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 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:

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

- Current value of max-length
- Current-Length + Previous-Length .

**result = return max-length+1**(// add 1 for flip bit count )

.

Below is C++ implementation of above idea.

// 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 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; }

Output :

4 8 5

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