Intuition

target is the majority of a subarray iff it appears strictly more than half the time. Replace every target with +1 and every other value with -1. Then a subarray has target as its majority element exactly when the sum of its ±1 values is strictly positive (more +1s than -1s).

Using prefix sums $P_0, P_1, \dots, P_n$ (with $P_0 = 0$), the subarray (i, j] is valid iff $P_j - P_i > 0$, i.e. $P_i < P_j$. So the answer is the number of index pairs $i < j$ with $P_i < P_j$.

Approach: Prefix Sum + Running Count of Smaller Prefixes

We sweep j from left to right and, for each j, add the number of earlier prefixes (including the empty prefix $P_0$) that are strictly smaller than the current prefix $P_j$.

Naively counting "how many earlier prefixes are smaller" looks like it needs a Fenwick tree, but the prefix sum changes by exactly ±1 at each step, so we can maintain that count incrementally in $O(1)$:

• Shift prefix values by n so they fit into an array of size 2n + 1. count tracks the current prefix sum (shifted), starting at n (value 0).
• prefix[v] = how many prefixes seen so far equal the value v. Seed prefix[n] = 1 for the empty prefix $P_0 = 0$.
• sum = how many earlier prefixes are strictly smaller than the current one.

Transitions when the prefix value moves:

• num == target → value goes up by 1. Everything that equalled the old value is now strictly smaller, so sum += prefix[count], then count++.
• num != target → value goes down by 1. After count--, everything that equals this new value is no longer strictly smaller, so sum -= prefix[count].

After updating, record the current prefix with prefix[count]++ and accumulate ans += sum. Because sum already counts how many strictly-smaller prefixes precede j, this directly sums the valid (i, j] subarrays ending at each j.

Complexity

• Time complexity: $O(n)$ — one pass with $O(1)$ work per element.
• Space complexity: $O(n)$ — the prefix frequency array of size $2n + 1$.

Code

Go

func countMajoritySubarrays(nums []int, target int) int64 {
    n := len(nums)
    prefix := make([]int, 2*n+1)
    prefix[n] = 1
    count := n
    ans, sum := 0, 0
    for _, num := range nums {
        if num == target {
            sum += prefix[count]
            count++
        } else {
            count--
            sum -= prefix[count]
        }
        prefix[count]++
        ans += sum
    }
    return int64(ans)
}

Rust

impl Solution {
    pub fn count_majority_subarrays(nums: Vec<i32>, target: i32) -> i64 {
        let n = nums.len();
        let mut prefix = vec![0; 2 * n + 1];
        prefix[n] = 1;
        let mut count = n;
        let (mut ans, mut sum) = (0_i64, 0_i64);
        for num in nums.into_iter() {
            if num == target {
                sum += prefix[count] as i64;
                count += 1;
            } else {
                count -= 1;
                sum -= prefix[count] as i64;
            }
            prefix[count] += 1;
            ans += sum;
        }
        ans
    }
}