Problem1:-
lets assume that we have an string ie “aaabbbccc”.
if the count of elements in the string are equal. we can say that it is an valid string else invalid string.
Let’s jump on the coding part:-
String ="aaabbbccc";
container=[]
element=[]
count=0
dict={}
for i in String:
if(container==[]):
container.append(i)
else:
if(i not in container):
container.append(i)
#The element are stored in container to get there count in the string
for i in container:
Counter=String.count(i);
dict.update( {i : Counter} )
print(dict)
bEqual=True
element.append(list(dict.values()))
print("dictionary values are saved in list:",element)
#nTemp is storing first value of the dictionary
nTemp = element[0][0]
print()
#so we are running an for loop for comparing the value of dictionary
for item in dict.values():
if nTemp != item:
bEqual = False
break;
if bEqual:
print("The given ",String,"is valid")
else:
print("The given ",String," is invalid")
Problem 2
Topic:-Longest common subsequence
Let’s jump on the coding part
import numpy as np
a=input("Enter string1:\t")
b=input("Enter string2:\t")
m=len(a)
n=len(b)
l = np.zeros([m+1,n+1],dtype=int)
#print("Matrix b : \n", l)
for i in range(m+1):
#print("string a",a[i-1])
for j in range(n+1):
#print("string b",b[j-1])
if i==0 or j==0:
l[i,j]=0
elif (a[i-1]==b[j-1]):
l[i,j]=1+l[i-1,j-1]
else:
l[i,j]=max(l[i-1,j],l[i,j-1])
index = l[m][n]
#print(index)
# Create a character array to store the lcs string
lcs = [""] * (index+1)
#print(lcs)
lcs[index] = ""
i = m
j = n
while i > 0 and j > 0:
# If current character in X[] and Y are same, then
# current character is part of LCS
if a[i-1] == b[j-1]:
lcs[index-1] = a[i-1]
i-=1
j-=1
index-=1
# If not same, then find the larger of two and
# go in the direction of larger value
elif l[i-1][j] > l[i][j-1]:
i-=1
else:
j-=1
print(l)
print ("The largest common subsequence is:\t","".join(lcs))
print("The letter in common are:\t ",l[-1][-1])
Harsh Rathod 