LeetCode 3480: Maximize Subarrays After Removing One Conflicting Pair Solution

Master LeetCode problem 3480 (Maximize Subarrays After Removing One Conflicting Pair), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3480. Maximize Subarrays After Removing One Conflicting Pair

Hard

Array Segment Tree Enumeration Prefix Sum

Problem Explanation

Explanation

To solve this problem, we can iterate through all possible pairs of conflicting elements, remove one pair at a time, and count the number of subarrays where both elements of the removed pair do not appear together. We can achieve this by keeping track of the positions of elements in the conflicting pairs and then counting the number of subarrays that do not contain both elements.

Create a map to store the positions of elements in conflicting pairs.
For each pair of conflicting elements:
- Remove one conflicting pair.
- Calculate the number of subarrays without both elements.
- Update the maximum count of subarrays.
Return the maximum count of subarrays.

Time Complexity: The time complexity of this approach is O(n^2), where n is the number of elements in the array nums.

Space Complexity: The space complexity is O(n) to store the positions of elements in conflicting pairs.

Solution Code

class Solution {
    public int maxNumOfSubarrays(int n, int[][] conflictingPairs) {
        int maxSubarrays = 0;
        Map<Integer, Integer> conflictingElements = new HashMap<>();
        
        for (int[] pair : conflictingPairs) {
            int a = pair[0];
            int b = pair[1];
            conflictingElements.put(a, conflictingElements.getOrDefault(a, 0) | 1);
            conflictingElements.put(b, conflictingElements.getOrDefault(b, 0) | 2);
        }
        
        for (int[] pair : conflictingPairs) {
            int a = pair[0];
            int b = pair[1];
            int count = 0;
            
            for (int i = 1; i <= n; i++) {
                if (((conflictingElements.getOrDefault(i, 0) & 3) != 3) || (a == i) || (b == i)) {
                    count++;
                } else {
                    maxSubarrays = Math.max(maxSubarrays, count);
                    count = 0;
                }
            }
            maxSubarrays = Math.max(maxSubarrays, count);
        }
        
        return maxSubarrays;
    }
}

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