In this guide, you will get 228. Summary Ranges LeetCode Solution with the best time and space complexity. The solution to Summary Ranges problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
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:
class Solution {
public:
vector<string> summaryRanges(vector<int>& nums) {
vector<string> ans;
for (int i = 0; i < nums.size(); ++i) {
const int begin = nums[i];
while (i + 1 < nums.size() && nums[i] == nums[i + 1] - 1)
++i;
const int end = nums[i];
if (begin == end)
ans.push_back(to_string(begin));
else
ans.push_back(to_string(begin) + "->" + to_string(end));
}
return ans;
}
};
/* code provided by PROGIEZ */
228. Summary Ranges LeetCode Solution in Java
class Solution {
public List<String> summaryRanges(int[] nums) {
List<String> ans = new ArrayList<>();
for (int i = 0; i < nums.length; ++i) {
final int begin = nums[i];
while (i + 1 < nums.length && nums[i] == nums[i + 1] - 1)
++i;
final int end = nums[i];
if (begin == end)
ans.add("" + begin);
else
ans.add("" + begin + "->" + end);
}
return ans;
}
}
// code provided by PROGIEZ
228. Summary Ranges LeetCode Solution in Python
class Solution:
def summaryRanges(self, nums: list[int]) -> list[str]:
ans = []
i = 0
while i < len(nums):
begin = nums[i]
while i < len(nums) - 1 and nums[i] == nums[i + 1] - 1:
i += 1
end = nums[i]
if begin == end:
ans.append(str(begin))
else:
ans.append(str(begin) + "->" + str(end))
i += 1
return ans
# code by PROGIEZ
