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:

- If the next bit is a 1, the previous Length should be set to the current Length.
- 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:

- The current value of max-length
- 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<bits/stdc++.h>` `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 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 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 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 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` ` ` `# 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 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 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

`<?php` `// PHP program to find maximum consecutive` `// 1's in binary representation of a number` `// after flipping one bit.` `function` `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` `++;` ` ` `// 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` ` ` `// 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

`<script>` ` ` `// Javascript program to` ` ` `// find maximum consecutive` ` ` `// 1's in binary representation` ` ` `// of a number` ` ` `// after flipping one bit.` ` ` ` ` `function` `flipBit(a)` ` ` `{` ` ` `/* If all bits are l,` ` ` `binary representation` ` ` `of 'a' has all 1s */` ` ` `if` `(~a == 0)` ` ` `{` ` ` `return` `8 * sizeof(int);` ` ` `}` ` ` ` ` `let 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 fliped bit` ` ` `return` `maxLen + 1;` ` ` `}` ` ` ` ` `// input 1` ` ` `document.write(flipBit(13) + ` `"</br>"` `);` ` ` `// input 2` ` ` `document.write(flipBit(1775) + ` `"</br>"` `);` ` ` `// input 3` ` ` `document.write(flipBit(15));` ` ` `</script>` |

**Output :**

4 8 5

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