LeetCode #3395 — HARD

Subsequences with a Unique Middle Mode I

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

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The Problem

Problem Statement

Given an integer array nums, find the number of subsequences of size 5 of nums with a unique middle mode.

Since the answer may be very large, return it modulo 109 + 7.

A mode of a sequence of numbers is defined as the element that appears the maximum number of times in the sequence.

A sequence of numbers contains a unique mode if it has only one mode.

A sequence of numbers seq of size 5 contains a unique middle mode if the middle element (seq[2]) is a unique mode.

Example 1:

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

Output: 6

Explanation:

[1, 1, 1, 1, 1] is the only subsequence of size 5 that can be formed, and it has a unique middle mode of 1. This subsequence can be formed in 6 different ways, so the output is 6. 

Example 2:

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

Output: 4

Explanation:

[1, 2, 2, 3, 4] and [1, 2, 3, 3, 4] each have a unique middle mode because the number at index 2 has the greatest frequency in the subsequence. [1, 2, 2, 3, 3] does not have a unique middle mode because 2 and 3 appear twice.

Example 3:

Input: nums = [0,1,2,3,4,5,6,7,8]

Output: 0

Explanation:

There is no subsequence of length 5 with a unique middle mode.

Constraints:

  • 5 <= nums.length <= 1000
  • -109 <= nums[i] <= 109

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: Given an integer array nums, find the number of subsequences of size 5 of nums with a unique middle mode. Since the answer may be very large, return it modulo 109 + 7. A mode of a sequence of numbers is defined as the element that appears the maximum number of times in the sequence. A sequence of numbers contains a unique mode if it has only one mode. A sequence of numbers seq of size 5 contains a unique middle mode if the middle element (seq[2]) is a unique mode.

Baseline thinking

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

Pattern signal: Array · Hash Map · Math

Example 1

[1,1,1,1,1,1]

Example 2

[1,2,2,3,3,4]

Example 3

[0,1,2,3,4,5,6,7,8]

Related Problems

  • Subsequences with a Unique Middle Mode II (subsequences-with-a-unique-middle-mode-ii)
Step 02

Core Insight

What unlocks the optimal approach

  • For each index, find the number of subsequences for which it is the unique middle mode. What combinations of values can the two numbers on the left and the right take?
  • For example, we can have exactly 1 element on the left equal to the middle and all other elements differ. What other combinations are acceptable?
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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 #3395: Subsequences with a Unique Middle Mode I
class Solution {
  public int subsequencesWithMiddleMode(int[] nums) {
    int n = nums.length;
    long ans = 0;
    Map<Integer, Integer> left = new HashMap<>();
    Map<Integer, Integer> right = new HashMap<>();

    for (int i = 0; i < 2; ++i)
      left.merge(nums[i], 1, Integer::sum);

    for (int i = 2; i < n; ++i)
      right.merge(nums[i], 1, Integer::sum);

    for (int i = 2; i < n - 2; ++i) {
      final int num = nums[i];
      if (right.merge(num, -1, Integer::sum) == 0)
        right.remove(num);

      final int leftCount = left.getOrDefault(num, 0);
      final int rightCount = right.getOrDefault(num, 0);
      final int leftOther = i - leftCount;
      final int rightOther = n - 1 - i - rightCount;

      // count[mode] = 5 -- [a a] a [a a]
      ans = (ans + nC2(leftCount) * nC2(rightCount)) % MOD;

      // count[mode] = 4 -- [a a] a [a ?]
      ans = (ans + nC2(leftCount) * rightCount % MOD * rightOther % MOD) % MOD;

      // count[mode] = 4 -- [a ?] a [a a]
      ans = (ans + leftCount * leftOther % MOD * nC2(rightCount) % MOD) % MOD;

      // count[mode] = 3 -- [a a] a [? ?]
      ans = (ans + nC2(leftCount) * nC2(rightOther) % MOD) % MOD;

      // count[mode] = 3 -- [? ?] a [a a]
      ans = (ans + nC2(leftOther) * nC2(rightCount) % MOD) % MOD;

      // count[mode] = 3 -- [a ?] a [a ?]
      ans = (ans + leftCount * leftOther % MOD * rightCount % MOD * rightOther % MOD) % MOD;

      // count[mode] = 2 -- [a ?] a [? ?]
      ans = (ans + leftCount * calc(num, leftOther, rightOther, left, right) % MOD) % MOD;

      // count[mode] = 2 -- [? ?] a [a ?]
      ans = (ans + rightCount * calc(num, rightOther, leftOther, right, left) % MOD) % MOD;

      // Update left map
      left.merge(num, 1, Integer::sum);
    }

    return (int) (ans % MOD);
  }

  private static final int MOD = 1_000_000_007;

  // Returns C(n, 2)
  private long nC2(long n) {
    return n * (n - 1) / 2 % MOD;
  }

  // Returns the count of subsequences that have 'a' as the middle number, where
  // invalid subsequences are excluded
  private long calc(int a, long other1, long other2, Map<Integer, Integer> count1,
                    Map<Integer, Integer> count2) {
    // [a ?] a [? ?]
    long res = other1 * nC2(other2) % MOD;

    for (Map.Entry<Integer, Integer> entry : count1.entrySet()) {
      final int b = entry.getKey();
      final long b1 = entry.getValue();
      if (b == a)
        continue;
      final long b2 = count2.getOrDefault(b, 0);
      // Exclude triples -- [a b] a [b b]
      res = (res - b1 * nC2(b2) % MOD + MOD) % MOD;
      // Exclude doubles -- [a b] a [b ?]
      res = (res - b1 * b2 % MOD * (other2 - b2) % MOD + MOD) % MOD;
    }

    for (Map.Entry<Integer, Integer> entry : count2.entrySet()) {
      final int b = entry.getKey();
      final long b2 = entry.getValue();
      if (b == a)
        continue;
      final long b1 = count1.getOrDefault(b, 0);
      // Exclude doubles -- [a ?] a [b b]
      res = (res - (other1 - b1) * nC2(b2) % MOD + MOD) % MOD;
    }

    return res;
  }
}
Step 06

Interactive Study Demo

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

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n)
Space
O(1)

Approach Breakdown

BRUTE FORCE
O(n²) time
O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED
O(n) time
O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.
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.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

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