Crack the IBM Coding Round in Just 20 Mins! 🚀 | Coding Series🔥
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- Опубліковано 5 лют 2025
- Did You Just Say Discounts? 🙌💸
📌𝗨𝘀𝗲 𝗖𝗼𝗱𝗲 - 𝗬𝗢𝗗𝗛𝗔 (𝗢𝗳𝗳𝗲𝗿 𝗼𝗻𝗹𝘆 𝗳𝗼𝗿 𝗙𝗶𝗿𝘀𝘁 𝟱𝟬 𝗦𝘁𝘂𝗱𝗲𝗻𝘁𝘀 𝗼𝗻𝗹𝘆)
𝐍𝐞𝐰 𝐘𝐨𝐝𝐡𝐚 𝐁𝐚𝐭𝐜𝐡 𝐟𝐨𝐫 𝟐𝟎𝟐𝟒,𝟐𝟑,𝟐𝟐,𝟐𝟏 & 𝐎𝐭𝐡𝐞𝐫 𝐏𝐚𝐬𝐬𝐢𝐧𝐠 𝐁𝐚𝐭𝐜𝐡:
bfwhv.courses....
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Did You Just Say Discounts? 🙌💸
𝗨𝘀𝗲 𝗖𝗼𝗱𝗲 - 𝗦𝗔𝗡𝗞𝗔𝗟𝗣 (𝗢𝗳𝗳𝗲𝗿 𝗼𝗻𝗹𝘆 𝗳𝗼𝗿 𝗙𝗶𝗿𝘀𝘁 𝟮𝟬 𝗦𝘁𝘂𝗱𝗲𝗻𝘁𝘀 𝗼𝗻𝗹𝘆)
𝐍𝐞𝐰 𝐒𝐚𝐧𝐤𝐚𝐥𝐩 𝟑.𝟎 𝐁𝐚𝐭𝐜𝐡 𝐟𝐨𝐫 𝟐𝟎𝟐𝟓:
bfwhv.courses....
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s = input("Enter a string: ")
a = [0] * 255
for i in range(0, len(s), 2):
a[ord(s[i])] += int(s[i + 1])
ans = ""
for i in range(255):
if a[i] != 0:
ans += chr(i)
ans += str(a[i])
print("Output:", ans)
Thanks a million
code for binary to decimal
#include
using namespace std;
int main(){
int binary=111,decimal=0,x=0;
while(binary>0){
int rem=binary%10;
decimal+=pow(2,x);
x++;
binary=binary/10;
}
cout
Integer.toBinaryString(64);
in Java. C++ has something similar?
question 1 in java
import java.util.*;
public class ibm2 {
public static void main(String []args){
Scanner sc=new Scanner(System.in);
String s=sc.nextLine();
int ch1[]=new int[255];
for(int i=0;i
** Ques-1 : JAVA Solution using single loop **
import java.util.HashMap;
import java.util.Map;
import java.util.stream.Collectors;
public class Test7 {
public static void main(String[] args) {
String compStr = "a3b2c5a1g9c2";
Map map = new HashMap();
for (int i = 0; i < compStr.length(); i+=2) {
if (!map.containsKey(compStr.charAt(i))) {
map.put(compStr.charAt(i), Character.getNumericValue(compStr.charAt(i+1)));
} else {
map.replace(compStr.charAt(i), map.get(compStr.charAt(i)) + Character.getNumericValue(compStr.charAt(i+1)));
}
}
String correctStr = map.entrySet().stream().map(e -> e.getKey()+""+e.getValue()).collect(Collectors.joining(""));
System.out.println(correctStr);
}
}
s = input("Enter: ")
chk_array = [0]*255
res = ""
for i in range(1,len(s),2):
chk_array[ord(s[i-1])] += int(s[i])
for i in range(len(chk_array)):
for j in range(chk_array[i]):
res += chr(i)
print(res)
What if the frequency is given in more than 1 digit? Like a5g14c6b1a12b3
Good question bhai👍
we have to parse the value like:
char key = s[i]; // The character key
i++; // Move to the next character which should be the start of the number
// Extract the number which may have multiple digits
int value = 0;
while (i < s.length() && isdigit(s[i])) {
value = value * 10 + (s[i] - '0');
i++;
}
Binary to decimal:
s = int(input('Enter: '))
res = 0
counter = 0
while s!=0:
res += (2**counter)*(s%10)
s = s//10
counter+=1
print("Output:", res)
My personal laptop so slow...can i give hackerrank exam in wipro laptop?
Hi Team, I had IBM coding round on 31st of August (ON_Campus) and questions were super huge and complex. Some required concepts from DP too.
which clg
@@20-ECE-145ShivamKumar Tier 3 college of Kolkata-
Can you please share the questions ?
Was the exam proctored? Can we refer syntax
Please tell me about the exam
I gave coding assessment for back-end developer I completed Java but while writing half of the sql my time over... So can I get selected for next English round
what was the question can u tell i am written today
Was the exam proctored? Can we refer syntax
Thank you sir
Kya hum javascript me code kr skte hai
Turner Corner
For 1st question we can use this solution :
import Foundation
var mainArr = ["g","2","b","4","a","2","c","5","a","1","Z","7","A","4"]
var ansArr = [String]()
for var i in 0..0 {
ansArr.append(name)
ansArr.append(String(count))
}
}
print("Hi:",ansArr )
print("Hello World")
//program to convert decimal number to binary number
import java.util.*;
public class ibm3 {
public static void main(String []args){
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
int rem=0;
StringBuffer sb=new StringBuffer();
while(n>0){
rem=n%2;
sb.append(rem);
n=n/2;
}
sb.reverse();
System.out.println(sb);
}
}
Bhaiya 1 ka ye dekha jara
s="a3b2c5a1g9c2"
d={}
for i in range(0,len(s),2):
char=s[i]
freq=int(s[i+1])
if char not in d:
d[char]=freq
else:
d[char]+=freq
res=""
for char in sorted(d.keys()):
res+=char
res+=str(d[char])
print(res)
Good morning Sir
Good Morning
** Ques -2 : JAVA Solution **
import java.util.Scanner;
public class Test8 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int num = sc.nextInt();
sc.close();
String binNum = "";
while (num != 1) {
binNum += num % 2;
num /= 2;
}
binNum += num;
System.out.println(new StringBuilder(binNum).reverse().toString());
}
}
Python solution using dict:
import string
st = input('Enter incorrectly compressed string: ')
dt = {}
for i in st:
if i.isalpha():
dt[i] = 0
for i in range(len(st)):
if st[i].isalpha():
dt[st[i]] += int(st[i+1])
print(dt)
ls_keys = list(dt.keys())
ls_keys.sort()
dt_sorted = {i:dt[i] for i in ls_keys}
ls_sorted_keys = list(dt_sorted.keys())
ls_vals = list(dt_sorted.values())
print(dt)
correct_st = ''
dt_len = len(dt)
print(dt_len)
for i in range(dt_len):
correct_st = correct_st + ls_keys[i] + str(ls_vals[i])
print(correct_st)
brother dont you think this code wont work when frequency is given in more than one digit like a10b5a2c7
st = 'a3b2c4a4d6b5'
l = list(st)
d={}
srr=''
lt=[]
l1=l[::2]
l2 = set(l1)
l3=list(l2)
l3.sort()
for i in l3:
d[i]=l1.count(i)
ll=list(d.items())
ln=len(ll)*2
for j in ll:
for i in j:
lt.append(i)
for t in lt:
srr=srr+str(t)
print(srr)
##python code
//program to convert binary number to decimal number
import java.util.*;
public class ibm4 {
public static void main(String []args){
Scanner sc=new Scanner(System.in);
String s=sc.nextLine();
StringBuffer sb=new StringBuffer(s);
sb.reverse();
s=new String(sb);
int n=Integer.parseInt(s);
int rem=0;
double dec=0;
int i=0;
while(n>0){
rem=n%10;
dec=dec+rem*Math.pow(2,i);
i++;
n=n/10;
}
int ans=(int)dec;
System.out.println(ans);
}
}