352. Data Stream as Disjoint Intervals #
Problem #
Given a data stream input of non-negative integers a1, a2, …, an, summarize the numbers seen so far as a list of disjoint intervals.
Implement the SummaryRanges class:
- SummaryRanges() Initializes the object with an empty stream.
- void addNum(int val) Adds the integer val to the stream.
- int[][] getIntervals() Returns a summary of the integers in the stream currently as a list of disjoint intervals [starti, endi].
Example 1:
Input
["SummaryRanges", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals"]
[[], [1], [], [3], [], [7], [], [2], [], [6], []]
Output
[null, null, [[1, 1]], null, [[1, 1], [3, 3]], null, [[1, 1], [3, 3], [7, 7]], null, [[1, 3], [7, 7]], null, [[1, 3], [6, 7]]]
Explanation
SummaryRanges summaryRanges = new SummaryRanges();
summaryRanges.addNum(1); // arr = [1]
summaryRanges.getIntervals(); // return [[1, 1]]
summaryRanges.addNum(3); // arr = [1, 3]
summaryRanges.getIntervals(); // return [[1, 1], [3, 3]]
summaryRanges.addNum(7); // arr = [1, 3, 7]
summaryRanges.getIntervals(); // return [[1, 1], [3, 3], [7, 7]]
summaryRanges.addNum(2); // arr = [1, 2, 3, 7]
summaryRanges.getIntervals(); // return [[1, 3], [7, 7]]
summaryRanges.addNum(6); // arr = [1, 2, 3, 6, 7]
summaryRanges.getIntervals(); // return [[1, 3], [6, 7]]
Constraints
- 0 <= val <= 10000
- At most 3 * 10000 calls will be made to addNum and getIntervals.
Problem Summary #
Given a data stream input consisting of non-negative integers a1, a2, …, an, summarize the numbers seen so far as a list of disjoint intervals.
Implement the SummaryRanges class:
- SummaryRanges() Initializes the object with an empty data stream.
- void addNum(int val) Adds the integer val to the data stream.
- int[][] getIntervals() Returns a summary of the integers in the data stream as a list of disjoint intervals [starti, endi].
Solution Approach #
- Use a dictionary to filter out duplicate numbers
- Put the filtered numbers into nums and sort them
- Use nums to construct non-overlapping intervals
Code #
package leetcode
import "sort"
type SummaryRanges struct {
nums []int
mp map[int]int
}
func Constructor() SummaryRanges {
return SummaryRanges{
nums: []int{},
mp : map[int]int{},
}
}
func (this *SummaryRanges) AddNum(val int) {
if _, ok := this.mp[val]; !ok {
this.mp[val] = 1
this.nums = append(this.nums, val)
}
sort.Ints(this.nums)
}
func (this *SummaryRanges) GetIntervals() [][]int {
n := len(this.nums)
var ans [][]int
if n == 0 {
return ans
}
if n == 1 {
ans = append(ans, []int{this.nums[0], this.nums[0]})
return ans
}
start, end := this.nums[0], this.nums[0]
ans = append(ans, []int{start, end})
index := 0
for i := 1; i < n; i++ {
if this.nums[i] == end + 1 {
end = this.nums[i]
ans[index][1] = end
} else {
start, end = this.nums[i], this.nums[i]
ans = append(ans, []int{start, end})
index++
}
}
return ans
}