Given some integers, find triplets where the product of two numbers equals the third number











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I am trying to solve a problem from coding challenge and the question is described as follows




Given N integers A1, A2, ..., AN, count the number of triplets (x, y,
z) (with 1 ≤ x < y < z ≤ N) such that at least one of the following is
true:



Ax = Ay × Az, and/or



Ay = Ax × Az, and/or



Az = Ax × Ay



5 2 4 6 3 1



In Sample Case #1, the only triplet satisfying the condition given in the problem statement is (2, 4, 5). The triplet is valid since the second, fourth, and fifth integers are 2, 6, and 3, and 2 × 3 = 6. thus the answer here is 1



2 4 8 16 32 64



the six triplets satisfying the condition given in the problem statement are: (1, 2, 3), (1, 3, 4), (1, 4, 5), (1, 5, 6), (2, 3, 5), (2, 4, 6). so the answer here is 6




My Code in python:



import itertools
count=0
for t in itertools.combinations(l,3):
if t[0]*t[1]==t[2] or t[1]*t[2]==t[0] or t[0]*t[2]==t[1]:
count+=1
print(count)


So this is the naive way of generating all possible 3 length combinations and checking for the condition. This works fine for smaller input but when the inout size increase time complexity increases. I am assuming for an example that has 1,2,3,6,8 the combinations generated are (2,3,6),(2,3,8) 2,3,6 satisfy the condition so the checking for 2,3,8 is unnecessary and can be avoided. How can I modify my code to take advantage of this observation ?










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    up vote
    0
    down vote

    favorite












    I am trying to solve a problem from coding challenge and the question is described as follows




    Given N integers A1, A2, ..., AN, count the number of triplets (x, y,
    z) (with 1 ≤ x < y < z ≤ N) such that at least one of the following is
    true:



    Ax = Ay × Az, and/or



    Ay = Ax × Az, and/or



    Az = Ax × Ay



    5 2 4 6 3 1



    In Sample Case #1, the only triplet satisfying the condition given in the problem statement is (2, 4, 5). The triplet is valid since the second, fourth, and fifth integers are 2, 6, and 3, and 2 × 3 = 6. thus the answer here is 1



    2 4 8 16 32 64



    the six triplets satisfying the condition given in the problem statement are: (1, 2, 3), (1, 3, 4), (1, 4, 5), (1, 5, 6), (2, 3, 5), (2, 4, 6). so the answer here is 6




    My Code in python:



    import itertools
    count=0
    for t in itertools.combinations(l,3):
    if t[0]*t[1]==t[2] or t[1]*t[2]==t[0] or t[0]*t[2]==t[1]:
    count+=1
    print(count)


    So this is the naive way of generating all possible 3 length combinations and checking for the condition. This works fine for smaller input but when the inout size increase time complexity increases. I am assuming for an example that has 1,2,3,6,8 the combinations generated are (2,3,6),(2,3,8) 2,3,6 satisfy the condition so the checking for 2,3,8 is unnecessary and can be avoided. How can I modify my code to take advantage of this observation ?










    share|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I am trying to solve a problem from coding challenge and the question is described as follows




      Given N integers A1, A2, ..., AN, count the number of triplets (x, y,
      z) (with 1 ≤ x < y < z ≤ N) such that at least one of the following is
      true:



      Ax = Ay × Az, and/or



      Ay = Ax × Az, and/or



      Az = Ax × Ay



      5 2 4 6 3 1



      In Sample Case #1, the only triplet satisfying the condition given in the problem statement is (2, 4, 5). The triplet is valid since the second, fourth, and fifth integers are 2, 6, and 3, and 2 × 3 = 6. thus the answer here is 1



      2 4 8 16 32 64



      the six triplets satisfying the condition given in the problem statement are: (1, 2, 3), (1, 3, 4), (1, 4, 5), (1, 5, 6), (2, 3, 5), (2, 4, 6). so the answer here is 6




      My Code in python:



      import itertools
      count=0
      for t in itertools.combinations(l,3):
      if t[0]*t[1]==t[2] or t[1]*t[2]==t[0] or t[0]*t[2]==t[1]:
      count+=1
      print(count)


      So this is the naive way of generating all possible 3 length combinations and checking for the condition. This works fine for smaller input but when the inout size increase time complexity increases. I am assuming for an example that has 1,2,3,6,8 the combinations generated are (2,3,6),(2,3,8) 2,3,6 satisfy the condition so the checking for 2,3,8 is unnecessary and can be avoided. How can I modify my code to take advantage of this observation ?










      share|improve this question















      I am trying to solve a problem from coding challenge and the question is described as follows




      Given N integers A1, A2, ..., AN, count the number of triplets (x, y,
      z) (with 1 ≤ x < y < z ≤ N) such that at least one of the following is
      true:



      Ax = Ay × Az, and/or



      Ay = Ax × Az, and/or



      Az = Ax × Ay



      5 2 4 6 3 1



      In Sample Case #1, the only triplet satisfying the condition given in the problem statement is (2, 4, 5). The triplet is valid since the second, fourth, and fifth integers are 2, 6, and 3, and 2 × 3 = 6. thus the answer here is 1



      2 4 8 16 32 64



      the six triplets satisfying the condition given in the problem statement are: (1, 2, 3), (1, 3, 4), (1, 4, 5), (1, 5, 6), (2, 3, 5), (2, 4, 6). so the answer here is 6




      My Code in python:



      import itertools
      count=0
      for t in itertools.combinations(l,3):
      if t[0]*t[1]==t[2] or t[1]*t[2]==t[0] or t[0]*t[2]==t[1]:
      count+=1
      print(count)


      So this is the naive way of generating all possible 3 length combinations and checking for the condition. This works fine for smaller input but when the inout size increase time complexity increases. I am assuming for an example that has 1,2,3,6,8 the combinations generated are (2,3,6),(2,3,8) 2,3,6 satisfy the condition so the checking for 2,3,8 is unnecessary and can be avoided. How can I modify my code to take advantage of this observation ?







      python performance algorithm python-3.x programming-challenge






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      edited 2 mins ago









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      Rohit

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