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Code Implementation

Special Subsequence Counter Implementation

Below is the implementation of the special subsequence counter:

solution.js
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/**
 * Count the number of special subsequences in the array.
 * 
 * @param {number[]} nums - The input array containing only 0s, 1s, and 2s
 * @return {number} - The number of special subsequences
 */
function countSpecialSubsequences(nums) {
  let count0 = 0;
  let count1 = 0;
  let count2 = 0;
  
  for (let i = 0; i < nums.length; i++) {
    if (nums[i] === 0) {
      // Double the existing count0 and add 1 for the new subsequence
      count0 = 2 * count0 + 1;
    } else if (nums[i] === 1) {
      // Double the existing count1 and add count0 for new subsequences
      count1 = 2 * count1 + count0;
    } else if (nums[i] === 2) {
      // Double the existing count2 and add count1 for new subsequences
      count2 = 2 * count2 + count1;
    }
  }
  
  return count2;
}
 
/**
 * Count the number of special subsequences in the array with modular arithmetic.
 * 
 * @param {number[]} nums - The input array containing only 0s, 1s, and 2s
 * @return {number} - The number of special subsequences
 */
function countSpecialSubsequencesMod(nums) {
  const MOD = 1000000007;
  let count0 = 0;
  let count1 = 0;
  let count2 = 0;
  
  for (let i = 0; i < nums.length; i++) {
    if (nums[i] === 0) {
      count0 = (2 * count0 + 1) % MOD;
    } else if (nums[i] === 1) {
      count1 = (2 * count1 + count0) % MOD;
    } else if (nums[i] === 2) {
      count2 = (2 * count2 + count1) % MOD;
    }
  }
  
  return count2;
}
 
/**
 * Count the number of special subsequences in the array using an array-based approach.
 * 
 * @param {number[]} nums - The input array containing only 0s, 1s, and 2s
 * @return {number} - The number of special subsequences
 */
function countSpecialSubsequencesArray(nums) {
  const MOD = 1000000007;
  const dp = [0, 0, 0];  // dp[i] = number of subsequences ending with i
  
  for (let i = 0; i < nums.length; i++) {
    const num = nums[i];
    dp[num] = (2 * dp[num] + (num === 0 ? 1 : dp[num - 1])) % MOD;
  }
  
  return dp[2];
}
 
// Test cases
console.log(countSpecialSubsequences([0, 1, 2, 2]));  // 3
console.log(countSpecialSubsequences([2, 2, 0, 0]));  // 0
console.log(countSpecialSubsequences([0, 1, 2, 0, 1, 2]));  // 7
 
console.log(countSpecialSubsequencesMod([0, 1, 2, 2]));  // 3
console.log(countSpecialSubsequencesMod([2, 2, 0, 0]));  // 0
console.log(countSpecialSubsequencesMod([0, 1, 2, 0, 1, 2]));  // 7
 
console.log(countSpecialSubsequencesArray([0, 1, 2, 2]));  // 3
console.log(countSpecialSubsequencesArray([2, 2, 0, 0]));  // 0
console.log(countSpecialSubsequencesArray([0, 1, 2, 0, 1, 2]));  // 7

Step-by-Step Explanation

Let's break down the implementation:

  1. Understand the Problem: First, understand that we need to count the number of special subsequences consisting of only 0s, followed by only 1s, followed by only 2s.
  2. Identify Key Insight: Recognize that we can use dynamic programming to keep track of the number of subsequences ending with 0, 1, and 2 at each position.
  3. Initialize Counters: Initialize three counters: count0, count1, and count2, to keep track of the number of subsequences ending with 0, 1, and 2, respectively.
  4. Update Counters: For each element in the array, update the appropriate counter based on its value (0, 1, or 2).
  5. Apply Modular Arithmetic: Since the answer may be very large, apply modular arithmetic to avoid overflow.
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