LeetCode #3739 — HARD

Count Subarrays With Majority Element II

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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

Return the number of 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

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums and an integer target. Return the number of 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Segment Tree

Example 1

[1,2,2,3]
2

Example 2

[1,1,1,1]
1

Example 3

[1,2,3]
4

Step 02

Core Insight

What unlocks the optimal approach

Convert to +1/-1: let <code>arr[i] = 1</code> if <code>nums[i] == target</code> else <code>-1</code>.
Build prefix sums: <code>pref[0]=0</code>, <code>pref[k] = pref[k - 1] + arr[k - 1]</code> for <code>k=1..n</code>.
Count pairs <code>(i < j)</code> with <code>pref[j] > pref[i]</code> (these correspond to subarrays where <code>target</code> is majority).
Use coordinate compression on all <code>pref</code> values and a Fenwick tree / ordered map: iterate <code>k</code> from <code>0..n</code>, query how many previous <code>pref</code> are < current, add to <code>ans</code>, then update.
If <code>target</code> never appears return <code>0</code>.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3739: Count Subarrays With Majority Element II
class BinaryIndexedTree {
    private int n;
    private int[] c;

    public BinaryIndexedTree(int n) {
        this.n = n;
        this.c = new int[n + 1];
    }

    public void update(int x, int delta) {
        for (; x <= n; x += x & -x) {
            c[x] += delta;
        }
    }

    public int query(int x) {
        int s = 0;
        for (; x > 0; x -= x & -x) {
            s += c[x];
        }
        return s;
    }
}

class Solution {
    public long countMajoritySubarrays(int[] nums, int target) {
        int n = nums.length;
        BinaryIndexedTree tree = new BinaryIndexedTree(2 * n + 1);
        int s = n + 1;
        tree.update(s, 1);
        long ans = 0;
        for (int x : nums) {
            s += x == target ? 1 : -1;
            ans += tree.query(s - 1);
            tree.update(s, 1);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3739: Count Subarrays With Majority Element II
type BinaryIndexedTree struct {
	n int
	c []int
}

func NewBinaryIndexedTree(n int) *BinaryIndexedTree {
	return &BinaryIndexedTree{
		n: n,
		c: make([]int, n+1),
	}
}

func (t *BinaryIndexedTree) update(x, delta int) {
	for x <= t.n {
		t.c[x] += delta
		x += x & -x
	}
}

func (t *BinaryIndexedTree) query(x int) int {
	s := 0
	for x > 0 {
		s += t.c[x]
		x -= x & -x
	}
	return s
}

func countMajoritySubarrays(nums []int, target int) int64 {
	n := len(nums)
	tree := NewBinaryIndexedTree(2*n + 1)
	s := n + 1
	tree.update(s, 1)
	var ans int64
	for _, x := range nums {
		if x == target {
			s++
		} else {
			s--
		}
		ans += int64(tree.query(s - 1))
		tree.update(s, 1)
	}
	return ans
}

# Accepted solution for LeetCode #3739: Count Subarrays With Majority Element II
class BinaryIndexedTree:
    __slots__ = "n", "c"

    def __init__(self, n: int):
        self.n = n
        self.c = [0] * (n + 1)

    def update(self, x: int, delta: int) -> None:
        while x <= self.n:
            self.c[x] += delta
            x += x & -x

    def query(self, x: int) -> int:
        s = 0
        while x:
            s += self.c[x]
            x -= x & -x
        return s


class Solution:
    def countMajoritySubarrays(self, nums: List[int], target: int) -> int:
        n = len(nums)
        tree = BinaryIndexedTree(n * 2 + 1)
        s = n + 1
        tree.update(s, 1)
        ans = 0
        for x in nums:
            s += 1 if x == target else -1
            ans += tree.query(s - 1)
            tree.update(s, 1)
        return ans

// Accepted solution for LeetCode #3739: Count Subarrays With Majority Element II
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3739: Count Subarrays With Majority Element II
// class BinaryIndexedTree {
//     private int n;
//     private int[] c;
// 
//     public BinaryIndexedTree(int n) {
//         this.n = n;
//         this.c = new int[n + 1];
//     }
// 
//     public void update(int x, int delta) {
//         for (; x <= n; x += x & -x) {
//             c[x] += delta;
//         }
//     }
// 
//     public int query(int x) {
//         int s = 0;
//         for (; x > 0; x -= x & -x) {
//             s += c[x];
//         }
//         return s;
//     }
// }
// 
// class Solution {
//     public long countMajoritySubarrays(int[] nums, int target) {
//         int n = nums.length;
//         BinaryIndexedTree tree = new BinaryIndexedTree(2 * n + 1);
//         int s = n + 1;
//         tree.update(s, 1);
//         long ans = 0;
//         for (int x : nums) {
//             s += x == target ? 1 : -1;
//             ans += tree.query(s - 1);
//             tree.update(s, 1);
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3739: Count Subarrays With Majority Element II
class BinaryIndexedTree {
    private n: number;
    private c: number[];

    constructor(n: number) {
        this.n = n;
        this.c = Array(n + 1).fill(0);
    }

    update(x: number, delta: number): void {
        for (; x <= this.n; x += x & -x) {
            this.c[x] += delta;
        }
    }

    query(x: number): number {
        let s = 0;
        for (; x > 0; x -= x & -x) {
            s += this.c[x];
        }
        return s;
    }
}

function countMajoritySubarrays(nums: number[], target: number): number {
    const n = nums.length;
    const tree = new BinaryIndexedTree(2 * n + 1);
    let s = n + 1;
    tree.update(s, 1);
    let ans = 0;
    for (const x of nums) {
        s += x === target ? 1 : -1;
        ans += tree.query(s - 1);
        tree.update(s, 1);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n × q) time

O(1) space

For each of q queries, scan the entire range to compute the aggregate (sum, min, max). Each range scan takes up to O(n) for a full-array query. With q queries: O(n × q) total. Point updates are O(1) but queries dominate.

SEGMENT TREE

O(n + q log n) time

O(n) space

Building the tree is O(n). Each query or update traverses O(log n) nodes (tree height). For q queries: O(n + q log n) total. Space is O(4n) ≈ O(n) for the tree array. Lazy propagation adds O(1) per node for deferred updates.

Shortcut: Build O(n), query/update O(log n) each. When you need both range queries AND point updates.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.