163. Missing Ranges

Description

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

Constraints

• -109 <= lower <= upper <= 109

• 0 <= nums.length <= 100

• lower <= nums[i] <= upper

• All the values of nums are unique.

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

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"]

Solutions

Solution 1

/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    public List<String> findMissingRanges(int[] nums, int lower, int upper) {
        List<String> result = new ArrayList();
        for(int i = 0; i < nums.length; i++) {
            if(lower == nums[i]) {
                lower++;
            } else {
                addRange(lower, (nums[i]-1), result);
                lower = nums[i]+1;
            }
        }
        if(lower <= upper) {
            addRange(lower, upper, result);
        }
        return result;
    }
    
    private void addRange(int low, int high, List<String> result) {
        if(low == high) {
            result.add(String.valueOf(low));
        } else {
            result.add(low  + "->" + high);
        }
    }
}

Solution 2

/**
 * Time complexity : O(N), where N is the length of the input array. 
 *    This is because we are only iterating over the array once.
 * Space complexity : O(N), where N is the length of the input array. 
 *    This is because we could have a missing range between each of the 
 *    consecutive element of the input array. Hence, our output list that 
 *    we need to return will be of size N.
 */
 
 class Solution {
    public List<String> findMissingRanges(int[] nums, int lower, int upper) {
        List<String> result = new ArrayList();
        int n = nums.length;
        
        if(n == 0) {
            addRange(lower, upper, result);
            return result;
        }
        
        if(nums[0] > lower) {
            addRange(lower, nums[0]-1, result);
        }
        
        for(int i = 1; i < n; i++) {
            if(nums[i]-nums[i-1] > 1) {
                addRange(nums[i-1]+1, nums[i]-1, result);
            }
        }
        
        if(nums[n-1] < upper) {
            addRange(nums[n-1]+1, upper, result);
        }
        
        return result;
    }
    
    private void addRange(int low, int high, List<String> result) {
        if(low == high) {
            result.add(String.valueOf(low));
        } else {
            result.add(low  + "->" + high);
        }
    }
}

Follow up