228. Summary Ranges 📊

Difficulty: Easy - Tags: Array

LeetCode Problem Link

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Problem Statement 📜

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

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Examples 🌟

🔹 Example 1:

Input:

nums = [0,1,2,4,5,7]

Output:

["0->2","4->5","7"]

Explanation:

• [0,2] --> "0->2"

• [4,5] --> "4->5"

• [7,7] --> "7"

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🔹 Example 2:

Input:

nums = [0,2,3,4,6,8,9]

Output:

["0","2->4","6","8->9"]

Explanation:

• [0,0] --> "0"

• [2,4] --> "2->4"

• [6,6] --> "6"

• [8,9] --> "8->9"

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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.

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Solution 💡

The goal is to identify consecutive ranges in the sorted array and format them as described.

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Java Solution

import java.util.ArrayList;
import java.util.List;

class Solution {
    public List<String> summaryRanges(int[] nums) {
        List<String> result = new ArrayList<>();
        if (nums.length == 0) {
            return result;
        }

        int start = nums[0];

        for (int i = 1; i <= nums.length; i++) {
            // Check for end of range or end of array
            if (i == nums.length || nums[i] != nums[i - 1] + 1) {
                if (start == nums[i - 1]) {
                    result.add(String.valueOf(start)); // Single number
                } else {
                    result.add(start + "->" + nums[i - 1]); // Range
                }
                if (i < nums.length) {
                    start = nums[i];
                }
            }
        }

        return result;
    }
}

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Explanation of the Solution

1. Initialize Start Point:

   • Use a start variable to track the beginning of the current range.

2. Iterate Over the Array:

   • Compare the current number with the previous one to detect the end of a range.

   • When a range ends or the array ends:

     • If the start is equal to the last number, it's a single element range.

     • Otherwise, format it as "start->end".

3. Update for the Next Range:

   • Reset start to the next number in the array.

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Time Complexity ⏳

• O(n): Each element is processed once.

Space Complexity 💾

• O(1): No additional space is used other than the result list.

You can find the full solution here.