Leetcode problem 129 - 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

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”

Constraints:

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

/**
 * @param {number[]} nums
 * @return {string[]}
 */
var summaryRanges = function (nums) {
  const result = [];
  let start = nums[0];

  for (let i = 1; i <= nums.length; i++) {
    if (nums[i - 1] + 1 === nums[i]) continue;

    if (start === nums[i - 1]) {
      result.push(`${start}`);
    } else {
      result.push(`${start}->${nums[i - 1]}`);
    }
    start = nums[i];
  }

  return result;
};

The time complexity of the provided code is O(n), where n is the number of elements in the input array nums. This is because the code iterates through the array once, and each iteration involves constant-time operations. Thus, the time complexity grows linearly with the size of the input array.

The space complexity of the code is also O(n). This is because the result array can potentially hold up to n elements, where each element represents a summary range. Additionally, the variables start and i require constant space, independent of the input size. Therefore, the space complexity is linearly proportional to the size of the input array.