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Count Triplets

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815 Discussions

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  • 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;
}

  • aggarwalaakash01
     + 1 comment

    can someone please help me understand the problem with my code??

    def countTriplets(arr, r):
    d = defaultdict(int)
    count = 0
    for i in arr:
        d[i] += 1
    for i in d:
        if i == i*r and i == i*r*r:
            count = count + (d[i] * (d[i]-1) * (d[i] - 2)) // 6
        else:
            if i*r in d and i*r*r in d:
                count = count + d[i] * d[i*r] * d[i*r*r]
        
    return count

  • Migueel0
     + 0 comments

    Recursive solution:

    def countTriplets(arr, r):
    sys.setrecursionlimit(10**6) #Avoid recursion error
    def rec_function(index, seconds, thirds, count):
        if index >= len(arr):
            return count
        i = arr[index]
        if i in thirds:
            count += thirds[i]
        if i in seconds:
            thirds[i * r] = thirds.get(i * r, 0) + seconds[i]
        seconds[i * r] = seconds.get(i * r, 0) + 1
        return rec_function(index + 1, seconds, thirds, count)
    
    return rec_function(0, {}, {}, 0)