我有以下问题:
你得到了N个计数器,最初设置为0,你有两个可能的操作:
increase(X) − counter X is increased by 1,
max counter − all counters are set to the maximum value of any counter.
给出了M个整数的非空零索引数组A。此数组表示连续操作:
if A[K] = X, such that 1 ≤ X ≤ N, then operation K is increase(X),
if A[K] = N + 1 then operation K is max counter.
例如,给定整数N=5和数组A,使得:
A[0] = 3
A[1] = 4
A[2] = 4
A[3] = 6
A[4] = 1
A[5] = 4
A[6] = 4
每次连续操作后的计数器值为:
(0, 0, 1, 0, 0)
(0, 0, 1, 1, 0)
(0, 0, 1, 2, 0)
(2, 2, 2, 2, 2)
(3, 2, 2, 2, 2)
(3, 2, 2, 3, 2)
(3, 2, 2, 4, 2)
目标是在所有操作之后计算每个计数器的值。
编写一个函数:
class Solution { public int[] solution(int N, int[] A); }
给定整数N和由M个整数组成的非空零索引数组a,返回表示计数器值的整数序列。
例如,给定:
A[0] = 3
A[1] = 4
A[2] = 4
A[3] = 6
A[4] = 1
A[5] = 4
A[6] = 4
函数应该返回[3,2,2,4,2],如上所述。
假设:
N and M are integers within the range [1..100,000];
each element of array A is an integer within the range [1..N + 1].
复杂性:
expected worst-case time complexity is O(N+M);
expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).
可以修改输入数组的元素。
我已经用以下代码回答了这个问题,但只得到了80%的性能,而不是100%的性能,尽管有O(N+M(的时间复杂性:
public class Solution {
public int[] solution(int N, int[] A) {
int highestCounter = N;
int minimumValue = 0;
int lastMinimumValue = 0;
int [] answer = new int[N];
for (int i = 0; i < A.length; i++) {
int currentCounter = A[i];
int answerEquivalent = currentCounter -1;
if(currentCounter >0 && currentCounter<=highestCounter){
answer[answerEquivalent] = answer[answerEquivalent]+1;
if(answer[answerEquivalent] > minimumValue){
minimumValue = answer[answerEquivalent];
}
}
if (currentCounter == highestCounter +1 && lastMinimumValue!=minimumValue){
lastMinimumValue = minimumValue;
Arrays.fill(answer, minimumValue);
}
}
return answer;
}
}
我在这里的表现在哪里?该代码给出了正确的答案,但尽管具有适当的时间复杂性,但性能并没有达到规范要求。
您不应该在遇到"最大计数器"操作时调用Arrays.fill(answer, minimumValue);
,该操作占用O(N)
,而应该跟踪由于"最大计数"操作而分配的最后一个最大值,并在处理完所有操作后只更新整个数组一次。这需要O(N+M(。
我将变量名称从min改为max,以减少混淆。
public class Solution {
public int[] solution(int N, int[] A) {
int highestCounter = N;
int maxValue = 0;
int lastMaxValue = 0;
int [] answer = new int[N];
for (int i = 0; i < A.length; i++) {
int currentCounter = A[i];
int answerEquivalent = currentCounter -1;
if(currentCounter >0 && currentCounter<=highestCounter){
if (answer[answerEquivalent] < lastMaxValue)
answer[answerEquivalent] = lastMaxValue +1;
else
answer[answerEquivalent] = answer[answerEquivalent]+1;
if(answer[answerEquivalent] > maxValue){
maxValue = answer[answerEquivalent];
}
}
if (currentCounter == highestCounter +1){
lastMaxValue = maxValue;
}
}
// update all the counters smaller than lastMaxValue
for (int i = 0; i < answer.length; i++) {
if (answer[i] < lastMaxValue)
answer[i] = lastMaxValue;
}
return answer;
}
}
以下操作为O(n)
时间:
Arrays.fill(answer, minimumValue);
现在,如果给你一个测试用例,其中max counter
操作经常重复(比如说总操作的n/3
(,你得到了一个O(n*m)
算法(最坏情况分析(,而不是O(n+m)
。
您可以优化它,使其在O(n+m)
时间内完成,方法是每次执行此操作时使用在O(1)
中初始化数组的算法。
这将把最坏情况下的时间复杂度从O(n*m)
降低到O(n+m)
1
(1(理论上,使用相同的想法,它甚至可以在O(m)
中完成-无论计数器数量的大小,但数组的第一次分配在java中需要O(n(时间
这有点像@Eran的解决方案,但将功能封装在对象中。本质上——跟踪max
值和atLeast
值,并让对象的功能来完成其余的工作
private static class MaxCounter {
// Current set of values.
final int[] a;
// Keeps track of the current max value.
int currentMax = 0;
// Min value. If a[i] < atLeast the a[i] should appear as atLeast.
int atLeast = 0;
public MaxCounter(int n) {
this.a = new int[n];
}
// Perform the defined op.
public void op(int k) {
// Values are one-based.
k -= 1;
if (k < a.length) {
// Increment.
inc(k);
} else {
// Set max
max(k);
}
}
// Increment.
private void inc(int k) {
// Get new value.
int v = get(k) + 1;
// Keep track of current max.
if (v > currentMax) {
currentMax = v;
}
// Set new value.
a[k] = v;
}
private int get(int k) {
// Returns eithe a[k] or atLeast.
int v = a[k];
return v < atLeast ? atLeast : v;
}
private void max(int k) {
// Record new max.
atLeast = currentMax;
}
public int[] solution() {
// Give them the solution.
int[] solution = new int[a.length];
for (int i = 0; i < a.length; i++) {
solution[i] = get(i);
}
return solution;
}
@Override
public String toString() {
StringBuilder s = new StringBuilder("[");
for (int i = 0; i < a.length; i++) {
s.append(get(i));
if (i < a.length - 1) {
s.append(",");
}
}
return s.append("]").toString();
}
}
public void test() {
System.out.println("Hello");
int[] p = new int[]{3, 4, 4, 6, 1, 4, 4};
MaxCounter mc = new MaxCounter(5);
for (int i = 0; i < p.length; i++) {
mc.op(p[i]);
System.out.println(mc);
}
int[] mine = mc.solution();
System.out.println("Solution = " + Arrays.toString(mine));
}
我的解决方案:100\100
class Solution
{
public int maxCounterValue;
public int[] Counters;
public void Increase(int position)
{
position = position - 1;
Counters[position]++;
if (Counters[position] > maxCounterValue)
maxCounterValue = Counters[position];
}
public void SetMaxCounter()
{
for (int i = 0; i < Counters.Length; i++)
{
Counters[i] = maxCounterValue;
}
}
public int[] solution(int N, int[] A)
{
if (N < 1 || N > 100000) return null;
if (A.Length < 1) return null;
int nlusOne = N + 1;
Counters = new int[N];
int x;
for (int i = 0; i < A.Length; i++)
{
x = A[i];
if (x > 0 && x <= N)
{
Increase(x);
}
if (x == nlusOne && maxCounterValue > 0) // this used for all maxCounter values in array. Reduces addition loops
SetMaxCounter();
if (x > nlusOne)
return null;
}
return Counters;
}
}
- (@molbdnilo:+1!(因为这只是一个算法测试,所以对变量过于冗长是没有意义的。基于零的数组索引调整的"answerEquivalent"?让我休息一下!只要回答[A[i]-1]就行了
- 测试表明假设A值总是在1和N+1之间。因此,不需要对此进行检查
- fillArray(.(是一个O(N(过程,它在O(M(过程中。当期望的最大复杂度为O(M+N(时,这使得整个代码成为O(M*N(过程。实现这一点的唯一方法是只结转计数器的当前最大值。这允许您在A[i]为N+1时始终保存正确的最大计数器值。后一个值是之后所有增量的一种基线值。在对所有A值进行操作之后,那些从未通过数组条目递增的计数器可以通过复杂性为O(N(的第二个for循环上升到所有计数器基线
看看Eran的解决方案。
这就是我们消除O(N*M)
复杂性的方法
在这个解决方案中,我没有为每个A[K]=N+1
填充结果数组,而是尝试保持所有元素的最小值,并在所有操作完成后更新结果数组。
如果有增加操作,则更新该位置:
if (counter[x - 1] < minVal) {
counter[x - 1] = minVal + 1;
} else {
counter[x - 1]++;
}
并跟踪结果数组中每个元素的minVal。
以下是完整的解决方案:
public int[] solution(int N, int[] A) {
int minVal = -1;
int maxCount = -1;
int[] counter = new int[N];
for (int i = 0; i < A.length; i++) {
int x = A[i];
if (x > 0 && x <= N) {
if (counter[x - 1] < minVal) {
counter[x - 1] = minVal + 1;
} else {
counter[x - 1]++;
}
if (maxCount < counter[x - 1]) {
maxCount = counter[x - 1];
}
}
if (x == N + 1 && maxCount > 0) {
minVal = maxCount;
}
}
for (int i = 0; i < counter.length; i++) {
if (counter[i] < minVal) {
counter[i] = minVal;
}
}
return counter;
}
这是我的swift 3解决方案(100/100(
public func solution(_ N : Int, _ A : inout [Int]) -> [Int] {
var counters = Array(repeating: 0, count: N)
var _max = 0
var _min = 0
for i in A {
if counters.count >= i {
let temp = max(counters[i-1] + 1, _min + 1)
_max = max(temp, _max)
counters[i-1] = temp
} else {
_min = _max
}
}
return counters.map { max($0, _min) }
}