Summary Ranges

LeetCode 228 | Difficulty: Easy

Easy

Problem Description

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`

- `-2^31 <= nums[i] <= 2^31 - 1`

- All the values of `nums` are **unique**.

- `nums` is sorted in ascending order.

Topics: Array

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Approach

Direct Approach

This problem can typically be solved with straightforward iteration or simple data structure usage. Focus on correctness first, then optimize.

When to use

Basic problems that test fundamental programming skills.

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Solutions

Solution 1: C# (Best: 136 ms)

Metric	Value
Runtime	136 ms
Memory	41.4 MB
Date	2022-01-11

Solution

public class Solution {
    public IList<string> SummaryRanges(int[] nums) {
        List<string> result = new List<string>();
    int i=0;
    while (i < nums.Length)
    {
        int j = i + 1;
        while (j < nums.Length && nums[j] == nums[j - 1] + 1)
        {
            j++;
        }

        if (j - 1 == i) result.Add($"{nums[i]}");
        else result.Add($"{nums[i]}->{nums[j-1]}");
        i=j;        
    }
    
    return result;
    }
}

📜 1 more C# submission(s)

Submission (2022-02-28) — 226 ms, 41.9 MB

public class Solution {
    public IList<string> SummaryRanges(int[] nums) {
        List<string> result = new List<string>();
    int i=0;
    while (i < nums.Length)
    {
        int j = i + 1;
        while (j < nums.Length && nums[j] == nums[j - 1] + 1)
        {
            j++;
        }

        if (j - 1 == i) result.Add($"{nums[i]}");
        else result.Add($"{nums[i]}->{nums[j-1]}");
        i=j;        
    }
    
    return result;
    }
}

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Complexity Analysis

Approach	Time	Space
Solution	$O(n)$	$O(1) to O(n)$

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Interview Tips

Key Points

• Start by clarifying edge cases: empty input, single element, all duplicates.