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@python_0
@python_0

Infosys TCS nqt solution
🔰🔰🔰🔰

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https://t.me/good_coders
https://t.me/good_coders
https://t.me/good_coders

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@python_0
@python_0

Python
49d

46b

32c

30 yes

45 d

28d

20c

39

39d

19d

12c

37a

13 b

fast

34 b

14 b

16c

17a

24 d

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20 relational

8order

15 d

14 c

18a

42d

43none

28img

48c

28img

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http://t.me/Coding_solution_0
http://t.me/Coding_solution_0
http://t.me/Coding_solution_0

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Revature Exam pattern

✅All solution Available at
http://t.me/Coding_solution_0

def minOps(A, B):
m = len(A)
n = len(B)

# This part checks whether conversion is possible or not
if n != m:
return -1

count = [0] * 256

for i in range(n): # count characters in A
count[ord(B[i])] += 1
for i in range(n): # subtract count for every char in B
count[ord(A[i])] -= 1
for i in range(256): # Check if all counts become 0
if count[i]:
return -1

# This part calculates the number of operations required
res = 0
i = n-1
j = n-1
while i >= 0:

# if there is a mismatch, then keep incrementing
# result 'res' until B[j] is not found in A[0..i]
while i>= 0 and A[i] != B[j]:
i -= 1
res += 1

# if A[i] and B[j] match
if i >= 0:
i -= 1
j -= 1

return res

# Driver program
A = "EACBD"
B = "EABCD"
print ("Minimum number of operations required is " + str(minOps(A,B)))


Python
Minimum number of operations required Code
✅
http://t.me/Coding_solution_0
http://t.me/Coding_solution_0
http://t.me/Coding_solution_0

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