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Why is the code not correct? Python




Write a program that finds the closest element in an array to a given number.

Input data

The first line contains a list of numbers – the elements of the array (integers not exceeding 1000 in absolute value).

The second line contains one integer x not exceeding 1000 in absolute value.


Print the value of the array element closest to x. If there are several such numbers, print any of them.


Input 1 2 3 4 5
Conclusion 5

Example of my code:

import sys
a = list (map (int, input (). split ()))
b = int (input ())
c = []
for i in a:
  c.append (b-i)
for i in a:
  if i == b:
    print (b)
    sys.exit ()
if min (c) & lt; 0:
  d = c.index (max (c))
  print (a [d])
  d = c.index (min (c))
  print (a [d])

Writes in the test that the code gives an incorrect answer

Answer 1, authority 100%

Here’s the code:

import sys
a = list (map (int, input (). split ()))
b = int (input ())
c = []
for i in a:
  c.append (abs (b-i))
d = c.index (min (c))
print (a [d])

abs – takes the value modulo

Answer 2, authority 33%

I can suggest this solution:


  1. Sort a list of numbers
  2. I iterate through the list and find the first superior number to the given b .
  3. I find the neighbors of the number b , based on the largest number from point 2, and compare them for closeness to a given number.
  4. I display the first number or the maximum in the list


a = sorted (list (map (int, input (). split ())))
b = int (input ())
res = []
for num, i in enumerate (a):
  if i & gt; = b:
    if b - a [num - 1] & gt; a [num] - b:
      res.append (a [num])
      res.append (a [num - 1])
if res:
  print (res [0])
  print (max (a))

Answer 3, authority 33%

The min built-in function has an optional key argument that accepts a function that determines the order in which elements are compared:

a = [1,2,3,4,5,7,6,15]
b = 19
m = min (a, key = lambda x: abs (x-b)) # 15

Here x elements are compared to each other by the absolute difference between them and the specified b value. That is, the smaller will be the one with the smaller difference.

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