LeetCode in Kotlin

228. Summary Ranges

Easy

You are given a sorted unique integer array nums.

Return the smallest sorted list of ranges that cover all the numbers in the array exactly. That is, each element of nums is covered by exactly one of the ranges, and there is no integer x such that x is in one of the ranges but not in nums.

Each range [a,b] in the list should be output as:

• "a->b" if a != b
• "a" if a == b

Example 1:

Input: nums = [0,1,2,4,5,7]

Output: [“0->2”,”4->5”,”7”]

Explanation: The ranges are: [0,2] –> “0->2” [4,5] –> “4->5” [7,7] –> “7”

Example 2:

Input: nums = [0,2,3,4,6,8,9]

Output: [“0”,”2->4”,”6”,”8->9”]

Explanation: The ranges are: [0,0] –> “0” [2,4] –> “2->4” [6,6] –> “6” [8,9] –> “8->9”

Example 3:

Input: nums = []

Output: []

Example 4:

Input: nums = [-1]

Output: [“-1”]

Example 5:

Input: nums = [0]

Output: [“0”]

Constraints:

• 0 <= nums.length <= 20
• -231 <= nums[i] <= 231 - 1
• All the values of nums are unique.
• nums is sorted in ascending order.

Solution

class Solution {
    fun summaryRanges(nums: IntArray): List<String> {
        val ranges: MutableList<String> = ArrayList()
        if (nums.size == 0) {
            return ranges
        }
        // size of array
        val n = nums.size
        // start of range
        var a = nums[0]
        // end of range
        var b = a
        val strB = StringBuilder()
        for (i in 1 until n) {
            // we need to make a decision if the next element
            // will expand the range
            // i starts at 1, not 0, because 1 is the next
            // candidate for expanding the range
            if (nums[i] != b + 1) {
                // only when our next element does not expand the range
                // do we add the range a->b to our list of ranges
                strB.append(a)
                if (a != b) {
                    strB.append("->").append(b)
                }
                ranges.add(strB.toString())
                // since nums[i] is not accounted for by our range a->b
                // because nums[i] is not b+1, we need to set a and b
                // to this new range start point of bigger than b+1
                // maybe it is b+2? b+3? b+4? all we know is it is not b+1
                a = nums[i]
                b = a
                // Reset string builder
                strB.setLength(0)
            } else {
                // if the next element expands our range we do so
                b++
            }
        }
        // the only range that is not accounted for at this point is the last range
        // if our a and b are not equal then we add the range accordingly
        strB.append(a)
        if (a != b) {
            strB.append("->").append(b)
        }
        ranges.add(strB.toString())
        return ranges
    }
}

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