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163. Missing Ranges

Description

You are given an inclusive range [lower, upper] and a sorted unique integer array nums, where all elements are within 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 shortest sorted list of ranges that exactly covers all the missing numbers. That is, no element of nums is included in any of the ranges, and each missing number is covered by one of the ranges.

 

 

Example 1:

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

Example 2:

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

 

Constraints:

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

Solutions

Solution 1: Simulation

We can simulate the problem directly according to the requirements.

The time complexity is $O(n)$, where $n$ is the length of the array $nums$. Ignoring the space consumption of the answer, the space complexity is $O(1)$.

  • class Solution {
    public List<List<Integer>> findMissingRanges(int[] nums, int lower, int upper) {
        int n = nums.length;
        if (n == 0) {
            return List.of(List.of(lower, upper));
        }
        List<List<Integer>> ans = new ArrayList<>();
        if (nums[0] > lower) {
            ans.add(List.of(lower, nums[0] - 1));
        }
        for (int i = 1; i < n; ++i) {
            if (nums[i] - nums[i - 1] > 1) {
                ans.add(List.of(nums[i - 1] + 1, nums[i] - 1));
            }
        }
        if (nums[n - 1] < upper) {
            ans.add(List.of(nums[n - 1] + 1, upper));
        }
        return ans;
    }
}

  • class Solution {
public:
    vector<vector<int>> findMissingRanges(vector<int>& nums, int lower, int upper) {
        int n = nums.size();
        if (n == 0) {
            return { {lower, upper} };
        }
        vector<vector<int>> ans;
        if (nums[0] > lower) {
            ans.push_back({lower, nums[0] - 1});
        }
        for (int i = 1; i < nums.size(); ++i) {
            if (nums[i] - nums[i - 1] > 1) {
                ans.push_back({nums[i - 1] + 1, nums[i] - 1});
            }
        }
        if (nums[n - 1] < upper) {
            ans.push_back({nums[n - 1] + 1, upper});
        }
        return ans;
    }
};

  • class Solution:
    def findMissingRanges(
        self, nums: List[int], lower: int, upper: int
    ) -> List[List[int]]:
        n = len(nums)
        if n == 0:
            return [[lower, upper]]
        ans = []
        if nums[0] > lower:
            ans.append([lower, nums[0] - 1])
        for a, b in pairwise(nums):
            if b - a > 1:
                ans.append([a + 1, b - 1])
        if nums[-1] < upper:
            ans.append([nums[-1] + 1, upper])
        return ans


  • func findMissingRanges(nums []int, lower int, upper int) (ans [][]int) {
	n := len(nums)
	if n == 0 {
		return [][]int{ {lower, upper} }
	}
	if nums[0] > lower {
		ans = append(ans, []int{lower, nums[0] - 1})
	}
	for i, b := range nums[1:] {
		if a := nums[i]; b-a > 1 {
			ans = append(ans, []int{a + 1, b - 1})
		}
	}
	if nums[n-1] < upper {
		ans = append(ans, []int{nums[n-1] + 1, upper})
	}
	return
}

  • function findMissingRanges(nums: number[], lower: number, upper: number): number[][] {
    const n = nums.length;
    if (n === 0) {
        return [[lower, upper]];
    }
    const ans: number[][] = [];
    if (nums[0] > lower) {
        ans.push([lower, nums[0] - 1]);
    }
    for (let i = 1; i < n; ++i) {
        if (nums[i] - nums[i - 1] > 1) {
            ans.push([nums[i - 1] + 1, nums[i] - 1]);
        }
    }
    if (nums[n - 1] < upper) {
        ans.push([nums[n - 1] + 1, upper]);
    }
    return ans;
}

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