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2 952
Thinkitive Job Application Form Year 2023 !!!
Batch 2022/2021 &Experience
https://docs.google.com/forms/d/e/1FAIpQLSfOiQt0I5G2J5pp_tFRPpYb9ZVZzGMwFqQfPDQq_6_Vak_o1w/viewform
Telegram:- https://t.me/It_7sem
2 952
OKCL OFF Campus Hiring
Batch: 2019-2023
Salary: 3.9-9LPA
http://okcl.org/careers/project-trainee-recruitment-2023
For MBA Hiring(Managament Trainee)
Batch : 2023-2021
http://okcl.org/careers/management-trainee-recruitment-2022-23
2 952
Paytm is Hiring
Role: Android Developer ( Kotlin )
Experience: Fresher's to 1 year of exp.
CTC: 8 to 10 LPA
Apply Link: https://bit.ly/3kAz3Ws
Job Location: Noida
Join https://t.me/It_7sem on telegram for more useful material and off campus updates 🔥🔥🔥
2 952
Trade Desk is Hiring
Role: Software Engineer Intern
Qualification: BE/BTech/ME/MTech/MSc/MCA
Branch: CS / IT or related program
Batch: 2023, 2024
Experience: Fresher's
Apply Link: https://bit.ly/3XNkk9e
Job Location: Bengaluru
Join https://t.me/It_7sem on telegram for more useful material and off campus updates 🔥🔥🔥
2 952
TSS CONSULTANCY is Hiring
Role: Associate Business Analyst
Qualification: BTech / BE / BCA / BSc
Branch: CS/IT/ ECE or related streams
Apply Link: https://bit.ly/406lIWz
Job Location: Mumbai
Join https://t.me/It_7sem on telegram for more useful material and off campus updates 🔥🔥🔥
2 952
Thinkitive is hiring for junior software engineer (20, 21 and 22 batch eligible)
https://www.thinkitive.com/company/trainee-software-engineer-job-description.html
2 952
Don't worry jiska nhi hua h me uska next company me krva dunga clear test
Koi bhi mayus mt hona
TCS Wipro Infotech bhut sari company h nagarrow vgera kisi me bhi krva dunga sare company k code available h
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public class XOR {
public static int findMaxOverlappingSegments(int[] A) {
int N = A.length;
int[][] dp = new int[N+1][N+1];
for (int i = 1; i <= N; i++) {
dp[i][i] = 0;
}
for (int i = 1; i <= N; i++) {
for (int j = i+1; j <= N; j++) {
dp[i][j] = -1;
}
}
for (int len = 2; len <= N; len++) {
for (int i = 1; i <= N-len+1; i++) {
int j = i+len-1;
int xor = 0;
for (int k = i; k <= j; k++) {
xor = xor ^ A[k-1];
}
if (xor == 0) {
dp[i][j] = Math.max(dp[i][j-1], dp[i][j-1] + dp[j][j] + 1);
} else {
dp[i][j] = dp[i][j-1];
}
}
}
return dp[1][N];
}
public static void main(String[] args) {
int[] A = {1, 2, 3, 4, 5, 6, 7, 8, 9};
System.out.println(findMaxOverlappingSegments(A));
}
}
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Two basket
C++
// Online C++ compiler to run C++ program online
#include <iostream>
using namespace std;
void splitting(int basket1[], int basket2[],int arr[], int n){
int j = 0, k=0;
for(int i=0; i<n; i++){
if(i%2 ==0){
basket1[j++] =arr[i];
}
else {
basket2[k++] = arr[i];
}
}
}
void remove_duplicate_elements(int A[], int n){
int i,j,k;
for(i=0; i<n; i++){
for(j = i+1; j<n; j++){
if(A[j] == A[i]){
A[j] = -1;
}
}
}
}
int main() {
// Write C++ code here
int n;
cin>>n;
int arr[n];
int basket1[n], basket2[n];
for(int i=0; i<n; i++){
basket2[i] = -1;
basket1[i] = -1;
}
for(int i=0; i<n; i++){
cin>>arr[i];
}
splitting(basket1, basket2, arr, n);
remove_duplicate_elements(basket1, n);
remove_duplicate_elements(basket2, n);
int count1 = 0, count2 = 0;
for(int i=0; i<n; i++){
if(basket1[i] != -1){
count1++;
}
}
for(int i=0; i<n; i++){
if(basket2[i] != -1){
count2++;
}
}
std::cout<<count1+count2;
return 0;
}
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def min_jumps(n, A):
if A[0] == 0:
return -1
jumps = [float('inf')] * n
jumps[0] = 0
for i in range(1, n):
for j in range(i):
if i <= j + A[j] and jumps[j] != float('inf'):
jumps[i] = min(jumps[i], jumps[j] + 1)
break
return jumps[-1] if jumps[-1] != float('inf') else -1
def main():
n = int(input().strip())
A = []
for _ in range(n):
A.append(int(input().strip()))
result = min_jumps(n, A)
print(result)
if name == "main":
main()
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import java.util.Scanner;
public class Main {
public static final int MOD = (int)1e9 + 7;
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int T = sc.nextInt();
long sum = 0;
while (T-- > 0) {
int N = sc.nextInt();
int M = 0;
for (int i = 30; i >= 0; i--) {
if (((N >> i) & 1) == 1) {
M = (M << 1) + 1;
} else {
M = (M << 1);
}
}
sum = (sum + M) % MOD;
}
System.out.println(sum);
}
}
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def longest_subsequence(A):
n = len(A)
dp = [1] * n # create an array to store the longest subsequence ending at each index
for i in range(1, n):
for j in range(i):
if A[j] > A[i] and dp[j] + 1 > dp[i]:
dp[i] = dp[j] + 1
return max(dp)
