0228 - Summary Ranges (Easy)

Problem Link

https://leetcode.com/problems/summary-ranges/

Problem Statement

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"

Constraints:

• 0 <= nums.length <= 20
• -2^31 <= nums[i] <= 2^31 - 1
• All the values of nums are unique.
• nums is sorted in ascending order.

Approach 1: Iteration

We take a number as an starting point at $i$, then search for the next ending point $j$ which satisfies $nums[j] + 1 \ne nums[j + 1]$. If both pointer is not same, then we can push $nums[i]$->$nums[j]$ to our answer array, else we just push $nums[i]$.

Written by @wingkwong

class Solution {
public:
    vector<string> summaryRanges(vector<int>& nums) {
        vector<string> ans;
        int n = (int) nums.size();
        for(int i = 0; i < n; i++) {
            // take nums[i] as a in range [a, b]
            int a = nums[i];
            // search for the next ending point
            while (i != n - 1 && nums[i] + 1 == nums[i + 1]) i++;
            // take nums[i] as b in range [a, b]
            int b = nums[i];
            // output as "a" if a == b
            if(a == b) ans.push_back(to_string(a));
            // output as "a->b" if a != b
            else ans.push_back(to_string(a) + "->" + to_string(b));
        }
        return ans;
    }
};