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Count Triplets

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  • karthikmeduri09
     + 0 comments

    First we import defaultdictionary from collections module. Then, use this code for better Understanding: def countTriplets(arr, r): total_pairs = 0 count2={} count3={} for num in arr: if num in count3: total_pairs+=count3[num] if num in count2: count3[num*r]=count3.get(num*r, 0) + count2[num] count2[num*r]=count2.get(num*r, 0) + 1 return total_pairs

  • hunterwei
     + 0 comments

    Easy to understand Python code

    from collections import Counter, defaultdict

def countTriplets(arr, r):
    left = defaultdict(int)
    right = Counter(arr)
    
    res = 0
    for a in arr:
        right[a] -= 1
        cl = left[a/r]
        cr = right[a * r]
        res += cl * cr
        left[a] += 1
        
    return res

  • sunny_jyrm
     + 0 comments

    Cleanest python solution:

    from collections import defaultdict
def countTriplets(nums, r):
    
    lefts = defaultdict(int)
    pairs = defaultdict(int)
    
    res = 0
    for n in nums:
        res += pairs[n]
                
        pairs[n*r] += lefts[n]
        
        lefts[n*r] += 1
        
    return res

    Can be made more succint using a key transform and one dictionary:

    from collections import defaultdict
def countTriplets(nums, r):
    lefts, res = defaultdict(int),0
    
    for n in nums:
        res += lefts[-n]
        lefts[-n*r] += lefts[n]
        lefts[n*r] += 1
        
    return res

  • Matthew_Garneau
     + 0 comments

    This one was a doozy! Took me way longer then I'd care to admit. JS solution (forgive my nesting 😅):

    function countTriplets(arr, r) {
    let result = 0;
    
    const values = {};
    const pairs = {};
    
    for(const v of arr) {
      const prev = v/r;
      
      if(prev in values) {
        if (prev in pairs) result += pairs[prev];
        
        pairs[v] = (pairs[v] ?? 0) + values[prev];
      }
      
      values[v] = (values[v] ?? 0) + 1;
    }
    
    return result;
}

  • khalid_hamdaan
     + 0 comments

    Why is this failing hidden test case 6 :(

    I came up with idea of moving forward and keeping a track of singles and doubles that have happened. This is passing all test cases except test case 6 :(

    static long countTriplets(List<long> arr, long r)
{
    Dictionary<long, long> singles = new();
    Dictionary<long, long> doubles = new();
    long count = 0;

    foreach (var num in arr)
    {
        long keyByr = num / r;

        // If num completes a triplet
        if (doubles.ContainsKey(keyByr))
        {
            count += doubles[keyByr];
        }

        // If num can be a middle of triplet
        if (singles.ContainsKey(keyByr))
        {
            if (doubles.ContainsKey(num))
                doubles[num] += singles[keyByr];
            else
                doubles[num] = singles[keyByr];
        }

        // Count num as potential start
        if (singles.ContainsKey(num))
            singles[num]++;
        else
            singles[num] = 1;
    }

    return count;
}