LeetCode 2276. Count Integers in Intervals Solution in Java, C++, Python & More | Explanation + Code

CoderIndeed
0
2276. Count Integers in Intervals

Description

Given an empty set of intervals, implement a data structure that can:

  • Add an interval to the set of intervals.
  • Count the number of integers that are present in at least one interval.

Implement the CountIntervals class:

  • CountIntervals() Initializes the object with an empty set of intervals.
  • void add(int left, int right) Adds the interval [left, right] to the set of intervals.
  • int count() Returns the number of integers that are present in at least one interval.

Note that an interval [left, right] denotes all the integers x where left <= x <= right.

 

Example 1:

Input
["CountIntervals", "add", "add", "count", "add", "count"]
[[], [2, 3], [7, 10], [], [5, 8], []]
Output
[null, null, null, 6, null, 8]

Explanation
CountIntervals countIntervals = new CountIntervals(); // initialize the object with an empty set of intervals. 
countIntervals.add(2, 3);  // add [2, 3] to the set of intervals.
countIntervals.add(7, 10); // add [7, 10] to the set of intervals.
countIntervals.count();    // return 6
                           // the integers 2 and 3 are present in the interval [2, 3].
                           // the integers 7, 8, 9, and 10 are present in the interval [7, 10].
countIntervals.add(5, 8);  // add [5, 8] to the set of intervals.
countIntervals.count();    // return 8
                           // the integers 2 and 3 are present in the interval [2, 3].
                           // the integers 5 and 6 are present in the interval [5, 8].
                           // the integers 7 and 8 are present in the intervals [5, 8] and [7, 10].
                           // the integers 9 and 10 are present in the interval [7, 10].

 

Constraints:

  • 1 <= left <= right <= 109
  • At most 105 calls in total will be made to add and count.
  • At least one call will be made to count.

Solutions

Solution 1: Segment Tree (Dynamic Opening)

According to the problem description, we need to maintain a set of intervals that supports adding intervals and querying operations. For adding intervals, we can use a segment tree to maintain the interval set.

The segment tree divides the entire interval into multiple non-contiguous sub-intervals, with the number of sub-intervals not exceeding log(width). To update the value of an element, we only need to update log(width) intervals, and these intervals are all contained within a larger interval that includes the element. When modifying intervals, we need to use lazy propagation to ensure efficiency.

  • Each node of the segment tree represents an interval;
  • The segment tree has a unique root node representing the entire range, such as [1, N];
  • Each leaf node of the segment tree represents a unit interval of length 1, [x, x];
  • For each internal node [l, r], its left child is [l, mid] and its right child is [mid+1, r], where mid = \lfloor (l + r) / 2 \rfloor (i.e., floor division).

Since the data range in the problem is large, we can use a dynamically opened segment tree. A dynamically opened segment tree means that we only open nodes when needed, rather than opening all nodes at the beginning, which saves space.

In terms of time complexity, each operation has a time complexity of O(log n). The space complexity is O(m × log n), where m is the number of operations and n is the data range.

PythonJavaC++GoTypeScript
class Node: __slots__ = ("left", "right", "l", "r", "mid", "v", "add") def __init__(self, l, r): self.left = None self.right = None self.l = l self.r = r self.mid = (l + r) // 2 self.v = 0 self.add = 0 class SegmentTree: def __init__(self): self.root = Node(1, int(1e9) + 1) def modify(self, l, r, v, node=None): if node is None: node = self.root if l > r: return if node.l >= l and node.r <= r: node.v = node.r - node.l + 1 node.add = v return self.pushdown(node) if l <= node.mid: self.modify(l, r, v, node.left) if r > node.mid: self.modify(l, r, v, node.right) self.pushup(node) def query(self, l, r, node=None): if node is None: node = self.root if l > r: return 0 if node.l >= l and node.r <= r: return node.v self.pushdown(node) v = 0 if l <= node.mid: v += self.query(l, r, node.left) if r > node.mid: v += self.query(l, r, node.right) return v def pushup(self, node): node.v = node.left.v + node.right.v def pushdown(self, node): if node.left is None: node.left = Node(node.l, node.mid) if node.right is None: node.right = Node(node.mid + 1, node.r) if node.add != 0: left, right = node.left, node.right left.add = node.add right.add = node.add left.v = left.r - left.l + 1 right.v = right.r - right.l + 1 node.add = 0 class CountIntervals: def __init__(self): self.tree = SegmentTree() def add(self, left, right): self.tree.modify(left, right, 1) def count(self): return self.tree.query(1, int(1e9)) # Your CountIntervals object will be instantiated and called as such: # obj = CountIntervals() # obj.add(left, right) # param_2 = obj.count()(code-box)

Post a Comment

0Comments

Post a Comment (0)

#buttons=(Accept !) #days=(20)

Our website uses cookies to enhance your experience. Check Now
Accept !