Minimum Element - arrays

Question

You are given an array A[]. You need to divide this array into exactly K non-empty segments and check whether the minimum element S amongst the maximum elements amongst all segments is less than Q or not.
In other words, if we store the maximum element of each of the segment in an array P[], then you have to check if minimum element in P[] is less than Q or not.
Input:
First line of input contains number of testcases, T
First line of each testcase contains three integers:
N denoting number of elements in array A[], K denoting number of segments and Q.
Second line of each testcase contains
N space-separated integers, A1 , A2, ... An.
Output:
For each testcase, if S is less than Q, print S, else print "NO" (without quotes).
Answer for each test case should come in a new line.
Constraints:
1 ≤ T ≤ 10
1 ≤ N ≤ 10^5
1 ≤ K ≤ N
0 ≤ Q ≤ 10^9
1 ≤ A[i] ≤ 10^9
Here's my solution:
On analyzing, I thought that if we keep the minimum element of the array in a segment of size 1, then the minimum of maximum of all segments would be the minimum element of the array. Doing so, we would just need to compare the minimum element and Q, to print the required output.
public static void main(String[] args)
{
Scanner s=new Scanner(System.in);
int t = s.nextInt();
int n, k, q, i, min;
while (t-- > 0)
{
n = s.nextInt();
k = s.nextInt();
q = s.nextInt();
int[] a = new int[n];
for(i = 0; i < n; i++)
a[i] = s.nextInt();
min = 0;
for(i = 1; i < n; i++) {
if(a[i] < a[min]) {
min = i;
}
}
if(a[min] < q)
System.out.println(a[min]);
else
System.out.println("NO");
}
}
This solution satisfies all the test cases except one, and which is hidden. Can anyone please help me out on this one.

Answer 1

I see, this question has been asked in Capillary Software Hackathon on Hackerearth.
Well, the question is ambiguous. No clear definition of "k", also is it supposed to be evenly spilt or randomly etc
This code passes one case which you couldnt pass, but fails in all other cases:
t = int(raw_input())
for _ in range(t):
n,k,q = map(int,raw_input().split())
arr = map(int,raw_input().split())
qt = n/k
arr_new = []
for i in range(k):
max_num = max(arr[i*qt : (i+1)*qt])
arr_new.append(max_num)
min_num = min(arr_new)
if(min_num < q):
print min_num
else:
print "NO"

Answer 2

The question clearly states "You need to divide this array into exactly K non-empty segments and check whether the minimum element S amongst the maximum elements amongst all segments is less than Q or not."
You have not made use of "k" at all in your program, as pointed out in the comment.
Example:
n = 6
A[] = [17, 3, 2, 5, 8, 19]
k = 3
Q = 7
According to your code, since the min element is 2, the output will be 2 (2 < 7). But, the question demands that you first segment the array in k equal parts (3 in this case) and then find the minimum among maximum of segments AND THEN compare it with Q.
So, if we divide the array A[] in 3 segments {17,3}, {2,5}, {8,19} we see the min element of all max is 5. Since, 5 < 7, output will be 5.
This is based on what I understood from the question. However, it's not quite clear how segmenting should be done based on 'k'. I have assumed a sequential segmenting.