TCS Digital Coding Question

13 comments on “TCS Digital Coding Question”

• 
  

  Velaga

  Q-6

  from itertools import permutations
  a,b=map(int,input().split())
  a=list(str(a))
  d=[]
  l=list(permutations((a),len(a)))
  for i in l:
  i=int(“”.join(i))
  if i>b:
  d.append(i)
  if len(d)>0:
  d.sort()
  print(d[0])
  else:
  print(-1)

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• 
  

  Velaga

  from itertools import combinations_with_replacement
  from itertools import permutations
  n1,n2=map(int,input().split())
  l=[i for i in range(n1,n2+1)]
  n=int(input())
  a=list(combinations_with_replacement(l,n))

  for i in a:
  for k in list(permutations(list(i),n)):
  if k not in a:
  a.append(k)
  c=0
  for ab in a:
  ab=list(ab)
  if sum(ab)%2==0:
  c=c+1
  print(c)

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• 
  

  Ankit

  Question 4 Solution :-
  Scanner sc= new Scanner(System.in);
  int n1=sc.nextInt();
  int n2=sc.nextInt();
  int k=sc.nextInt();

  int num=n2-n1+1;
  int totalComb= (int)(Math.pow(num, k));

  //out of all combinations half will give even sum
  int evenSum=(int)Math.ceil(totalComb/2.0);
  System.out.print(evenSum);

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• 
  

  shounakpurnapatre1

  Question 6 Can be get solve with this way with minimum number of lines:
  def toString(List):
  return ”.join(List)

  # a =String , l = Starting index , r = Ending index = n-1.
  def permute(a, l, r):
  if l == r:
  ls.append(toString(a))
  else:
  for i in range(l, r+1):
  a[l], a[i] = a[i], a[l]
  permute(a, l+1, r)
  a[l], a[i] = a[i], a[l]

  x, y = map(int, input().split())
  x = str(x)
  n = len(x)
  a = list(x)
  ls = []
  permute(a, 0, n-1)

  test_list = list(map(int, ls)) # Converting all string into integer
  test_list.sort() # Sort the list

  for i in range(len(test_list)):
  if test_list[i] > y:
  print(test_list[i])
  break

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• 
  

  shounakpurnapatre1

  # ************************* UPDATED ******************
  # a =String , l = Starting index , r = Ending index = n-1.
  def permute(a, l, r, y):
  if l == r:

  temp2 = ”.join(a)
  if int(temp2) > y:
  flag = 1
  print(temp2)
  exit()
  else:
  for i in range(l, r+1):
  a[l], a[i] = a[i], a[l]
  permute(a, l+1, r, y)
  a[l], a[i] = a[i], a[l]

  def getSortedNumber(n):
  number = str(n)
  number = ”.join(sorted(number))
  number = int(number)
  return number

  x, y = map(int, input().split())
  x = getSortedNumber(x)
  x = str(x)
  n = len(x)
  a = list(x)
  flag = 0
  permute(a, 0, n-1, y)
  if flag == 0:
  print(-1)

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• 
  

  Priyanka

  Thank you prepinsta for the questions this type of questions got repeated in the exam.Can you please tell me Where can I prepare for codevita or competitive coding?

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• 
  

  avik394451

  Solutions are really helpful. But where can I get more practice questions for Codevita and other Hackathons? Can you please guide me?

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• 
  

  Shruti

  Thankyou PrepInsta. I am highly grateful to you for the questions you provided. I had my test recently and the coding questions were very similar to these. Please help me with the codevita questions too. Where can I find them?

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• 
  

  sarina

  Thank You so much! These questions really helped me out

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• 
  

  Swastik

  Question 4
  public class Test8 {

  public static int fact(int n) {
  int fac = 1;
  for(int i=1;i<=n;i++) {
  fac = fac * i;
  }
  return fac;
  }
  public static int comb(int a, int b) {
  return fact(a) / (fact(a-b) * fact(b));
  }
  public static int sum(int a, int b, int c) {
  int odd = 0, even = 0,total_outcome,odd_outcome = 0;
  for(int i = a;i<=b;i++) {
  if(i%2 == 0) {
  even++;
  }else {
  odd++;
  }
  }
  for(int j = 0;j<=c;j++) {
  if(j%2!=0) {
  odd_outcome = odd_outcome + comb(c,j) * (int)Math.pow(odd, j) * (int)Math.pow(even, c-j);
  }
  }
  total_outcome = (int) Math.pow(odd+even, c);
  return total_outcome – odd_outcome;
  }
  public static void main(String[] args) {
  Scanner s = new Scanner(System.in);
  int a = s.nextInt();
  int b = s.nextInt();
  int c = s.nextInt();

  System.out.println(sum(a, b, c));
  }

  }

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• 
  

  abhitejkumarvoruganti

  import itertools
  l,h=map(int,input().split())
  k=int(input())
  x=range(l,h+1)
  perm=[p for p in itertools.product(x, repeat=k)]
  cnt=0
  for i in list(perm):
  if sum(i)%2==0:
  cnt+=1
  print(cnt)

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• 
  

  abhitejkumarvoruganti

  # Python3 program for the above approach

  # Function to return the number
  # of all permutations such that
  # sum of K numbers in range is even
  def countEvenSum(low, high, k):

  # Find total count of even and
  # odd number in given range
  even_count = high / 2 – (low – 1) / 2
  print(even_count,’even_count’)
  odd_count = (high + 1) / 2 – low / 2
  print(odd_count,’odd_count’)

  even_sum = 1
  odd_sum = 0

  # Iterate loop k times and update
  # even_sum & odd_sum using
  # previous values
  for i in range(0, k):

  # Update the prev_even and
  # odd_sum
  prev_even = even_sum
  print(prev_even,’prev_even’)
  prev_odd = odd_sum
  print(prev_odd,’prev_odd’)

  # Even sum
  even_sum = ((prev_even * even_count) +
  (prev_odd * odd_count))
  print(even_sum,’even_sum’)

  # Odd sum
  odd_sum = ((prev_even * odd_count) +
  (prev_odd * even_count))
  print(odd_sum,’odd_suma’)

  # Return even_sum
  print(int(even_sum))

  # Driver Code

  # Given ranges
  low = int(input());
  high = int(input());

  # Length of permutation
  K = int(input());

  # Function call
  countEvenSum(low, high, K);

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• 
  

  Niranjana

  Number series with twist-2 Python program solution is wrong. The correct answer should be:

  n = int(input(‘enter the number: ‘))

  a= 1

  b= 1

  for i in range(1, n+1):

  if(i%2!=0):

  a = a*2

  else:

  b = b*3

  if(n%2!=0):

  print(‘\n{} term of series is {}\t’.format(n,a/2))

  else:

  print(‘\n{} term of series is {}\t’.format(n,b/3))

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