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Code to find out the number of triplets which lead to a sum less than or equal to threshold is giving a wrong output


Code to find out the number of triplets which lead to a sum less than or equal to threshold is giving a wrong output

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Code to find out the number of triplets which lead to a sum less than or equal to threshold is giving a wrong output
Tag : cpp , By : Porta
Date : November 24 2020, 05:44 AM

I wish did fix the issue. Every element of nums is greater equal zero.
Therfor a = nums[i] is greater equal zero.
if(a > 0) break;
if(a > threshold) break;

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Array.equal() giving wrong output


Tag : java , By : Daniel Halsey
Date : March 29 2020, 07:55 AM
To fix the issue you can do This has nothing to do with arrays really. Your comparison is equivalent to:
Object x = Integer.valueOf(3);
Object y = Byte.valueOf((byte) 3);
boolean equal = x.equals(y);

Program to find trailing number of zeroes is giving wrong output


Tag : c , By : unfool
Date : March 29 2020, 07:55 AM
To fix the issue you can do You're never initializing number, and then you're printing a pointer to it instead of the number itself. Of course you're going to print garbage results.
Also, I don't understand how your algorithm is supposed to work. Dividing by five and adding that to the number of zeroes? If I did that with the number 100, that would add 20, but 100 doesn't have 20 trailing zeroes.

Find the number of triplets i,j,k in an array such that the xor of elements indexed i to j-1 is equal to the xor of elem


Tag : arrays , By : Giles
Date : March 29 2020, 07:55 AM
will help you Here's an O(n) solution based on CiaPan's comment under the question description:
import random

def brute_force(A):
  res = 0

  for i in xrange(len(A) - 1):
    left = A[i]
    for j in xrange(i + 1, len(A)):
      if j > i + 1:
        left ^= A[j - 1]
      right = A[j]
      for k in xrange(j, len(A)):
        if k > j:
          right ^= A[k]
        if left == right:
          res += 1

  return res

def f(A):
  ps = [A[0]] + [0] * (len(A) - 1)
  for i in xrange(1, len(A)):
    ps[i] = ps[i- 1] ^ A[i]

  res = 0
  seen = {0: (-1, 1, 0)}

  for  i in xrange(len(A)):
    if ps[i] in seen:
      prev_i, i_count, count = seen[ps[i]]
      new_count = count + i_count * (i - prev_i) - 1
      res += new_count
      seen[ps[i]] = (i, i_count + 1, new_count)
    else:
      seen[ps[i]] = (i, 1, 0)

  return res

for i in xrange(100):
  A = [random.randint(1, 10) for x in xrange(200)]
  f_A, brute_force_A = f(A), brute_force(A)
  assert f_A == brute_force_A
print "Done"

Find ways an Integer can be expressed as sum of n-th power of unique natural numbers.code giving wrong output


Tag : cpp , By : Peter Leung
Date : March 29 2020, 07:55 AM
I wish this helpful for you I inserted a simple debug output in func. For the given input "1 10 2" x sometimes gets negative. This causes UB when accessing the array, but does not necessarily crash.
You already check if x is less than 0, but after using x. Move the if(x < 0) up and you are done.

Find the total number of triplets when summed are less than a given threshold


Tag : python , By : cynix
Date : March 29 2020, 07:55 AM
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