Maximize Subarrays After Removing One Conflicting Pair solution and explanation
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The problem
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]fromconflictingPairs. Now,conflictingPairs = [[1, 4]]. - There are 9 subarrays in
numswhere[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
conflictingPairsis 9.
Example 2:
Input: n = 5, conflictingPairs = [[1,2],[2,5],[3,5]]
Output: 12
Explanation:
- Remove
[1, 2]fromconflictingPairs. Now,conflictingPairs = [[2, 5], [3, 5]]. - There are 12 subarrays in
numswhere[2, 5]and[3, 5]do not appear together. - The maximum number of subarrays we can achieve after removing one element from
conflictingPairsis 12.
Constraints:
2 <= n <= 1051 <= conflictingPairs.length <= 2 * nconflictingPairs[i].length == 21 <= conflictingPairs[i][j] <= nconflictingPairs[i][0] != conflictingPairs[i][1]
Approach overview
Problem Overview: You are given n elements and a list of conflicting pairs (a, b). Any subarray containing both values of a conflicting pair is invalid. You may remove exactly one conflicting pair from the list. The task is to maximize the number of valid subarrays that remain.
Approach 1: Brute Force Pair Removal + Subarray Validation (O(n2 * m) time, O(1) space)
Try removing each conflicting pair one at a time. For the remaining pairs, enumerate every possible subarray and check whether it contains both elements of any conflict. This requires scanning the subarray or tracking membership, making each validation expensive. The approach demonstrates the problem structure but becomes infeasible when n and the number of pairs grow. It mainly helps you reason about how conflicting pairs limit valid subarray boundaries.
Approach 2: Enumeration + Maintaining Minimum and Second Minimum Values (O(n + m) time, O(n) space)
The key observation: for a subarray ending at position r, the earliest valid start depends on the largest conflicting left endpoint among all pairs whose right endpoint is ≤ r. If the maximum left value is max1, any subarray must start after max1. That gives r - max1 valid subarrays ending at r.
Group conflicts by their right endpoint and iterate r from 1..n. While processing, maintain the largest and second-largest left endpoints seen so far. The largest value defines the current restriction for valid subarrays. The second-largest becomes relevant if the pair causing the largest restriction is removed.
For every index r, add r - max1 to the base answer. The improvement from removing the dominant pair equals max1 - max2, because the boundary shifts from max1 to max2. Accumulate this potential gain for the pair responsible for max1. After scanning the array, add the maximum gain to the base count.
This technique combines ideas from Array enumeration and prefix tracking similar to Prefix Sum style accumulation. The ordering of constraints acts like a simplified structure compared to a full Segment Tree, which would otherwise maintain range restrictions dynamically.
Recommended for interviews: The enumeration with maintained maximum and second maximum constraints is the expected solution. It reduces the problem to a linear scan and demonstrates strong reasoning about how conflicts restrict subarray boundaries. Mentioning the brute force idea first shows you understand the constraint interactions before optimizing.
Solution
We store all conflicting pairs (a, b) (assuming a < b) in a list g, where g[a] represents the set of all numbers b that conflict with a.
If no deletion occurs, we can enumerate each subarray's left endpoint a in reverse order. The upper bound of its right endpoint is the minimum value b_1 among all g[x geq a] (excluding b_1), and the contribution to the answer is b_1 - a.
If we delete a conflicting pair containing b_1, then the new b_1 becomes the second minimum value b_2 among all g[x geq a], and its additional contribution to the answer is b_2 - b_1. We use an array cnt to record the additional contribution for each b_1.
The final answer is the sum of all b_1 - a contributions plus the maximum value of cnt[b_1].
The time complexity is O(n), and the space complexity is O(n), where n is the number of conflicting pairs.
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 ansclass 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;
}
}class Solution {
public:
long long maxSubarrays(int n, vector<vector<int>>& conflictingPairs) {
vector<vector<int>> g(n + 1);
for (auto& pair : conflictingPairs) {
int a = pair[0], b = pair[1];
if (a > b) {
swap(a, b);
}
g[a].push_back(b);
}
vector<long long> cnt(n + 2, 0);
long 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 = max(add, cnt[b1]);
}
ans += add;
return ans;
}
};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;
}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
}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
}
}Complexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Brute Force Pair Removal + Subarray Check | O(n² * m) | O(1) | Useful for understanding the constraint behavior in small inputs |
| Enumeration with Max and Second-Max Conflict Tracking | O(n + m) | O(n) | Optimal solution for large inputs where subarray boundaries depend on prefix conflict limits |
Frequently asked questions
Is Maximize Subarrays After Removing One Conflicting Pair easy or hard?
The problem is classified as Hard because it requires recognizing how conflicting pairs restrict subarray start positions and how removing one pair changes those constraints. The final implementation is linear, but deriving the max and second-max prefix logic requires careful reasoning.
Maximize Subarrays After Removing One Conflicting Pair Python/Java solution
Python and Java implementations follow the same structure: group pairs by right endpoint, iterate r from 1..n, maintain the largest and second-largest left endpoints, accumulate base valid subarrays, and track the best improvement from removing a pair. The algorithm runs in O(n + m) time and uses O(n) auxiliary space.
How to solve Maximize Subarrays After Removing One Conflicting Pair in O(n)?
Store conflicts grouped by their right endpoint and iterate through the array from left to right. Track the largest and second-largest left endpoints among all processed conflicts. For each position r, add r - max1 valid subarrays and record the improvement max1 - max2 if the dominant pair were removed. The final answer is the base count plus the maximum improvement.
What is the best approach for Maximize Subarrays After Removing One Conflicting Pair?
The optimal approach is enumeration while maintaining the largest and second-largest conflicting left endpoints encountered so far. For each right boundary r, the largest left value determines the earliest valid subarray start. Removing the pair that causes this largest restriction yields a gain of (max1 - max2). This produces an O(n + m) time solution.
Is Maximize Subarrays After Removing One Conflicting Pair asked at Google/Amazon/Meta?
Hard array and interval-constraint problems like this frequently appear in interviews at large tech companies such as Google, Amazon, and Meta. The problem tests reasoning about constraints, prefix tracking, and optimizing from quadratic enumeration to linear-time counting.
What data structure is used in Maximize Subarrays After Removing One Conflicting Pair?
The optimal solution mainly uses arrays or adjacency lists to group conflicts by their right endpoint and tracks two running maximum values. Conceptually it resembles prefix constraint tracking; heavier structures like segment trees are unnecessary for the linear-time solution.
What is the time complexity of Maximize Subarrays After Removing One Conflicting Pair?
The optimal algorithm runs in O(n + m) time, where n is the number of elements and m is the number of conflicting pairs. Each pair is processed once when grouped by its right endpoint, and the array is scanned once while maintaining two prefix constraints.