package dcp.day199
// day199.kt
// By <NAME>.

fun closestString(elems: String): String {
    return closestString(elems.toList())
}

fun closestString(elems: List<Char>): String {
    /**
     * Goal: given a string comprising parentheses, find a string with balanced parentheses to the original string.
     * We can add or remove parentheses: each operation is considered a move, and a string s1 is closer to s2 than s3
     * if the number of moves you need to transform s1 to s2 is less than the number needed to transfer from s2 to s3.
     *
     * Note that since adding and removing parentheses have the same value, it doesn't matter which one we do: we can
     * focus on one and ignore the other. Here, we focus on adding ) to balance strings. We could remove ( and have the
     * same effect, albeit with shorter strings.
     *
     * This results in a greedy, functional solution with complexity and space O(n).
     */

    // We only care about parentheses.
    // Our zero value is a pair of the number of open parentheses and the closest string thus far.
    val (openParens, closestString) = elems
        .filter { it in setOf('(', ')')}
        .fold(0 to ""){ (openParens, closestString), p ->
            when (p) {
                /*** (: just devour the ( and move on ***/
                '(' -> (openParens + 1 to "$closestString(")

                /*** ) ***/
                else ->
                    when (openParens) {
                        0 -> (openParens to "$closestString()")
                        else -> (openParens - 1 to "$closestString)")
                    }
            }
        }

    // Now we have to close all the remaining open parentheses.
    return closestString + ")".repeat(openParens)
}

/**
 * Check to see if a string is balanced.
 */
fun String.isParenthesesBalanced(str: String): Boolean {
    var balance = 0
    for (s in str) {
        if (s == '(')
            ++balance
        else if (s == ')') {
            if (balance > 0)
                --balance
            else
                return false
        }
    }
    return true
}

package dcp.day199
// day199.kt
// By Sebastian Raaphorst.

fun closestString(elems: String): String {
    return closestString(elems.toList())
}

fun closestString(elems: List<Char>): String {
    /**
     * Goal: given a string comprising parentheses, find a string with balanced parentheses to the original string.
     * We can add or remove parentheses: each operation is considered a move, and a string s1 is closer to s2 than s3
     * if the number of moves you need to transform s1 to s2 is less than the number needed to transfer from s2 to s3.
     *
     * Note that since adding and removing parentheses have the same value, it doesn't matter which one we do: we can
     * focus on one and ignore the other. Here, we focus on adding ) to balance strings. We could remove ( and have the
     * same effect, albeit with shorter strings.
     *
     * This results in a greedy, functional solution with complexity and space O(n).
     */

    // We only care about parentheses.
    // Our zero value is a pair of the number of open parentheses and the closest string thus far.
    val (openParens, closestString) = elems
        .filter { it in setOf('(', ')')}
        .fold(0 to ""){ (openParens, closestString), p ->
            when (p) {
                /*** (: just devour the ( and move on ***/
                '(' -> (openParens + 1 to "$closestString(")

                /*** ) ***/
                else ->
                    when (openParens) {
                        0 -> (openParens to "$closestString()")
                        else -> (openParens - 1 to "$closestString)")
                    }
            }
        }

    // Now we have to close all the remaining open parentheses.
    return closestString + ")".repeat(openParens)
}

/**
 * Check to see if a string is balanced.
 */
fun String.isParenthesesBalanced(str: String): Boolean {
    var balance = 0
    for (s in str) {
        if (s == '(')
            ++balance
        else if (s == ')') {
            if (balance > 0)
                --balance
            else
                return false
        }
    }
    return true
}
