arr = [1, 3, 2, 1], m = 3, k = 2
m represents a group size
k represents the kth minimum element in the group of size m
so find the 2nd minimum element in a group of size 3.
There are 2 contiguous groups possible:
[1, 3, 2] and [3, 2, 1]
k = 2 , 2nd minimum element in above groups is [2,2]
public static int[] solve(int[] arr, int m, int k) {
List result = new ArrayList();
int n = arr.length;
for(int i=0; i Integer.compare(b, a));
for (int i = 0; i < m; i++) {
q.add(arr[i]);
if (q.size() > k) {
q.poll();
}
}
result[0] = q.peek();
for (int i = 1; i
Подробнее здесь: [url]https://stackoverflow.com/questions/78180982/find-min-items-in-each-group-of-array[/url]
Дайте массив чисел размера n, выберите группу размера m, найдите k-й минимальный элемент в этой группе. [b]Пример:[/b] [code]arr = [1, 3, 2, 1], m = 3, k = 2 m represents a group size k represents the kth minimum element in the group of size m so find the 2nd minimum element in a group of size 3.
There are 2 contiguous groups possible: [1, 3, 2] and [3, 2, 1]
k = 2 , 2nd minimum element in above groups is [2,2] [/code] [b]Ограничения:[/b] [code]k, m , n are of size 1 to 10^5 array item is of size 1 to 10^9 [/code] Вот мой код: [b]Подход 1:[/b] Это мой наивный подход: [code]public static int[] solve(int[] arr, int m, int k) { List result = new ArrayList(); int n = arr.length; for(int i=0; i Integer.compare(b, a));
for (int i = 0; i < m; i++) { q.add(arr[i]); if (q.size() > k) { q.poll(); } } result[0] = q.peek();
Мобильная версия