LeetCode-in-Java

3739. Count Subarrays With Majority Element II

Hard

You are given an integer array nums and an integer target.

Return the number of non-empty subarrays of nums in which target is the majority element.

The majority element of a subarray is the element that appears strictly more than half of the times in that subarray.

Example 1:

Input: nums = [1,2,2,3], target = 2

Output: 5

Explanation:

Valid subarrays with target = 2 as the majority element:

• nums[1..1] = [2]
• nums[2..2] = [2]
• nums[1..2] = [2,2]
• nums[0..2] = [1,2,2]
• nums[1..3] = [2,2,3]

So there are 5 such subarrays.

Example 2:

Input: nums = [1,1,1,1], target = 1

Output: 10

Explanation:

All 10 subarrays have 1 as the majority element.

Example 3:

Input: nums = [1,2,3], target = 4

Output: 0

Explanation:

target = 4 does not appear in nums at all. Therefore, there cannot be any subarray where 4 is the majority element. Hence the answer is 0.

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 109
• 1 <= target <= 109

Solution

public class Solution {
    public long countMajoritySubarrays(int[] nums, int target) {
        int n = nums.length;
        int pre = n + 1;
        long[] count = new long[2 * n + 2];
        long[] acc = new long[2 * n + 2];
        long res = 0;
        count[pre] = acc[pre] = 1;
        for (int a : nums) {
            pre += (a == target ? 1 : -1);
            acc[pre] = ++count[pre] + acc[pre - 1];
            res += acc[pre - 1];
        }
        return res;
    }
}

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