Algorithm
Problem Name: 352. Data Stream as Disjoint Intervals
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 value)
Adds the integervalue
to the stream.int[][] getIntervals()
Returns a summary of the integers in the stream currently as a list of disjoint intervals[starti, endi]
. The answer should be sorted bystarti
.
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 <= value <= 104
- At most
3 * 104
calls will be made toaddNum
andgetIntervals
.
Code Examples
#1 Code Example with Javascript Programming
Code -
Javascript Programming
var SummaryRanges = function() {
this.intervals = []
}
SummaryRanges.prototype.addNum = function(val) {
const current = [val, val]
const intervals = this.intervals
const less = []
const more = []
for (let vals of intervals) {
if (vals[0] > current[1] + 1) {
more.push(vals)
} else if (vals[1] + 1 < current[0]) {
less.push(vals)
} else {
current[0] = Math.min(current[0], vals[0])
current[1] = Math.max(current[1], vals[1])
}
}
this.intervals = [...less, current, ...more]
}
SummaryRanges.prototype.getIntervals = function() {
return this.intervals
}
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#2 Code Example with Python Programming
Code -
Python Programming
class SummaryRanges:
def __init__(self):
self.starts, self.ends, self.used = [-float("inf"), float("inf")], [-float("inf"), float("inf")], set()
def addNum(self, val):
if val not in self.used:
self.used.add(val)
i = bisect.bisect_left(self.starts, val) - 1
if self.ends[i] + 1 == val and val + 1 == self.starts[i + 1]: # if val is the gap btw 2 intervals
del self.starts[i + 1]
del self.ends[i]
elif self.ends[i] + 1 == val: #if val is adjacent to current end
self.ends[i] += 1
elif self.starts[i + 1] == val + 1: # if val is adjacent to next start
self.starts[i + 1] -= 1
elif val > self.ends[i]: # if val is independent of those 2 intervals
self.starts.insert(i + 1, val)
self.ends.insert(i + 1, val)
def getIntervals(self):
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#3 Code Example with C# Programming
Code -
C# Programming
using System;
using System.Collections.Generic;
using System.Linq;
namespace LeetCode
{
public class _0352_DataStreamAsDisjointIntervals
{
private readonly SortedDictionary < int, int[]> map;
/** Initialize your data structure here. */
public _0352_DataStreamAsDisjointIntervals()
{
map = new SortedDictionary<int, int[]>();
}
public void AddNum(int val)
{
if (map.ContainsKey(val)) return;
var keys = map.Keys.ToArray();
var indexHi = Array.BinarySearch(keys, val);
indexHi = ~indexHi;
var indexLo = indexHi - 1;
if (indexLo >= 0 && indexLo < keys.Length && indexHi < keys.Length && map[keys[indexLo]][1] + 1 == val && keys[indexHi] == val + 1)
{
map[keys[indexLo]][1] = map[keys[indexHi]][1];
map.Remove(keys[indexHi]);
}
else if (indexLo >= 0 && indexLo < keys.Length && map[keys[indexLo]][1] + 1 >= val)
map[keys[indexLo]][1] = Math.Max(map[keys[indexLo]][1], val);
else if (indexHi < keys.Length && keys[indexHi] == val + 1)
{
map[val] = new int[] { val, map[keys[indexHi]][1] };
map.Remove(keys[indexHi]);
}
else
map[val] = new int[] { val, val };
}
public int[][] GetIntervals()
{
return map.Values.ToArray();
}
}
}
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