# Lucky alive person in a circle | Set – 2

Given that N person (numbered 1 to N) standing as to form a circle. They all have the gun in their hand which is pointed to their leftmost Partner.
Every one shoots such that 1 shoot 2, 3 shoots 4, 5 shoots 6 …. (N-1)the shoot N (if N is even otherwise N shoots 1).
Again on the second iteration, they shoot the rest of remains as above mentioned logic (now for n as even, 1 will shoot to 3, 5 will shoot to 7 and so on).

The task is to find which person is the luckiest(didn’t die)?

Examples:

Input: N = 3
Output: 3
As N = 3 then 1 will shoot 2, 3 will shoot 1 hence 3 is the luckiest person.

Input: N = 8
Output: 1
Here as N = 8, 1 will shoot 1, 3 will shoot 4, 5 will shoot 6, 7 will shoot 8, Again 1 will shoot 3, 5 will shoot 7, Again 1 will shoot 5 and hence 1 is the luckiest person.

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

This problem has already been discussed in Lucky alive person in a circle | Code Solution to sword puzzle. In this post, a different approach is discussed.

Approach:

1. Take the Binary Equivalent of N.
2. Find its 1’s compliment and convert its equal decimal number N`.
3. find |N – N`|.

Below is the implementation of the above approach:

 `// C++ implementation of the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to convert string to number ` `int` `stringToNum(string s) ` `{ ` ` `  `    ``// object from the class stringstream ` `    ``stringstream geek(s); ` ` `  `    ``// The object has the value 12345 and stream ` `    ``// it to the integer x ` `    ``int` `x = 0; ` `    ``geek >> x; ` ` `  `    ``return` `x; ` `} ` ` `  `// Function to convert binary to decimal ` `int` `binaryToDecimal(string n) ` `{ ` `    ``int` `num = stringToNum(n); ` `    ``int` `dec_value = 0; ` ` `  `    ``// Initializing base value to 1, i.e 2^0 ` `    ``int` `base = 1; ` ` `  `    ``int` `temp = num; ` `    ``while` `(temp) { ` `        ``int` `last_digit = temp % 10; ` `        ``temp = temp / 10; ` ` `  `        ``dec_value += last_digit * base; ` ` `  `        ``base = base * 2; ` `    ``} ` ` `  `    ``return` `dec_value; ` `} ` ` `  `string itoa(``int` `num, string str, ``int` `base) ` `{ ` `    ``int` `i = 0; ` `    ``bool` `isNegative = ``false``; ` ` `  `    ``/* Handle 0 explicitly, otherwise  ` `    ``empty string is printed for 0 */` `    ``if` `(num == 0) { ` `        ``str[i++] = ``'0'``; ` `        ``return` `str; ` `    ``} ` ` `  `    ``// In standard itoa(), negative numbers ` `    ``// are handled only with base 10. ` `    ``// Otherwise numbers are considered unsigned. ` `    ``if` `(num < 0 && base == 10) { ` `        ``isNegative = ``true``; ` `        ``num = -num; ` `    ``} ` ` `  `    ``// Process individual digits ` `    ``while` `(num != 0) { ` `        ``int` `rem = num % base; ` `        ``str[i++] = (rem > 9) ? (rem - 10) + ``'a'` `: rem + ``'0'``; ` `        ``num = num / base; ` `    ``} ` ` `  `    ``// If the number is negative, append '-' ` `    ``if` `(isNegative) ` `        ``str[i++] = ``'-'``; ` ` `  `    ``// Reverse the string ` `    ``reverse(str.begin(), str.end()); ` ` `  `    ``return` `str; ` `} ` ` `  `char` `flip(``char` `c) ` `{ ` `    ``return` `(c == ``'0'``) ? ``'1'` `: ``'0'``; ` `} ` ` `  `// Function to find the ones complement ` `string onesComplement(string bin) ` `{ ` `    ``int` `n = bin.length(), i; ` ` `  `    ``string ones = ``""``; ` ` `  `    ``// for ones complement flip every bit ` `    ``for` `(i = 0; i < n; i++) ` `        ``ones += flip(bin[i]); ` ` `  `    ``return` `ones; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``// Taking the number of a person ` `    ``// standing in a circle. ` `    ``int` `N = 3; ` `    ``string arr = ``""``; ` ` `  `    ``// Storing the binary equivalent in a string. ` `    ``string ans(itoa(N, arr, 2)); ` ` `  `    ``// taking one's compelement and ` `    ``// convert it to decimal value ` `    ``int` `N_dash = binaryToDecimal(onesComplement(ans)); ` ` `  `    ``int` `luckiest_person = N - N_dash; ` ` `  `    ``cout << luckiest_person; ` ` `  `    ``return` `0; ` `} `

Output:

```3
```

Alternate Shorter Implementation :

The approach used here is same.

 `// C++ implementation of the above approach ` `#include ` `using` `namespace` `std; ` ` `  `int` `luckiest_person(``int` `n) ` `{ ` `    ``// to calculate the number of bits in ` `    ``// the binary equivalent of n ` `    ``int` `len = log2(n) + 1;  ` ` `  `    ``// Finding complement by inverting the ` `    ``// bits one by one from last ` `    ``int` `n2 = n; ` `    ``for` `(``int` `i = 0; i < len; i++) { ` ` `  `        ``// XOR of n2 with (1<

Output:

```3
```

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