Maximize Subarrays After Removing One Conflicting Pair

Maximize Subarrays After Removing One Conflicting Pair - Problem

Maximize Subarrays After Removing One Conflicting Pair

You're given an array containing consecutive integers from 1 to n, and a list of conflicting pairs. Your goal is to strategically remove exactly one conflicting pair to maximize the number of valid subarrays.

🎯 The Challenge: A subarray is considered valid if it doesn't contain both elements of any remaining conflicting pair. For example, if [2, 5] is a conflicting pair, then any subarray containing both 2 and 5 is invalid.

Input:
• An integer n representing the array [1, 2, 3, ..., n]
• A 2D array conflictingPairs where each pair [a, b] represents conflicting elements

Output:
• The maximum number of valid non-empty subarrays after removing exactly one conflicting pair

Example: With n = 4 and pairs [[1,3], [2,4]], removing [1,3] might give us more valid subarrays than removing [2,4].

Input & Output

example_1.py — Basic Case

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

› Output: 8

💡 Note: If we remove [1,3], remaining conflict is [2,4]. Subarrays containing both 2 and 4 are invalid: [1,2,3,4], [2,3,4], [2,4]. So 10-3=7 valid subarrays. If we remove [2,4], remaining conflict is [1,3]. Invalid subarrays: [1,2,3], [1,3]. So 10-2=8 valid subarrays. Maximum is 8.

example_2.py — Single Conflict

$ Input: n = 3, conflictingPairs = [[1,3]]

› Output: 6

💡 Note: We must remove the only conflicting pair [1,3]. After removal, no conflicts remain, so all 6 possible subarrays are valid: [1], [2], [3], [1,2], [2,3], [1,2,3].

example_3.py — Overlapping Conflicts

$ Input: n = 5, conflictingPairs = [[1,3], [3,5], [2,4]]

› Output: 12

💡 Note: We need to try removing each pair and see which gives the maximum valid subarrays. The optimal choice depends on how conflicts overlap and affect the total count of invalid subarrays.

Visualization

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Understanding the Visualization

Setup Exhibition

You have 4 artifacts (1,2,3,4) to arrange in showcases

Identify Conflicts

Artifacts [1,3] and [2,4] cannot be in the same showcase

Try Removing Rules

Remove one conflict rule and count valid showcase arrangements

Maximize Displays

Choose the rule removal that allows maximum showcase configurations

Key Takeaway

🎯 Key Insight: By strategically removing the rule that causes the most restrictions, we maximize the number of valid exhibition arrangements, just like maximizing valid subarrays by removing the optimal conflicting pair.

Time & Space Complexity

Time Complexity

⏱️

O(k²)

k iterations for each pair removal, each requiring O(k) analysis of remaining conflicts

✓ Linear Growth

Space Complexity

O(k)

Space for storing conflicts and temporary calculations

✓ Linear Space

Constraints

2 ≤ n ≤ 1000
1 ≤ conflictingPairs.length ≤ 100
conflictingPairs[i].length == 2
1 ≤ conflictingPairs[i][0], conflictingPairs[i][1] ≤ n
conflictingPairs[i][0] ≠ conflictingPairs[i][1]
No duplicate conflicting pairs

Asked in

G Google 42 f Meta 38 a Amazon 31 ⊞ Microsoft 24

The optimal approach uses mathematical formulas to calculate valid subarrays efficiently. Instead of generating all O(n²) subarrays explicitly, we calculate the total count n*(n+1)/2 and subtract invalid ones. For each potential conflict removal, we analyze the remaining conflicts and use set operations to count overlapping invalid subarrays in O(k²) time, where k is the number of conflicts.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Try All Removals)	O(k * n³)	O(k)	Remove each conflicting pair and count valid subarrays explicitly
Mathematical Formula (Optimal)	O(k²)	O(k)	Use mathematical formulas to calculate valid subarrays efficiently

Brute Force (Try All Removals) — Algorithm Steps

For each conflicting pair to potentially remove:
Create a temporary set of remaining conflicts
Generate all possible subarrays of the array [1,2,...,n]
For each subarray, check if it contains both elements of any remaining conflict
Count valid subarrays and track the maximum

Visualization

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Step-by-Step Walkthrough

Remove First Pair

Temporarily remove first conflicting pair

Check All Subarrays

Generate and validate each possible subarray

Count Valid Ones

Count subarrays that don't violate remaining conflicts

Repeat Process

Repeat for each conflicting pair

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>

int countValidSubarrays(int n, int conflicts[][2], int conflictCount) {
    int count = 0;
    // Check all possible subarrays
    for (int i = 1; i <= n; i++) {
        for (int j = i; j <= n; j++) {
            bool valid = true;
            // Check against all remaining conflicts
            for (int k = 0; k < conflictCount; k++) {
                int a = conflicts[k][0], b = conflicts[k][1];
                if (a >= i && a <= j && b >= i && b <= j) {
                    valid = false;
                    break;
                }
            }
            if (valid) count++;
        }
    }
    return count;
}

int maxSubarrays(int n, int conflictingPairs[][2], int pairCount) {
    int maxCount = 0;
    
    // Try removing each conflicting pair
    for (int i = 0; i < pairCount; i++) {
        int remaining[100][2]; // Assuming max 100 pairs
        int remainingCount = 0;
        
        for (int j = 0; j < pairCount; j++) {
            if (j != i) {
                remaining[remainingCount][0] = conflictingPairs[j][0];
                remaining[remainingCount][1] = conflictingPairs[j][1];
                remainingCount++;
            }
        }
        
        int count = countValidSubarrays(n, remaining, remainingCount);
        if (count > maxCount) maxCount = count;
    }
    
    return maxCount;
}

int main() {
    int n;
    scanf("%d", &n);
    int conflictingPairs[100][2]; // Parse input
    int pairCount = 0;
    printf("%d\n", maxSubarrays(n, conflictingPairs, pairCount));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(k * n³)

k iterations for each pair removal, n² subarrays, each checked against O(n) remaining conflicts

⚠ Quadratic Growth

Space Complexity

O(k)

Space for storing conflicts and temporary removal sets

✓ Linear Space

Constraints

2 ≤ n ≤ 1000
1 ≤ conflictingPairs.length ≤ 100
conflictingPairs[i].length == 2
1 ≤ conflictingPairs[i][0], conflictingPairs[i][1] ≤ n
conflictingPairs[i][0] ≠ conflictingPairs[i][1]
No duplicate conflicting pairs

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force (Try All Removals) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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