longest sequence of 1’s in binary representation with one flip

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:

1. If the next bit is a 1, the previous Length should be set to the current Length.
2. If the next bit is a 0, then we can’t merge these sequences together. So, set the previous Length to 0.

We update max length by comparing the following two:

1. The current value of max-length
2. Current-Length + Previous-Length .
• Result = return max-length+1 (// add 1 for flip bit count )

Below is the implementation of the above idea :

## C++

 `// C++ program to find maximum consecutive``// 1's in binary representation of a number``// after flipping one bit.``#include``using` `namespace` `std;` `int` `flipBit(unsigned a)``{``    ``/* If all bits are l, binary representation``       ``of 'a' has all 1s */``    ``if` `(~a == 0)``        ``return` `8*``sizeof``(``int``);` `    ``int` `currLen = 0, prevLen = 0, maxLen = 0;``    ``while` `(a!= 0)``    ``{``        ``// If Current bit is a 1 then increment currLen++``        ``if` `((a & 1) == 1)``            ``currLen++;` `        ``// If Current bit is a 0 then check next bit of a``        ``else` `if` `((a & 1) == 0)``        ``{``            ``/* Update prevLen to 0 (if next bit is 0)``            ``or currLen (if next bit is 1). */``            ``prevLen = (a & 2) == 0? 0 : currLen;` `            ``// If two consecutively bits are 0``            ``// then currLen also will be 0.``            ``currLen = 0;``        ``}` `        ``// Update maxLen if required``        ``maxLen = max(prevLen + currLen, maxLen);` `        ``// Remove last bit (Right shift)``        ``a >>= 1;``    ``}` `    ``// We can always have a sequence of``    ``// at least one 1, this is flipped bit``    ``return` `maxLen+1;``}` `// Driver code``int` `main()``{``    ``// input 1``    ``cout << flipBit(13);``    ``cout << endl;` `    ``// input 2``    ``cout << flipBit(1775);``    ``cout << endl;` `    ``// input 3``    ``cout << flipBit(15);``    ``return` `0;``}`

## Java

 `// Java program to find maximum consecutive``// 1's in binary representation of a number``// after flipping one bit.` `class` `GFG``{` `    ``static` `int` `flipBit(``int` `a)``    ``{``        ``/* If all bits are l, binary representation``        ``of 'a' has all 1s */``        ``if` `(~a == ``0``)``        ``{``            ``return` `8` `* sizeof();``        ``}` `        ``int` `currLen = ``0``, prevLen = ``0``, maxLen = ``0``;``        ``while` `(a != ``0``)``        ``{``            ``// If Current bit is a 1``            ``// then increment currLen++``            ``if` `((a & ``1``) == ``1``)``            ``{``                ``currLen++;``            ``}``            ` `            ``// If Current bit is a 0 then``            ``// check next bit of a``            ``else` `if` `((a & ``1``) == ``0``)``            ``{``                ``/* Update prevLen to 0 (if next bit is 0)``                ``or currLen (if next bit is 1). */``                ``prevLen = (a & ``2``) == ``0` `? ``0` `: currLen;` `                ``// If two consecutively bits are 0``                ``// then currLen also will be 0.``                ``currLen = ``0``;``            ``}` `            ``// Update maxLen if required``            ``maxLen = Math.max(prevLen + currLen, maxLen);` `            ``// Remove last bit (Right shift)``            ``a >>= ``1``;``        ``}` `        ``// We can always have a sequence of``        ``// at least one 1, this is flipped bit``        ``return` `maxLen + ``1``;``    ``}` `    ``static` `byte` `sizeof()``    ``{``        ``byte` `sizeOfInteger = ``8``;``        ``return` `sizeOfInteger;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``// input 1``        ``System.out.println(flipBit(``13``));` `        ``// input 2``        ``System.out.println(flipBit(``1775``));` `        ``// input 3``        ``System.out.println(flipBit(``15``));``    ``}``}` `// This code is contributed by PrinciRaj1992`

## Python3

 `# Python3 program to find maximum``# consecutive 1's in binary``# representation of a number``# after flipping one bit.``def` `flipBit(a):``    ` `    ``# If all bits are l,``    ``# binary representation``    ``# of 'a' has all 1s``    ``if` `(~a ``=``=` `0``):``        ``return` `8` `*` `sizeof();` `    ``currLen ``=` `0``;``    ``prevLen ``=` `0``;``    ``maxLen ``=` `0``;``    ``while` `(a > ``0``):``        ` `        ``# If Current bit is a 1``        ``# then increment currLen++``        ``if` `((a & ``1``) ``=``=` `1``):``            ``currLen ``+``=` `1``;` `        ``# If Current bit is a 0``        ``# then check next bit of a``        ``elif` `((a & ``1``) ``=``=` `0``):``            ` `            ``# Update prevLen to 0``            ``# (if next bit is 0)``            ``# or currLen (if next``            ``# bit is 1). */``            ``prevLen ``=` `0` `if``((a & ``2``) ``=``=` `0``) ``else` `currLen;` `            ``# If two consecutively bits``            ``# are 0 then currLen also``            ``# will be 0.``            ``currLen ``=` `0``;` `        ``# Update maxLen if required``        ``maxLen ``=` `max``(prevLen ``+` `currLen, maxLen);` `        ``# Remove last bit (Right shift)``        ``a >>``=` `1``;` `    ``# We can always have a sequence``    ``# of at least one 1, this is``    ``# flipped bit``    ``return` `maxLen ``+` `1``;` `# Driver code``# input 1``print``(flipBit(``13``));` `# input 2``print``(flipBit(``1775``));` `# input 3``print``(flipBit(``15``));``    ` `# This code is contributed by mits`

## C#

 `// C# program to find maximum consecutive``// 1's in binary representation of a number``// after flipping one bit.``using` `System;` `class` `GFG``{`` ` `    ``static` `int` `flipBit(``int` `a)``    ``{``        ``/* If all bits are l, binary representation``        ``of 'a' has all 1s */``        ``if` `(~a == 0)``        ``{``            ``return` `8 * ``sizeof``(``int``);``        ``}`` ` `        ``int` `currLen = 0, prevLen = 0, maxLen = 0;``        ``while` `(a != 0)``        ``{``            ``// If Current bit is a 1``            ``// then increment currLen++``            ``if` `((a & 1) == 1)``            ``{``                ``currLen++;``            ``}``             ` `            ``// If Current bit is a 0 then``            ``// check next bit of a``            ``else` `if` `((a & 1) == 0)``            ``{``                ``/* Update prevLen to 0 (if next bit is 0)``                ``or currLen (if next bit is 1). */``                ``prevLen = (a & 2) == 0 ? 0 : currLen;`` ` `                ``// If two consecutively bits are 0``                ``// then currLen also will be 0.``                ``currLen = 0;``            ``}`` ` `            ``// Update maxLen if required``            ``maxLen = Math.Max(prevLen + currLen, maxLen);`` ` `            ``// Remove last bit (Right shift)``            ``a >>= 1;``        ``}`` ` `        ``// We can always have a sequence of``        ``// at least one 1, this is flipped bit``        ``return` `maxLen + 1;``    ``}`` ` `    ``// Driver code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``// input 1``        ``Console.WriteLine(flipBit(13));`` ` `        ``// input 2``        ``Console.WriteLine(flipBit(1775));`` ` `        ``// input 3``        ``Console.WriteLine(flipBit(15));``    ``}``}`` ` `// This code contributed by Rajput-Ji`

## PHP

 `>= 1;``    ``}` `    ``// We can always have a sequence of``    ``// at least one 1, this is flipped bit``    ``return` `\$maxLen``+1;``}` `    ``// Driver code``    ``// input 1``    ``echo` `flipBit(13);``    ``echo` `"\n"``;` `    ``// input 2``    ``echo` `flipBit(1775);``    ``echo` `"\n"``;` `    ``// input 3``    ``echo` `flipBit(15);``    ` `// This code is contributed by aj_36``?>`

## Javascript

 ``

Output

```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:

## C++

 `// C++ program to find maximum consecutive``// 1's in binary representation of a number``// after flipping one bit.``#include``using` `namespace` `std;` `int` `findMaxConsecutiveOnes(``int` `num) {``    ``int` `left = 0, right = 0, flips = 0, max_len = 0;``    ``string binary = bitset<32>(num).to_string(); ``// convert integer to binary string``    ``while` `(right < binary.size()) {``        ``if` `(binary[right] == ``'0'``) {``            ``flips++;``        ``}``        ``while` `(flips > 1) {``            ``if` `(binary[left] == ``'0'``) {``                ``flips--;``            ``}``            ``left++;``        ``}``        ``max_len = max(max_len, right - left + 1);``        ``right++;``    ``}``    ``return` `max_len;``}` `// Driver code``int` `main()``{``    ``// input 1``    ``cout << findMaxConsecutiveOnes(13);``    ``cout << endl;` `    ``// input 2``    ``cout << findMaxConsecutiveOnes(1775);``    ``cout << endl;` `    ``// input 3``    ``cout << findMaxConsecutiveOnes(15);``    ``return` `0;``}`

Output

```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. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.