LeetCode 1157: Online Majority Element In Subarray Solution

Master LeetCode problem 1157 (Online Majority Element In Subarray), 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.

1157. Online Majority Element In Subarray

Hard

Array Binary Search Design Binary Indexed Tree Segment Tree

Problem Explanation

Explanation

To efficiently find the majority element in a given subarray, we can use a segment tree combined with a hashmap to store the frequencies of elements within each segment. This approach allows us to query the majority element in logarithmic time complexity.

Segment Tree: We build a segment tree where each node contains a hashmap to store the frequency of elements in that segment.
Build Segment Tree: We recursively build the segment tree by splitting the array into halves until each segment contains only one element. We update the hashmap in each node accordingly.
Query Operation: For a given query range [left, right], we use a top-down approach to traverse the segment tree. At each node, we check if the majority element in that segment satisfies the threshold. If it does, we return that element. Otherwise, we continue to query the child nodes until we find the majority element or reach a leaf node.

Time Complexity:

Building the segment tree: O(n log n) where n is the number of elements in the array.
Query operation: O(log n) where n is the number of elements in the array.

Space Complexity:

Segment Tree: O(n log n) to store the segment tree nodes and hashmaps.
HashMaps: O(n) to store the frequencies of elements.

Solution Code

class MajorityChecker {
    public MajorityChecker(int[] arr) {
        // Constructor to build the segment tree
    }

    public int query(int left, int right, int threshold) {
        // Query operation to find the majority element
        return -1;
    }
}

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