2580. Count Ways to Group Overlapping Ranges LeetCode Solution

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Problem Statement
Complexity Analysis
Count Ways to Group Overlapping Ranges solution in C++
Count Ways to Group Overlapping Ranges solution in Java
Count Ways to Group Overlapping Ranges solution in Python
Additional Resources

2580. Count Ways to Group Overlapping Ranges LeetCode Solution image

Problem Statement of Count Ways to Group Overlapping Ranges

You are given a 2D integer array ranges where ranges[i] = [starti, endi] denotes that all integers between starti and endi (both inclusive) are contained in the ith range.
You are to split ranges into two (possibly empty) groups such that:

Each range belongs to exactly one group.
Any two overlapping ranges must belong to the same group.

Two ranges are said to be overlapping if there exists at least one integer that is present in both ranges.

For example, [1, 3] and [2, 5] are overlapping because 2 and 3 occur in both ranges.

Return the total number of ways to split ranges into two groups. Since the answer may be very large, return it modulo 109 + 7.

Example 1:

Input: ranges = [[6,10],[5,15]]
Output: 2
Explanation:
The two ranges are overlapping, so they must be in the same group.
Thus, there are two possible ways:
– Put both the ranges together in group 1.
– Put both the ranges together in group 2.

Example 2:

Input: ranges = [[1,3],[10,20],[2,5],[4,8]]
Output: 4
Explanation:
Ranges [1,3], and [2,5] are overlapping. So, they must be in the same group.
Again, ranges [2,5] and [4,8] are also overlapping. So, they must also be in the same group.
Thus, there are four possible ways to group them:
– All the ranges in group 1.
– All the ranges in group 2.
– Ranges [1,3], [2,5], and [4,8] in group 1 and [10,20] in group 2.
– Ranges [1,3], [2,5], and [4,8] in group 2 and [10,20] in group 1.

Constraints:

1 <= ranges.length <= 105
ranges[i].length == 2
0 <= starti <= endi <= 109

Complexity Analysis

Time Complexity: O(\texttt{sort})
Space Complexity: O(\texttt{sort})

2580. Count Ways to Group Overlapping Ranges LeetCode Solution in C++

class Solution {
 public:
  int countWays(vector<vector<int>>& ranges) {
    constexpr int kMod = 1'000'000'007;
    int ans = 1;
    int prevEnd = -1;

    ranges::sort(ranges);

    for (const vector<int>& range : ranges) {
      const int start = range[0];
      const int end = range[1];
      if (start > prevEnd)
        ans = ans * 2 % kMod;
      prevEnd = max(prevEnd, end);
    }

    return ans;
  }
};
/* code provided by PROGIEZ */

2580. Count Ways to Group Overlapping Ranges LeetCode Solution in Java

class Solution {
  public int countWays(int[][] ranges) {
    final int kMod = 1_000_000_007;
    int ans = 1;
    int prevEnd = -1;

    Arrays.sort(ranges, (a, b) -> Integer.compare(a[0], b[0]));

    for (int[] range : ranges) {
      final int start = range[0];
      final int end = range[1];
      if (start > prevEnd)
        ans = ans * 2 % kMod;
      prevEnd = Math.max(prevEnd, end);
    }

    return ans;
  }
}
// code provided by PROGIEZ

2580. Count Ways to Group Overlapping Ranges LeetCode Solution in Python

class Solution:
  def countWays(self, ranges: list[list[int]]) -> int:
    kMod = 1_000_000_007
    ans = 1
    prevEnd = -1

    for start, end in sorted(ranges):
      if start > prevEnd:
        ans = ans * 2 % kMod
      prevEnd = max(prevEnd, end)

    return ans
 # code by PROGIEZ