package lesson1

/**
 * A binary gap within a positive integer N is any maximal sequence of consecutive zeros that is surrounded by ones at both ends in the binary representation of N.
 *
 * For example:
 * - The number 9 has binary representation 1001 and contains a binary gap of length 2.
 * - The number 529 has binary representation 1000010001 and contains two binary gaps: one of length 4 and one of length 3.
 * - The number 20 has binary representation 10100 and contains one binary gap of length 1.
 * - The number 15 has binary representation 1111 and has no binary gaps.
 * - The number 32 has binary representation 100000 and has no binary gaps.
 *
 * Write a function:
 *
 *     class Solution { public int solution(int N); }
 *
 * that, given a positive integer N, returns the length of its longest binary gap.
 * The function should return 0 if N doesn't contain a binary gap.
 *
 * For example, given N = 1041 the function should return 5, because N has binary representation 10000010001 and so its longest binary gap is of length 5.
 * Given N = 32 the function should return 0, because N has binary representation '100000' and thus no binary gaps.
 *
 * Write an efficient algorithm for the following assumptions:
 *
 * N is an integer within the range [1..2,147,483,647].
 */
class BinaryGap {

    fun solution(n: Int): Int {
        var longestBinaryGap = 0
        var currentBinaryGap = 0
        var isCounting = false

        // Get binary representation
        val binaryString = Integer.toBinaryString(n)

        for (char in binaryString) {
            when {
                // If we were already counting means binary gap ended
                char == '1' && isCounting -> {
                    if (currentBinaryGap > longestBinaryGap) {
                        longestBinaryGap = currentBinaryGap
                    }
                    currentBinaryGap = 0
                }

                // When finding a 1 start counting
                char == '1' -> isCounting = true

                // When counting -> increment our binary gap
                char == '0' && isCounting -> currentBinaryGap++
            }
        }

        return longestBinaryGap
    }

}


package lesson1

/**
 * A binary gap within a positive integer N is any maximal sequence of consecutive zeros that is surrounded by ones at both ends in the binary representation of N.
 *
 * For example:
 * - The number 9 has binary representation 1001 and contains a binary gap of length 2.
 * - The number 529 has binary representation 1000010001 and contains two binary gaps: one of length 4 and one of length 3.
 * - The number 20 has binary representation 10100 and contains one binary gap of length 1.
 * - The number 15 has binary representation 1111 and has no binary gaps.
 * - The number 32 has binary representation 100000 and has no binary gaps.
 *
 * Write a function:
 *
 *     class Solution { public int solution(int N); }
 *
 * that, given a positive integer N, returns the length of its longest binary gap.
 * The function should return 0 if N doesn't contain a binary gap.
 *
 * For example, given N = 1041 the function should return 5, because N has binary representation 10000010001 and so its longest binary gap is of length 5.
 * Given N = 32 the function should return 0, because N has binary representation '100000' and thus no binary gaps.
 *
 * Write an efficient algorithm for the following assumptions:
 *
 * N is an integer within the range [1..2,147,483,647].
 */
class BinaryGap {

    fun solution(n: Int): Int {
        var longestBinaryGap = 0
        var currentBinaryGap = 0
        var isCounting = false

        // Get binary representation
        val binaryString = Integer.toBinaryString(n)

        for (char in binaryString) {
            when {
                // If we were already counting means binary gap ended
                char == '1' && isCounting -> {
                    if (currentBinaryGap > longestBinaryGap) {
                        longestBinaryGap = currentBinaryGap
                    }
                    currentBinaryGap = 0
                }

                // When finding a 1 start counting
                char == '1' -> isCounting = true

                // When counting -> increment our binary gap
                char == '0' && isCounting -> currentBinaryGap++
            }
        }

        return longestBinaryGap
    }

}

