# 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.

```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A simple solution is to store binary representation of given number in a binary array. Once we have elements in binary array, we can apply methods discussed here.

An efficient solution is to walk through the bits in binary representation of given number. We keep track of current 1’s sequence length and the previous 1’s sequence length. When we see a zero, update previous Length:

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

We update max length by comparing following two:

1. 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 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 increament 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 fliped 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 increament 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 fliped 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 increament 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  ` `    ``# fliped 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 increament 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 fliped 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 fliped bit ` `    ``return` `\$maxLen``+1; ` `} ` ` `  `    ``// Driver code ` `    ``// input 1 ` `    ``echo` `flipBit(13); ` `    ``echo` ```" "````; ` ` `  `    ``// input 2 ` `    ``echo` `flipBit(1775); ` `    ``echo` ```" "````; ` ` `  `    ``// input 3 ` `    ``echo` `flipBit(15); ` `     `  `// This code is contributed by aj_36 ` `?> `

Output :

```4
8
5
```

## tags:

Bit Magic binary-representation Bit Magic