Accumulating with Fold

Part of the Logic & Flow section of Coddy's Kotlin journey. Lesson 34 of 63.

fold starts with an explicit initial accumulator and processes elements in order. Its lambda receives the accumulator and the next element, and returns the next accumulator. The initial value is also the result for an empty input.

Inside main:

println(listOf(2, 3, 4).fold(1) { product, value -> product * value })

The accumulator starts at one and becomes two, six and twenty-four.

fold uses its initial accumulator as the empty-input result.

Challenge

Medium

Complete sumWithFold with parameters values: List<Int>. Use fold with initial value zero to return the sum of the values.

Unless stricter bounds are stated above, collections contain at most 100 elements at each level, and integer arguments and integer collection values are between -1000 and 1000. Text supplied for parsing can include invalid or out-of-range representations as described.

Return a value of type Int. Keep the supplied input/output code. It reads scalar arguments one per line; a list starts with its count followed by its elements, and a map starts with its entry count followed by each key and value. Nested lists repeat the count-and-elements format for each row. The supplied main prints the return value followed by one newline. Lists use Kotlin's standard bracketed format; print no additional labels.

Try it yourself

fun sumWithFold(values: List<Int>): Int {
    // Write your solution here.
    return 0
}
fun main() {
    val valuesCount = readln().toInt()
    val values = mutableListOf<Int>()
    for (i0 in 0 until valuesCount) {
        val valuesItem = readln().toInt()
        values.add(valuesItem)
    }
    println(sumWithFold(values))
}