LeetCode #2580 — MEDIUM

Count Ways to Group Overlapping Ranges

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given a 2D integer array ranges where ranges[i] = [start_i, end_i] denotes that all integers between start_i and end_i (both inclusive) are contained in the i^th 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 <= start_i <= end_i <= 10⁹

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[[6,10],[5,15]]

Example 2

[[1,3],[10,20],[2,5],[4,8]]

Core Insight

What unlocks the optimal approach

Can we use sorting here?
Sort the ranges and merge the overlapping ranges. Then count number of non-overlapping ranges.
How many ways can we group these non-overlapping ranges?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2580: Count Ways to Group Overlapping Ranges
class Solution {
    public int countWays(int[][] ranges) {
        Arrays.sort(ranges, (a, b) -> a[0] - b[0]);
        int cnt = 0, mx = -1;
        for (int[] e : ranges) {
            if (e[0] > mx) {
                ++cnt;
            }
            mx = Math.max(mx, e[1]);
        }
        return qpow(2, cnt, (int) 1e9 + 7);
    }

    private int qpow(long a, int n, int mod) {
        long ans = 1;
        for (; n > 0; n >>= 1) {
            if ((n & 1) == 1) {
                ans = ans * a % mod;
            }
            a = a * a % mod;
        }
        return (int) ans;
    }
}

// Accepted solution for LeetCode #2580: Count Ways to Group Overlapping Ranges
func countWays(ranges [][]int) int {
	sort.Slice(ranges, func(i, j int) bool { return ranges[i][0] < ranges[j][0] })
	cnt, mx := 0, -1
	for _, e := range ranges {
		if e[0] > mx {
			cnt++
		}
		if mx < e[1] {
			mx = e[1]
		}
	}
	qpow := func(a, n, mod int) int {
		ans := 1
		for ; n > 0; n >>= 1 {
			if n&1 == 1 {
				ans = ans * a % mod
			}
			a = a * a % mod
		}
		return ans
	}
	return qpow(2, cnt, 1e9+7)
}

# Accepted solution for LeetCode #2580: Count Ways to Group Overlapping Ranges
class Solution:
    def countWays(self, ranges: List[List[int]]) -> int:
        ranges.sort()
        cnt, mx = 0, -1
        for start, end in ranges:
            if start > mx:
                cnt += 1
            mx = max(mx, end)
        mod = 10**9 + 7
        return pow(2, cnt, mod)

// Accepted solution for LeetCode #2580: Count Ways to Group Overlapping Ranges
/**
 * [2580] Count Ways to Group Overlapping Ranges
 */
pub struct Solution {}

// submission codes start here
struct Range {
    start: i32,
    end: i32,
}

impl Solution {
    pub fn count_ways(ranges: Vec<Vec<i32>>) -> i32 {
        let mut ranges: Vec<Range> = ranges
            .iter()
            .map(|a| {
                return Range {
                    start: a[0],
                    end: a[1],
                };
            })
            .collect();

        ranges.sort_unstable_by_key(|x| x.start);

        let mut merged_ranges = Vec::with_capacity(ranges.len());

        let mut i = 0;
        let mut last: Option<&mut Range> = None;

        while i < ranges.len() {
            if let Some(last_range) = last {
                while i < ranges.len() && last_range.end >= ranges[i].start {
                    last_range.end = last_range.end.max(ranges[i].end);
                    i += 1;
                }

                if i >= ranges.len() {
                    break;
                }
            }

            merged_ranges.push(Range {
                start: ranges[i].start,
                end: ranges[i].end,
            });
            last = merged_ranges.last_mut();
            i += 1;
        }

        let mut result = 1;

        for _ in 0..merged_ranges.len() as i32 {
            result = result * 2 % 1_000_000_007;
        }

        result
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_2580() {
        assert_eq!(2, Solution::count_ways(vec![vec![6, 10], vec![5, 15]]));
        assert_eq!(
            4,
            Solution::count_ways(vec![vec![1, 3], vec![10, 20], vec![2, 5], vec![4, 8]])
        );
    }
}

// Accepted solution for LeetCode #2580: Count Ways to Group Overlapping Ranges
function countWays(ranges: number[][]): number {
    ranges.sort((a, b) => a[0] - b[0]);
    let mx = -1;
    let ans = 1;
    const mod = 10 ** 9 + 7;
    for (const [start, end] of ranges) {
        if (start > mx) {
            ans = (ans * 2) % mod;
        }
        mx = Math.max(mx, end);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(log n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.