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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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Rooted tree with N nodes code
An array A of N integer code
Bob received n job invitations code
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Java
Maximum possibility of unique digits in k divided subarrray of given arraylist
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n=int(input())
m=int(input())
s=input()
t=input()
def subString(s,n):
c=0
for i in range(n):
for len in range(i+1,n+1):
a=s[i: len]
c+=1
return c-2
if s==t:
print(1)
else:
if t in s:
print(subString(s,n))
else:
print(0)
Python
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Python
Rooted tree with N nodes code
An array A of N integer code
Bob received n job invitations code
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https://t.me/Coding_solution_0
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https://t.me/Coding_solution_0
n = int(input())
a = list(map(int,input().split()))
c = 0
for i in range(n-1):
for j in range(i+1:n):
s1 = sum(list(map(int,bin(i)[2:].split())))
s2 = sum(list(map(int,bin(j)[2:].split())))
if s1 == s2:
c += 1
print(c)
Python
Bob and Numbers Code
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n = int(input())
a = list(map(int,input().split()))
c = 0
for i in range(n-1):
for j in range(i+1:n):
s1 = sum(list(map(int,bin(i)[2:].split())))
s2 = sum(list(map(int,bin(j)[2:].split())))
if s1 == s2:
c += 1
print(c)
Python
Bob and Numbers Code
Telegram- https://t.me/Coding_solution_0
