# Simple Java O(N) solution

• ``````public class Solution {

//if both start and end are same, then we just need one entry, else get the range
public String getRangeString(int start, int end){
return (start==end)?String.valueOf(start):(start+"->"+end);
}

public List<String> findMissingRanges(int[] nums, int lower, int upper) {
List<String> list = new ArrayList<>();
//if input array is empty, then just generate the range between lower and upper
if(nums.length==0){
list.add(getRangeString(lower, upper));
return list;
}

//if the first element and the lower vary by more than 0, then it means that the input array starts
//atleast by +1 of lower, so we need to at the minimum add the lower as an entry for output. So we
//generate the range of lower, first_element-1
if(nums[0]-lower>0){
list.add(getRangeString(lower, nums[0]-1));
//set the lower as first element
lower = nums[0];
}

//now just loop over the array finding the range if the difference is >=2. If the difference is just 1 then it
//means that the two elements are contiguous, so no need to worry. If its 2 then it means we atleast miss one element,
//so we find the range of lower+1, current_element-1
for(int i=1;i<nums.length;i++){
if(nums[i]-lower >=2){
list.add(getRangeString(lower+1, nums[i]-1));
}
lower = nums[i];
}

//for the last element, we need to get the range between last_element+1 and the upper
if(lower+1 <= upper){
list.add(getRangeString(lower+1, upper));
}

//return the list
return list;
}
``````

}

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