163. Missing Ranges

163. Missing Ranges (Easy)

You are given an inclusive range [lower, upper] and a sorted unique integer array nums, where all elements are in the inclusive range.

A number x is considered missing if x is in the range [lower, upper] and x is not in nums.

Return the smallest sorted list of ranges that cover every missing number exactly. That is, no element of nums is in any of the ranges, and each missing number is in one of the ranges.

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

• "a->b" if a != b
• "a" if a == b

 

Example 1:

Input: nums = [0,1,3,50,75], lower = 0, upper = 99
Output: ["2","4->49","51->74","76->99"]
Explanation: The ranges are:
[2,2] --> "2"
[4,49] --> "4->49"
[51,74] --> "51->74"
[76,99] --> "76->99"

Example 2:

Input: nums = [], lower = 1, upper = 1
Output: ["1"]
Explanation: The only missing range is [1,1], which becomes "1".

Example 3:

Input: nums = [], lower = -3, upper = -1
Output: ["-3->-1"]
Explanation: The only missing range is [-3,-1], which becomes "-3->-1".

Example 4:

Input: nums = [-1], lower = -1, upper = -1
Output: []
Explanation: There are no missing ranges since there are no missing numbers.

Example 5:

Input: nums = [-1], lower = -2, upper = -1
Output: ["-2"]

 

Constraints:

• -109 <= lower <= upper <= 109
• 0 <= nums.length <= 100
• lower <= nums[i] <= upper
• All the values of nums are unique.

Companies:
Facebook, ByteDance

Related Topics:
Array

Similar Questions:

• Summary Ranges (Easy)

Solution 1.

// OJ: https://leetcode.com/problems/missing-ranges/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
    string formatRange(int lower, int upper) {
        if (lower == upper) return to_string(lower);
        return to_string(lower) + "->" + to_string(upper);
    }
public:
    vector<string> findMissingRanges(vector<int>& A, int lower, int upper) {
        vector<string> ans;
        int i = lower, j = 0, N = A.size();
        while (j < N) {
            if (i == A[j]) {
                ++i;
                ++j;
            } else {
                int first = i, last = A[j] - 1;
                ans.push_back(formatRange(first, last));
                i = A[j++] + 1;
            }
        }
        if (i <= upper) ans.push_back(formatRange(i, upper));
        return ans;
    }
};

Or

// OJ: https://leetcode.com/problems/missing-ranges/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
    string formatRange(int lower, int upper) {
        if (lower == upper) return to_string(lower);
        return to_string(lower) + "->" + to_string(upper);
    }
public:
    vector<string> findMissingRanges(vector<int>& A, int lower, int upper) {
        vector<string> ans;
        int prev = lower - 1, N = A.size();
        for (int i = 0; i <= N; ++i) {
            int cur = i < N ? A[i] : upper + 1;
            if (prev + 1 <= cur - 1) ans.push_back(formatRange(prev + 1, cur - 1));
            prev = cur;
        }
        return ans;
    }
};