Уже доступно! Исследование Telegram 2025 — ключевые инсайты года 
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*IBM Hiring for Associate Systems Engineer| 2024 - BE/BTECH/MCA/MTech/MSC-CS,IT,ECE,EEE Only |Location: Bangalore |Package: 4.5LPA + 25K*
*Only for 2024 Graduates…🎓*
*Official Links 🔗:*
Registration Process:
• Step 1. Apply to the job opening below on
IBM career portal:
Link - https://ibm.biz/BdGs58
• Step 2. Note the Person ID which will be shared in an email from IBM Talent Acquisition < talent@ibm.com>. Ensure you keep this ID handy for the next step.
• Step 3. Go to the link below and complete the radancy form.
•
Link - https://ibm.biz/BdGs5s
Kindly ensure that you provide accurate academic information in the Radancy form.
*Apply Asap 🔥*
*Note: Apply asap. Don’t Delay*
n=int(input())
arr=[int(i) for i in input().split()]
k=int(input())
print(*(arr[-k:]+arr[:-k]))
+3
WIPRO 12PM SLOT DONE SUCCESSFULLY ✅
Contact: @MLCODER2
x,y,z=map(int,input().split())
print(sum(sorted([x,y,z])[1:]))
#sum_of_maximum_two_numbers
print(sum(int(i) for i in input().split() if int(i)<0))
#sum of negative numbers
n=int(input())
arr=[int(i) for i in input().split()]
p=[i for i in arr if i>0]
mid=len(p)//2
if mid%2==0:print(p[mid-1])
else:print(p[mid])
#positviemidindex
def solve(s):
for i in range(1, len(s)):
s1, s2 = s[:i], s[i:]
if s1.count('a') == len(s1) and s2.count('b') == len(s2):
return True
return False
n=int(input())
ans=0
for _ in range(n):
s=input()
if solve(s):
ans+=1
print(ans)
#special string
#wipro
+1
IBM OFFLINE CODING + RECORDED VIDER ROUND + GD Cleared ✅✅
NOW FINAL INTERVIEW HAS TO BE CLEARED ✅✅✅
100% CLEARANCE FOR ANY PLACEMENT EXAM & INTERVIEW ✅
Contact: @MLCODER2
IBM SVAR ROUND CLEARED & GOT MAIL FOR OFFLINE CODING ✅✅
100% clearance for any exam ✅
Contact: @MLCODER2
IBM SVAR ROUND CLEARED & GOT MAIL FOR OFFLINE CODING ✅✅
100% clearance for any exam ✅
Contact: @MLCODER2
