Summary Ranges

You are given a sorted unique integer array nums.

A range [a,b] is the set of all integers from a to b (inclusive).

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

LeetCode Problem - 228

class Solution {
    public List<String> summaryRanges(int[] nums) {
        int n = nums.length;  // Get the length of the input array
        List<String> ans = new ArrayList<>();  // List to store the summary ranges
        if (n == 0) return ans;  // Return an empty list if the input is empty

        int start = 0;  // Initialize the start index of the current range

        // Loop through the array, using 'end' to explore potential ranges
        for (int end = 1; end <= n; end++) {
            // Check if we've reached the end of the array or if the current number
            // does not continue the consecutive range
            if (end == n || nums[end] != nums[end - 1] + 1) {
                // If the start and end indices are the same, it's a single number
                if (start == end - 1) {
                    ans.add(String.valueOf(nums[start]));  // Add the single number to the result
                } else {
                    // If there's a range, format it as "start->end"
                    ans.add(nums[start] + "->" + nums[end - 1]);
                }
                start = end;  // Move start to the current end index for the next range
            }
        }

        return ans;  // Return the list of summary ranges
    }
}