If O(k*logk) solution is acceptable then it is very simple.
maintain a min heap.
push a[0][0],
i = 0;
while ( i < k)
{
pop an element. ---> O(log(i)) --> coz number of elements in heap is i;
push the two adjacent element that is one right and right below. --->
O(log(i));
i++;
}
last element popped will be answer.
size of heap at every iteration will increase by one coz we are inserting
two elements and removing one element at every iteration.
If anyone has doubt on correctness of this solution can ask.
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