LeetCode #3480 — HARD

Maximize Subarrays After Removing One Conflicting Pair

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given an integer n which represents an array nums containing the numbers from 1 to n in order. Additionally, you are given a 2D array conflictingPairs, where conflictingPairs[i] = [a, b] indicates that a and b form a conflicting pair.

Remove exactly one element from conflictingPairs. Afterward, count the number of non-empty subarrays of nums which do not contain both a and b for any remaining conflicting pair [a, b].

Return the maximum number of subarrays possible after removing exactly one conflicting pair.

Example 1:

Input: n = 4, conflictingPairs = [[2,3],[1,4]]

Output: 9

Explanation:

Remove [2, 3] from conflictingPairs. Now, conflictingPairs = [[1, 4]].
There are 9 subarrays in nums where [1, 4] do not appear together. They are [1], [2], [3], [4], [1, 2], [2, 3], [3, 4], [1, 2, 3] and [2, 3, 4].
The maximum number of subarrays we can achieve after removing one element from conflictingPairs is 9.

Example 2:

Input: n = 5, conflictingPairs = [[1,2],[2,5],[3,5]]

Output: 12

Explanation:

Remove [1, 2] from conflictingPairs. Now, conflictingPairs = [[2, 5], [3, 5]].
There are 12 subarrays in nums where [2, 5] and [3, 5] do not appear together.
The maximum number of subarrays we can achieve after removing one element from conflictingPairs is 12.

Constraints:

2 <= n <= 10⁵
1 <= conflictingPairs.length <= 2 * n
conflictingPairs[i].length == 2
1 <= conflictingPairs[i][j] <= n
conflictingPairs[i][0] != conflictingPairs[i][1]

Step 01

Brute Force Baseline

Problem summary: You are given an integer n which represents an array nums containing the numbers from 1 to n in order. Additionally, you are given a 2D array conflictingPairs, where conflictingPairs[i] = [a, b] indicates that a and b form a conflicting pair. Remove exactly one element from conflictingPairs. Afterward, count the number of non-empty subarrays of nums which do not contain both a and b for any remaining conflicting pair [a, b]. Return the maximum number of subarrays possible after removing exactly one conflicting pair.

Baseline thinking

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

Pattern signal: Array · Segment Tree

Example 1

4
[[2,3],[1,4]]

Example 2

5
[[1,2],[2,5],[3,5]]

Step 02

Core Insight

What unlocks the optimal approach

Let <code>f[i]</code> (where <code>i = 1, 2, 3, ..., n</code>) be the end index of the longest valid subarray (without any conflicting pair) starting at index <code>i</code>.
The answer is: <code>sigma(f[i] - i + 1) for i in [1..n]</code>, which simplifies to: <code>sigma(f[i]) - n * (n + 1) / 2 + n</code>.
Focus on maintaining <code>f[i]</code>.
If we have a conflicting pair <code>(x, y)</code> with <code>x < y</code>: 1. Sort the conflicting pairs by <code>y</code> values in non-increasing order. 2. Update each prefix of the <code>f</code> array accordingly.
Use a segment tree or another suitable data structure to maintain the range update and sum query efficiently.

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

Largest constraint values

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 #3480: Maximize Subarrays After Removing One Conflicting Pair
class Solution {
    public long maxSubarrays(int n, int[][] conflictingPairs) {
        List<Integer>[] g = new List[n + 1];
        Arrays.setAll(g, k -> new ArrayList<>());
        for (int[] pair : conflictingPairs) {
            int a = pair[0], b = pair[1];
            if (a > b) {
                int c = a;
                a = b;
                b = c;
            }
            g[a].add(b);
        }
        long[] cnt = new long[n + 2];
        long ans = 0, add = 0;
        int b1 = n + 1, b2 = n + 1;
        for (int a = n; a > 0; --a) {
            for (int b : g[a]) {
                if (b < b1) {
                    b2 = b1;
                    b1 = b;
                } else if (b < b2) {
                    b2 = b;
                }
            }
            ans += b1 - a;
            cnt[b1] += b2 - b1;
            add = Math.max(add, cnt[b1]);
        }
        ans += add;
        return ans;
    }
}

// Accepted solution for LeetCode #3480: Maximize Subarrays After Removing One Conflicting Pair
func maxSubarrays(n int, conflictingPairs [][]int) (ans int64) {
	g := make([][]int, n+1)
	for _, pair := range conflictingPairs {
		a, b := pair[0], pair[1]
		if a > b {
			a, b = b, a
		}
		g[a] = append(g[a], b)
	}

	cnt := make([]int64, n+2)
	var add int64
	b1, b2 := n+1, n+1

	for a := n; a > 0; a-- {
		for _, b := range g[a] {
			if b < b1 {
				b2 = b1
				b1 = b
			} else if b < b2 {
				b2 = b
			}
		}
		ans += int64(b1 - a)
		cnt[b1] += int64(b2 - b1)
		if cnt[b1] > add {
			add = cnt[b1]
		}
	}

	ans += add
	return ans
}

# Accepted solution for LeetCode #3480: Maximize Subarrays After Removing One Conflicting Pair
class Solution:
    def maxSubarrays(self, n: int, conflictingPairs: List[List[int]]) -> int:
        g = [[] for _ in range(n + 1)]
        for a, b in conflictingPairs:
            if a > b:
                a, b = b, a
            g[a].append(b)
        cnt = [0] * (n + 2)
        ans = add = 0
        b1 = b2 = n + 1
        for a in range(n, 0, -1):
            for b in g[a]:
                if b < b1:
                    b2, b1 = b1, b
                elif b < b2:
                    b2 = b
            ans += b1 - a
            cnt[b1] += b2 - b1
            add = max(add, cnt[b1])
        ans += add
        return ans

// Accepted solution for LeetCode #3480: Maximize Subarrays After Removing One Conflicting Pair
impl Solution {
    pub fn max_subarrays(n: i32, conflicting_pairs: Vec<Vec<i32>>) -> i64 {
        let mut g: Vec<Vec<i32>> = vec![vec![]; (n + 1) as usize];
        for pair in conflicting_pairs {
            let mut a = pair[0];
            let mut b = pair[1];
            if a > b {
                std::mem::swap(&mut a, &mut b);
            }
            g[a as usize].push(b);
        }

        let mut cnt: Vec<i64> = vec![0; (n + 2) as usize];
        let mut ans = 0i64;
        let mut add = 0i64;
        let mut b1 = n + 1;
        let mut b2 = n + 1;

        for a in (1..=n).rev() {
            for &b in &g[a as usize] {
                if b < b1 {
                    b2 = b1;
                    b1 = b;
                } else if b < b2 {
                    b2 = b;
                }
            }
            ans += (b1 - a) as i64;
            cnt[b1 as usize] += (b2 - b1) as i64;
            add = std::cmp::max(add, cnt[b1 as usize]);
        }

        ans += add;
        ans
    }
}

// Accepted solution for LeetCode #3480: Maximize Subarrays After Removing One Conflicting Pair
function maxSubarrays(n: number, conflictingPairs: number[][]): number {
    const g: number[][] = Array.from({ length: n + 1 }, () => []);
    for (let [a, b] of conflictingPairs) {
        if (a > b) {
            [a, b] = [b, a];
        }
        g[a].push(b);
    }

    const cnt: number[] = Array(n + 2).fill(0);
    let ans = 0,
        add = 0;
    let b1 = n + 1,
        b2 = n + 1;

    for (let a = n; a > 0; a--) {
        for (const b of g[a]) {
            if (b < b1) {
                b2 = b1;
                b1 = b;
            } else if (b < b2) {
                b2 = b;
            }
        }
        ans += b1 - a;
        cnt[b1] += b2 - b1;
        add = Math.max(add, cnt[b1]);
    }

    ans += add;
    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)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n × q) time

O(1) space

For each of q queries, scan the entire range to compute the aggregate (sum, min, max). Each range scan takes up to O(n) for a full-array query. With q queries: O(n × q) total. Point updates are O(1) but queries dominate.

SEGMENT TREE

O(n + q log n) time

O(n) space

Building the tree is O(n). Each query or update traverses O(log n) nodes (tree height). For q queries: O(n + q log n) total. Space is O(4n) ≈ O(n) for the tree array. Lazy propagation adds O(1) per node for deferred updates.

Shortcut: Build O(n), query/update O(log n) each. When you need both range queries AND point updates.

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.