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#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>




int main(){
int t,count=0;
scanf("%d",&t);
for(int a0 = 0; a0 < t; a0++){
count=0;
int n;
int k;
scanf("%d %d",&n,&k);
int a[n];
for(int a_i = 0; a_i < n; a_i++){
scanf("%d",&a[a_i]);
}

for(int a_i = 0; a_i < n; a_i++){
if(a[a_i]<=0)
count++;
}
if(count<k)
printf("YES\n");
else
printf("NO\n");
//count=0;
}
return 0;
}


C language
Angry professor
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int main() {
char s[100];
int i;
printf("\nEnter a string : ");
gets(s);
for (i = 0; s[i]!='\0'; i++) {
if(s[i] >= 'a' && s[i] <= 'z') {
s[i] = s[i] - 32;
}
}
printf("\nString in Upper Case = %s", s);
return 0;
}

Warehouse Code
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int main() {
char s[100];
int i;
printf("\nEnter a string : ");
gets(s);
for (i = 0; s[i]!='\0'; i++) {
if(s[i] >= 'a' && s[i] <= 'z') {
s[i] = s[i] - 32;
}
}
printf("\nString in Upper Case = %s", s);
return 0;
}

Warehouse Code
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def findSubArray(arr, n):
sum = 0
maxsize = -1
for i in range(0, n-1):
sum = -1 if(arr[i] == 0) else 1
for j in range(i + 1, n):
sum = sum + (-1) if (arr[j] == 0) else sum + 1
if (sum == 0 and maxsize < j-i + 1):
maxsize = j - i + 1
startindex = i
if (maxsize == -1):
print(0);
else:
print(maxsize);

size = int(input())
arr = list(map(int, input().split()))
findSubArray(arr, size)

Python
Largest Sub Array
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58. All
59. 1024
60. Web page
61. All

62 Secure
63 Remote
64 Internet Ex

73 None
74 1 and 2
75 %temp%
TCS Interview questions:

1.Tell me about your self
2. About Project
3. Are you used any programming language in your project.
4. And explain about your project in depth.
5. Tell the syntax to print He'll world
6. How to print 4th index value from a string.
7. Can we write static public instant of public static in the main method why
8. JDK, JRE, JVM
9. Why JVM is platform dependant
10. What is DBMS
11. Upto where they reached in teaching
12. Why u choose TCS ?
13. are you able to work under high pressure
14. Any questions for the interviewer.

Questions are asked by civil engineer students.

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https://t.me/Accenture_lti_Accolite_exam