https://meet.google.com/ydq-hfwq-gdf
Be patient, Everything is coming Together.
Monday, October 14, 2024
Monday, August 5, 2024
Consider an automated Cancer Detector application(CDP).CDP performs 600 tests per day.Among 600 tests 400 tests were consider as positive but original positives are 370 and from the detected negatives 150 negatives were original negatives. Calculate the accuracy, precison, specificity, sensitivity, kappa score and error value.
Friday, July 19, 2024
Sunday, June 30, 2024
Saturday, June 29, 2024
Task
task1: until single digit occurred sum of all digits in the given number
task2: non reputed digits count in the given number
task3: difference between sum of even index values and sum of odd index values.
task4: non occurrence of digits in the given number
taks5: reputed occurrence of digits in the given number
task6: reputed occurrence of digits count in the given number
task6: find the factorial of the given number
task7: find the nth prime number
task8: find the nth fibnocci number.
taks9: find the factorial of the each digit in the given number
Tuesday, June 25, 2024
Monday, March 18, 2024
Tuesday, February 13, 2024
https://classroom.google.com/c/NjYzMTQwMzM1MTYx?cjc=4jfcjvo
George is very much interested in finding anagrams. One day, his professor gave him two strings S and C. George's task is to count the occurrences of anagrams of the string C in string S.
Write a program to help George to complete this task.
Input Format
The First line of input contains the string S.
The Next line contains the string C.
Refer to the sample input for formatting specifications.
Output Format
The Output prints the count of the occurrences of anagrams of the string C in string S.
Refer to the sample output for formatting specifications.
Constraints
1 <= |S| <= |C| <= 50
Sample 1 Input
forxxorfxdofr
for
Sample 1 Output
3
-----------------------------------------------------------------------------------------------------------------------------
IP Addresses
A valid IP address consists of exactly four integers separated by single dots. Each integer is between 0 and 255 (inclusive) and cannot have leading zeros.
For example, "0.1.2.201" and "192.168.1.1" are valid IP addresses, but "0.011.255.245", "192.168.1.312" and "192.168@1.1" are invalid IP addresses.
Given a string s containing only digits, return all possible valid IP addresses that can be formed by inserting dots into s. You are not allowed to reorder or remove any digits in s. You may return the valid IP addresses in any order.
Example 1:
Input: s = "25525511135"
Output: ["255.255.11.135","255.255.111.35"]
Example 2:
Input: s = "0000"
Output: ["0.0.0.0"]
Example 3:
Input: s = "101023"
Output: ["1.0.10.23","1.0.102.3","10.1.0.23","10.10.2.3","101.0.2.3"]
----------------------------------------------------------------------------------------------------------------------------
Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.
Symbol Value
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
For example, 2 is written as II in Roman numeral, just two one's added together. 12 is written as XII, which is simply X + II. The number 27 is written as XXVII, which is XX + V + II.
Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:
I can be placed before V (5) and X (10) to make 4 and 9.
X can be placed before L (50) and C (100) to make 40 and 90.
C can be placed before D (500) and M (1000) to make 400 and 900.
Given an integer, convert it to a roman numeral.
Example 1:
Input: num = 3
Output: "III"
Explanation: 3 is represented as 3 ones.
Example 2:
Input: num = 58
Output: "LVIII"
Explanation: L = 50, V = 5, III = 3.
Example 3:
Input: num = 1994
Output: "MCMXCIV"
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
---------------------------------------------------------------------------------------------------------
Balanced Paranthesis
Given an
expression string exp, write a program to examine whether the pairs
and the orders of “{“, “}”, “(“, “)”, “[“, “]” are correct in the given
expression.
Code:
#include <stdio.h>
#include <stdbool.h>
#include <string.h>
bool areBracketsBalanced(char s[],int n)
{
int i = -1;
char s1[n];
for (int j = 0; j < n; j++) {
if (s[j] == '(' || s[j] == '{' || s[j] == '[')
s1[++i] = s[j];
else {
if (i >= 0
&& ((s1[i] == '('
&& s[j] == ')')
|| (s1[i] == '{' &&
s[j] == '}')
|| (s1[i] == '[' &&
s[j] == ']')))
i--;
else
return false;
}
}
return i == -1;
}
int main()
{
char s[100];
scanf("%s",s);
int n=strlen(s);
if (areBracketsBalanced(s,n))
printf("Balanced\n");
else
printf("Not Balanced\n");
return 0;
}
----------------------------------------------------------------------------------------------------------------------------------------
Balanced Paranthesis
Given an
expression string exp, write a program to examine whether the pairs
and the orders of “{“, “}”, “(“, “)”, “[“, “]” are correct in the given
expression.
Code:
#include <stdio.h>
#include <stdbool.h>
#include <string.h>
bool areBracketsBalanced(char s[],int n)
{
int i = -1;
char s1[n];
for (int j = 0; j < n; j++) {
if (s[j] == '(' || s[j] == '{' || s[j] == '[')
s1[++i] = s[j];
else {
if (i >= 0
&& ((s1[i] == '('
&& s[j] == ')')
|| (s1[i] == '{' &&
s[j] == '}')
|| (s1[i] == '[' &&
s[j] == ']')))
i--;
else
return false;
}
}
return i == -1;
}
int main()
{
char s[100];
scanf("%s",s);
int n=strlen(s);
if (areBracketsBalanced(s,n))
printf("Balanced\n");
else
printf("Not Balanced\n");
return 0;
}
---------------------------------------------------------------------------------------------------------------------------------------
String
Replacement
Program Description
Take
a string as input. The input string consists of numbers also. Generate a new
string from the input string in such a way that the numbers in the string
should be replaced by a character which is generated by considering the
previous character in the following manner:
The
new character is generated by adding the ascii value of previous character with
the number and getting the character for the present location.
Example:
If the previous character is a and the number is 9 then take the 9th
character after a which is j and replace the number by j.
Sample Input 1
a4k3b2
Sample Output 1
aeknbd
Sample Input 2
a9x4z1
Sample Output 2
ajxbza
Sample Test Case 1
Input
a4k3b2
Output
aeknbd
Sample Test Case 2
Input
a9x4z1
Output
ajxbza
Hidden Test Case 1
Input
x4y3z2
Output
xbybzb
Hidden Test
Case 2
Input
a2b5c9d8
Output
acbgcldl
Hidden Test Case 3
Input
a5d9i3t6y2a1
Output
afdmiltzyaab
Hidden Test Case 4
Input
c7r9a5z8y1
Output
cjraafzhyz
C Program
#include<stdio.h>
#include<string.h>
int
main(){
char s[1000];
char s1[100];
int i,n,j=0;
s1[j]=0;
scanf("%s",s);
for(i=0;s[i]!='\0';i+=2){
if(isalpha(s[i])){
s1[j++]=s[i];
s1[j++]=(char)((s[i]-'a'+s[i+1]-'0')%26+97);
}
}
s1[j]='\0';
printf("%s",s1);
return 0;
}
JAVA Program
import java.io.*;
import java.util.*;
class Pattern
{
public static
void main(String args[])
{
Scanner
sc=new Scanner(System.in);
String
s=sc.next();
char
ch[]=s.toCharArray();
String
result="";
for(int
i=0;i<ch.length;i=i+2)
{
result=result+ch[i]+(char)((ch[i]-'a'+ch[i+1]-'0')%26+97);
}
System.out.println(result);
}
}
Python Program
s=input()
res=""
for
i in s:
if i.isalpha():
res+=i
x=i
else:
d=chr((ord(x)-ord('a')+ord(i)-ord('0'))%26+97)
res+=d
print(res)
-----------------------------------------------------------------------------------------------------------------------------
Run Length Decoding
Take
a string as input. The string consists of a character succeeded by a number.
Now you need to generate a new string as output in such a way that it is formed
by repeating each character for the number of times it is succeeded by the
given number.
Sample Input 1
a4b3c2
Sample Output 1
aaaabbbcc
Sample Input 2
a3b5
Sample Output 2
aaabbbbb
Hidden Test Case 1
Input
a5b3c2d1e2
Output
aaaaabbbccdee
Hidden Test Case 2
Input
j1a2g3
Output
jaaggg
Hidden Test Case 3
Input
T1r2a3i1n1i1n2g3
Output
Trraaaininnggg
C Program
#include<stdio.h>
#include<string.h>
int
main(){
char s[1000];
char s1[100];
int i,n,j=0;
scanf("%s",s);
for(i=0;s[i]!='\0';i+=2){
n=s[i+1]-'0';
while(n>0){
s1[j++]=s[i];
n--;
}
}
s1[j]='\0';
printf("%s",s1);
return 0;
}
JAVA
Program
import java.util.*;
class Pattern1
{
public static
void main(String args[])
{
Scanner
sc=new Scanner(System.in);
String
s=sc.next();
char
ch[]=s.toCharArray();
String
result="";
for(int
i=0;i<ch.length;i=i+2)
{
int
n=Integer.parseInt(ch[i+1]+"");
while(n>0)
{
result=result+ch[i];
n--;
}
}
System.out.println(result);
}
}
Python Program
s=input()
result=""
for
i in s:
if i.isalpha():
result+=i
x=i
else:
d=int(i)
newc=x*(d-1)
result+=newc
print(result)
---------------------------------------------------------------------------------------------------------------------
Run Length Encoding
Take
a string as input. The string consists of a character in succession. Now you
need to generate a new string as output in such a way that it is formed by
repeating each character followed by number of times it is repeated in the
string in the order.
Sample Input 1
a4b3c2
aaaabbbcc
Sample Output 1
a4b3c2
Sample Input 2
aaabbbbb
Sample Output 2
a3b5
Hidden Test Case 1
Input
aaaaabbbccdee
Output
a5b3c2d1e2
Hidden Test Case 2
Input
jaaggg
Output
j1a2g3
Hidden Test Case 3
Input
Trraaaininnggg
Output
T1r2a3i1n1i1n2g3
Code:
#include <stdio.h>
#include<string.h>
#include<stdlib.h>
int main() {
char str[100];
scanf("%s",str);
int len = strlen(str);
char current_char = str[0];
int count = 1;
for (int i = 1; i <= len; i++) {
if (str[i] == current_char) {
count++;
} else {
printf("%c%d",
current_char, count);
current_char = str[i];
count = 1;
}
}
return 0;
}
---------------------------------------------------------------------------------------------------------------------------------------
Reverse String
Problem Description
Given
a string as input you need to print the reverse of the words without reversing
the characters. Example: If the given string is Aditya Engineering College then
the program should print College Engineering Aditya
Sample Input 1
Technical
Training
Sample Output 1
Training
Technical
Sample Input 2
Aditya
Sample Output 2
Aditya
Hidden Test Case 1
Input
Information
Technology
Output
Technology
Information
Hidden Test Case 2
Input
Computer
Science & Engineering
Output
Engineering
& Science Computer
Hidden Test Case 3
Input
Lock
down
Output
down
Lock
Hidden Test Case 4
Input
A wise
man once said not to go with intuition go with guts
Output
guts
with go intuition with go to not said once man wise A
C Program
#include<stdio.h>
#include<string.h>
int
main(){
char str[100],text[100];
int i=0,j=0;
gets(str);
while(str[i]!='\0')
i++;
while(i>0){
text[j]=str[--i];
++j;
}
text[j]='\0';
for(i=0;text[i]!='\0';i++){
if(text[i+1]==' ' || text[i+1]==NULL){
for(j=i;j>=0 &&
text[j]!=' ';j--)
printf("%c",text[j]);
printf("
");
}
}
return
0;
}
JAVA Program
import
java.util.regex.Pattern;
import
java.util.*;
public
class Exp {
static
String reverseWords(String str)
{
Pattern
pattern = Pattern.compile("\\s");
String[]
temp = pattern.split(str);
String
result = "";
for
(int i = 0; i < temp.length; i++) {
if
(i == temp.length - 1)
result
= temp[i] + result;
else
result
= " " + temp[i] + result;
}
return
result;
}
public
static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
String
s1 = sc.next();
System.out.println(reverseWords(s1));
}
}
Python Program
S=input()
Words=S.split(“
“)
Sentence=”
“.join(reversed(words))
print(Sentence)
---------------------------------------------------------------------------------------------------------------------------------------
Chain of
pearls
Program Description
Saif
likes kareena and he presented her a chain of pearls. The pearls are red and
green in colors. Saif likes chains in which the red pearls and green pearls alternate
each other i.e., R followed by G or G followed by R. So he wants to replace
some of the pearls in the chain to suit his liking.
Let the red pearl be
denoted as 'R' and let the green pearl be denoted as 'G'. For example 'RGRGRGR' is a Chain of Saif’s
liking but 'RGTGRRG' is not. Help Saif to calculate the minimum number of
pearls he need to replace (ex. 'R' to 'G' or 'G' to 'R') to get a chain that he
would like.
Input Format
Input consists of a single string consisting of only the letters 'R' and 'G'.
Assume that the maximum length of the input string is 75.
Output Format
Output consists of a single interger - the minimal number of pearls that Saif
need to replace.
Sample Input 1
RRRGRGRGGG
Sample Output 1
2
Sample Input 2
GGGGGGG
Sample Output 2:
3
Explanation
Example case 1.
We can change symbol 2 from 'R' to 'G' and symbol 9 from 'G' to 'R' and receive
'RGRGRGRGRG'.
Example case 2.
We can change symbols 2, 4 and 6 from 'G' to 'R' and receive 'GRGRGRG'.
Sample
Test Case 1
Input
RRRGRGRGGG
Output
2
Sample Test Case 2
Input
GGGGGG
Output
3
Hidden Test Case 1
Input
RGRGRGRG
Output
0
Hidden Test Case 2
Input
GGGRRR
Output
2
Hidden Test Case 3
Input
RRRRGRGRGRRGRG
Output
6
Hidden Test Case 4
Input
RGRGRGGRGRGRGRGRGRGRGRGRRGRGRGGRGRRG
Output
14
C Program
#include<stdio.h>
#include<string.h>
int
main(){
char s[75];
int i,n,count=0,x;
scanf("%s",s);
n=strlen(s);
for(i=0;i<n;i++){
if(i%2==0 && s[i]=='R'){
count++;
}
else if(i%2==1 && s[i]=='G'){
count++;
}
}
if(count<n-count)
x=count;
else
x=n-count;
printf("%d\n",x);
return 0;
}
JAVA Program
import
java.io.*;
import
java.util.*;
class
Pearls{
public static void main(String args[]){
int i,n,count=0,x;
Scanner sc=new Scanner(System.in);
String s=sc.nextLine();
n=s.length;
for(i=0;i<n;i++){
if(i%2==0 && s[i]=='R'){
count++;
}
else if(i%2==1 &&
s[i]=='G'){
count++;
}
}
if(count<n-count)
x=count;
else
x=n-count;
System.out.println(x);
}
}
Python Program
s=input()
count=0
n=len(s)
for i in range(n):
if(i%2==0 && s[i]=='R'):
count+=1
else if(i%2==1 && s[i]=='G'):
count+=1
if(count<n-count):
x=count
else:
x=n-count
print(x)
------------------------------------------------------------------------------------------------------------------
Panagram
String
Program Description
Given a string as input you need to
determine whether the string is panagram or not. If the string is panagram
print the string as “Panagram String.” Otherwise print “Not a Panagram
String.”. A string is called panagram if it contains all the alphabets in
english as the characters in it.
Note: The input string contains boh uppercase and lowercase
characters in it.
Sample Input 1
ABcde fghIJkl MnopQr StuvWxyz
Sample Output 1
Panagram String.
Sample input 2
The five members of the team are so
annoying.
Sample Output 2
Not a Panagram String.
Sample Test Case 1
Input
ABcde fghIJkl MnopQr StuvWxyz
Output
Panagram
String.
Sample Test Case 2
Input
The five members of the team are so
annoying.
Output
Not
a Panagram String.
Hidden Test Case 1
Input
The
five boxing wizards jump quickly.
Output
Panagram
String.
Hidden Test Case 2
Input
The
quick brown fox jumps over the lazy dog.
Output
Panagram
String.
Hidden Test Case 3
Input
John quickly
extemporized five tow bags.
Output
Panagram String
Hidden Test Case 4
Input
The
Vixens of the hazardious grimp.
Output
Not
a Panagram String.
C Program
#include<stdio.h>
#include<string.h>
void
main()
{
char s[100];
int i,used[26]={0},total=0;
gets(s);
for(i=0;s[i]!='\0';i++)
{
if('a'<=s[i] && s[i]<='z')
{
total+=!used[s[i]-'a'];
used[s[i]-'a']=1;
}
else if('A'<=s[i] &&
s[i]<='Z')
{
total+=!used[s[i]-'A'];
used[s[i]-'A']=1;
}
}
if(total==26)
{
printf("Panagram String.");
}
else
{
printf("Not a Panagram
String.");
}
return 0;
}
JAVA Program
import
java.io.*;
import
java.util.*;
import
java.text.*;
import
java.math.*;
import
java.util.regex.*;
public
class Panagram {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
String sentence = scan.nextLine();
sentence = sentence.toUpperCase();
sentence =
sentence.replaceAll("[^A-Z]", "");
char[] chars = sentence.toCharArray();
Set<Character> set = new
HashSet<Character>();
for( int i = 0; i < chars.length;
i++ ) set.add(chars[i]);
System.out.println(set.size() == 26 ?
"Panagram String" : "Not a Pangram String");
}
}
Python Program
import
string
alpha=set(string.ascii_lowercase)
s=input()
if(set(s.lower())>=alpha):
print("Panagram string")
else:
print("Not a Panagram String")
------------------------------------------------------------------------------------------------------------------
String Occurence
Raja
has been given 2 sentences. He needs to do a task with those 2 sentences. He
needs to tell how many times the first sentence appeared in the second sentence
which is called as the target along with the fist sentence. Help Raja in
completing the task.
Input:
2
lines consists of 2 sentences
Constraints:
1<=s1,s2<=100
Output:
A
single line consists of first sentence and the count of occurences of it in
second sentence.
Sample Testcase1:
Input:
are
youareveryrarepicturearenowthare
Output:
are 4
Sample Testcase2:
Input:
is
thisisveryclearideaishowisyourhealthandisforthisnowtothis
Output:
is 7
Hidden Testcase1:
Input:
is
thisishowis
Output:
is 3
Hidden Testcase2:
Input:
are
therearearenoware
Output:
are 3
Hidden Testcase3:
Input:
for
formeforyouforeveryone
Output:
for 3
Hidden Testcase4:
Input:
ride
ridersgoingforarideanditsalongestrideofallrides
Output:
ride 4
C Code:
#include
<stdio.h>
#include
<string.h>
char
str[100], sub[100];
int
count = 0, count1 = 0;
void
main()
{
int i, j, l, l1, l2;
scanf("%s%s", sub,str);
l2 = strlen(sub);
l1 = strlen(str);
for (i = 0; i < l1;)
{
j = 0;
count = 0;
while ((str[i] == sub[j]))
{
count++;
i++;
j++;
}
if (count == l2)
{
count1++;
count = 0;
}
else
i++;
}
printf("%s %d", sub, count1);
}
Java Code:
import
java.util.*;
class
StringOccur
{
public static void main(String args[])
{
Scanner sc=new Scanner(System.in);
String s1=sc.next();
String s2=sc.next();
int l=s2.length();
int m=l-
s2.replaceAll(s1,"").length();
System.out.println(s1+"
"+(int)m/s1.length());
}
}
Python
Code:
pat=input()
text=input()
M = len(pat)
N = len(text)
res = 0
for i in range(N - M + 1):
j = 0
for j in range(M):
if (text[i + j] != pat[j]):
break
if (j == M - 1):
res += 1
j = 0
print(pat,res)
------------------------------------------------------------------------------------------------------------------
String Anagrams
Sudeep
is well known for his ability in coding in his college. A district wise coding
competition is being held in the city. The competition allows the participants
to take help from others if they are unable to solve it. Sudeep has taken his
friend vijay. The task given in competition is like this: read two strings as
an input. you need to check the number of times every character repeated in
first string is exactly same as the number of times the same characters
repeated in other string. Then you need
to print “YES” otherwise we need to print “NO”.
Now Sudeep asked vijay to do the task. Vijay is dazzled by seeing the
code. Now you need to help vijay in solving the code so that they can go
through further rounds.
Sample Testcase:
Input:
3
listen
silent
liril
river
top
pot
Output:
YES
NO
YES
Sample Input:
2
Raja
raj
ravi
arvi
Output:
NO
YES
Testcase1:
Input:
4
funeral
realfun
theeyes
thaysee
agentleman
elegantman
convers
voices
Output:
YES
NO
YES
NO
Testcase2:
Input:
2
schoolmaster
theclassroom
astronomer
moonstarer
Output:
YES
YES
Testcase3:
Input:
3
ramu
murali
raju
arju
siva
vikas
Output:
NO
YES
NO
Test Case 4:
Input:
4
sari
rasi
send
dens
settle
seattle
jeera
rajee
Output:
YES
YES
NO
YES
C Code:
#include<stdio.h>
#include<string.h>
int
isnagram(char[],char[]);
int
isanagram(char ch[],char ch1[]){
int i,j,temp=0,temp1=0;
int n1=strlen(ch);
int n2=strlen(ch1);
if(n1!=n2){
return 0;
}
for(i=0;i<n1;i++){
for(j=i+1;j<n1;j++){
if(ch[i]>ch[j]){
temp=ch[i];
ch[i]=ch[j];
ch[j]=temp;
}
if(ch1[i]>ch1[j]){
temp1=ch1[i];
ch1[i]=ch1[j];
ch1[j]=temp1;
}
}
}
for(i=0;i<n1;i++){
if(ch[i]!=ch1[i]){
return 0;
}
}
return 1;
}
int
main(){
int t,i,j;
scanf("%d",&t);
for(i=0;i<t;i++){
char
ch[100],ch1[100],temp=0,temp1=0;
scanf("%s%s",ch,ch1);
if(isanagram(ch,ch1)){
printf("YES\n");
}
else{
printf("NO\n");
}
}
}
Java Code:
import
java.util.*;
class
Anagrams
{
public static void main(String args[])
{
Scanner sc=new
Scanner(System.in);
int testcases;
testcases=sc.nextInt();
while(testcases-->0)
{
String s1,s2;
s1=sc.next();
s2=sc.next();
char ch1[]=s1.toCharArray();
char ch2[]=s2.toCharArray();
Arrays.sort(ch1);
Arrays.sort(ch2);
if(Arrays.equals(ch1,ch2))
System.out.println("YES");
else
System.out.println("NO");
}
sc.close();
}
}
Python Code:
def areAnagram(str1, str2):
n1 = len(str1)
n2 = len(str2)
if n1 != n2:
return 0
str1 = sorted(str1)
str2 = sorted(str2)
for i in range(0, n1):
if str1[i] != str2[i]:
return 0
return 1
t=int(input())
for i in range(t):
str1 = input()
str2 = input()
if areAnagram(str1, str2):
print ("YES",end="\n")
else:
print ("NO",end="\n")
------------------------------------------------------------------------------------------------------------------
Vowel Count
Raju is good at handling strings. Now
Sasi has given a task to him. Sasi gives n number of strings to Raju. Now Raju
has to find out the count of the number of vowels present in each string. Help
Raju in completing the task.
Note:
Characters will be in both upper and lowercase. So irrespective of case the
count of vowel should be taken
Input:
N: Number of strings
Each Ni lines consists of a string
Constraints:
1<=N<=10
Output:
N lines each consists of count of vowels
in each Nth string
Sample Test Case 1:
Input:
3
srinu
ramesh
aditya
Output:
2
2
3
Sample Test Case 2:
Input:
2
AdityA
SriAditya
Output:
3
4
Hidden Test Case 1:
Input:
5
adityaengineeringcollege
RaOgopAlRAO
MangO
Apple
Banana
Output:
11
6
2
2
3
Hidden Test Case 2:
Input:
2
adityagroupofengineeringcollegesarelocatedatsurampalemCAMPUSANDHRAPRADESH
YES
Output:
30
1
Hidden Test Case 3:
Input:
6
JagadEesh
SrInU
UDAYbhaskAr
VinayKumar
Vanathi
PrAvEeN
Output:
4
2
4
4
3
3
Hidden Test Case 4:
Input:
3
Arya
Ajith
Arun
Output:
2
2
2
C Code:
#include <stdio.h>
#include<string.h>
#include<stdlib.h>
int main()
{
int c = 0,
count = 0,n,i,j;
scanf("%d",&n);
char
ch[n][100];
for(i=0;i<n;i++){
scanf("%s",ch[i]);
}
for(i=0;i<n;i++){
count=0;
for(j=0;j<100;j++){
if(ch[i][j]=='a'||ch[i][j]=='A'||ch[i][j]=='e'||ch[i][j]=='E'||ch[i][j]=='i'||ch[i][j]=='I'||ch[i][j]=='o'||ch[i][j]=='O'||ch[i][j]=='u'||ch[i][j]=='U'){
count++;
}
}
printf("%d\n",count);
}
return 0;
}
Java Code:
import java.util.*;
class VowelCount
{
public
static void main(String args[])
{
Scanner
sc=new Scanner(System.in);
int
num,i,j,count=0;
String
word,vowels="aeiouAEIOU";
char
ch[];
num=sc.nextInt();
for(i=0;i<num;i++)
{ count=0;
word=sc.next();
ch=word.toCharArray();
for(j=0;j<ch.length;j++)
{
if(vowels.contains(ch[j]+""))
count++;
}
System.out.println(count);
}
}
}
Python Code:
n=int(input())
a="aeiouAEIOU"
for i in range(n):
s=input()
count=0
for i in
s:
if i
in a:
count+=1
print(count)
------------------------------------------------------------------------------------------------------------------
String Equality
Sanath is a well versed student in the college. A new principal came to college and he gets to know about the knowledge of Sanath. He want’s to test Sanath and he has given a task like this: Given a sentence he needs to determine whether the string contains equal number of lowercase alphabets, upper case alphabets, digits and symbols. If so he needs to say “Equality For Everyone” otherwise he needs to say “No Equality”. Sanath wants to prove his ability to the new principal. So help Sanath.
Input:
A single line consisting of input sentence
Output:
A single line either display “Equality For Everyone” or “No Equality”
Sample Input 1:
aB1$
Sample Output 1:
Equality For Everyone
Sample Input 2:
ab23$%
Sample Output 2:
No Equality
Hidden Testcase 1:
Input:
a program to find the given string contains equal number of lowercase alphabets, upper case alphabets, digits and symbol123434023489274@#$%^!@#$%^(*&12345667897654333444
Output:
No Equality
Hidden Testcase 2:
Input:
abcd23$#,*46
Output:
No Equality
Hidden Testcase 3:
Input:
12ab32bc$#CB*^BC
Output:
Equality For Everyone
Hidden Test Case 4:
Input:
11223344aabbccddAABBCCDD$#$#$#$#
Output:
Equality For Everyone
C Code:
#include<stdio.h>
#include<string.h>
int main(){
int l=0,u=0,d=0,s=0,i;
char x[1000];
scanf("%s",x);
for(i=0;x[i]!='\0';i++)
{
if(islower(x[i]))
l++;
else if(isupper(x[i]))
u++;
else if(x[i]>='0' && x[i]<='9')
d++;
else
s++;
}
if(l==u && u==d && d==s)
printf("Equality For Everyone");
else
printf("No Equality");
}
Java Code:
import java.util.*;
class Equality
{
public static void main(String args[])
{
Scanner sc=new Scanner(System.in);
int l=0,u=0,d=0,s=0;
String str=sc.next();
char x[]=str.toCharArray();
for(int i=0;i<x.length;i++)
{
if(Character.isLowerCase(x[i]))
l++;
else if(Character.isUpperCase(x[i]))
u++;
else if(Character.isDigit(x[i]))
d++;
else
s++;
}
if(l==u && u==d && d==s)
System.out.println("Equality For Everyone");
else
System.out.println("No Equality");
}
}
Python Code:
s=input()
u=l=d=t=0
for i in s:
if i.isupper():
u+=1
elif i.islower():
l+=1
elif i.isdigit():
d+=1
else:
t+=1
if(l==u and u==d and d==t):
print("Equality For Everyone")
else:
print("No Equality")
https://meet.google.com/ydq-hfwq-gdf
-
Design LALR Bottom up Parser. <parser.l> %{ #include<stdio.h> #include "y.tab.h" %} %% [0-9]+ {yylval...
-
Lucky Gifts "Planet Kids Entertainment Fair" is back to delight kids and parents. The Fair will have non-stop entertainment w...