You are given an integer array nums. A special triplet is defined as a triplet of indices (i, j, k) such that: 0 <= i < j < k < n, where n = nums.length nums[i] == nums[j] * 2 nums[k] == nums[j] * 2 Return the total number of special triplets in th

specialTriplets.js

/**
 * @param {number[]} nums
 * @return {number}
 */
var specialTriplets = function(nums) {
    const MOD = 1e9 + 7; // modulus to avoid overflow
    const n = nums.length;
    
    // Frequency maps
    let leftCount = new Map();   // counts of numbers before j
    let rightCount = new Map();  // counts of numbers after j
    
    // Initialize rightCount with all numbers
    for (let num of nums) {
        rightCount.set(num, (rightCount.get(num) || 0) + 1);
    }
    
    let result = 0;
    
    // Iterate through each possible j
    for (let j = 0; j < n; j++) {
        let val = nums[j];
        
        // Remove nums[j] from rightCount since j is now the "middle"
        rightCount.set(val, rightCount.get(val) - 1);
        if (rightCount.get(val) === 0) {
            rightCount.delete(val);
        }
        
        // Condition: nums[i] == 2 * nums[j]
        let target = val * 2;
        
        // Count how many i < j satisfy nums[i] == target
        let countLeft = leftCount.get(target) || 0;
        
        // Count how many k > j satisfy nums[k] == target
        let countRight = rightCount.get(target) || 0;
        
        // Each pair (i, k) with this j forms a valid triplet
        result = (result + (countLeft * countRight) % MOD) % MOD;
        
        // Add nums[j] to leftCount for future iterations
        leftCount.set(val, (leftCount.get(val) || 0) + 1);
    }
    
    return result;
};