A non-empty zero-indexed array A consisting of N integers is given.
A?peak?is an array element which is larger than its neighbors. More precisely, it is an index P such that 0 < P < N ? 1,? A[P ? 1] < A[P] and A[P] > A[P + 1].
For example, the following array A:
?
A[0] = 1 A[1] = 2 A[2] = 3 A[3] = 4 A[4] = 3 A[5] = 4 A[6] = 1 A[7] = 2 A[8] = 3 A[9] = 4 A[10] = 6 A[11] = 2
has exactly three peaks: 3, 5, 10.
We want to divide this array into blocks containing the same number of elements. More precisely, we want to choose a number K that will yield the following blocks:
- A[0], A[1], ..., A[K ? 1],
- A[K], A[K + 1], ..., A[2K ? 1],
...- A[N ? K], A[N ? K + 1], ..., A[N ? 1].
What‘s more, every block should contain at least one peak. Notice that extreme elements of the blocks (for example A[K ? 1] or A[K]) can also be peaks, but only if they have both neighbors (including one in an adjacent blocks).
The goal is to find the maximum number of blocks into which the array A can be divided.
Array A can be divided into blocks as follows:
- one block (1, 2, 3, 4, 3, 4, 1, 2, 3, 4, 6, 2). This block contains three peaks.
- two blocks (1, 2, 3, 4, 3, 4) and (1, 2, 3, 4, 6, 2). Every block has a peak.
- three blocks (1, 2, 3, 4), (3, 4, 1, 2), (3, 4, 6, 2). Every block has a peak. Notice in particular that the first block (1, 2, 3, 4) has a peak at A[3], because A[2] < A[3] > A[4], even though A[4] is in the adjacent block.
However, array A cannot be divided into four blocks, (1, 2, 3), (4, 3, 4), (1, 2, 3) and (4, 6, 2), because the (1, 2, 3) blocks do not contain a peak. Notice in particular that the (4, 3, 4) block contains two peaks: A[3] and A[5].
The maximum number of blocks that array A can be divided into is three.
Write a function:
int solution(vector<int> &A);
that, given a non-empty zero-indexed array A consisting of N integers, returns the maximum number of blocks into which A can be divided.
If A cannot be divided into some number of blocks, the function should return 0.
For example, given:
?
A[0] = 1 A[1] = 2 A[2] = 3 A[3] = 4 A[4] = 3 A[5] = 4 A[6] = 1 A[7] = 2 A[8] = 3 A[9] = 4 A[10] = 6 A[11] = 2
the function should return 3, as explained above.
Assume that:
- N is an integer within the range [1..100,000];
- each element of array A is an integer within the range [0..1,000,000,000].
Complexity:
- expected worst-case time complexity is O(N*log(log(N)));
- expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).
Elements of input arrays can be modified.
思路
题目大意是希望将序列等分成c片，每片都要至少有一个peak，peak就是比左右都大的数(原序列首尾不能算)，求c的最大值 //Codility题目描述还真是啰嗦啊喂
- 用O(n)得空间统计从开始到目前为止得peak数sum[]，以及最远两peak间坐标差D
- 求最大分片数c，即求最小分片长度k，可行解k的可能范围在(D/2,min(D,n/2)]，要等分首先 n % k == 0，其次sum[k - 1], sum[2 * k - 1], sum[3 * k - 1],....sum[n - 1]这个数列有n/k项。所以外层循环k的次数等于(D/2,D]间n的约数个数（小于D/2），内层判断是否可行需要n/k的操作(小于2n/D)。于是第2步时间复杂度也是O(n)。
- 编程上要注意计算D的时候要考虑第一个和最后一个peak到首尾的距离，还有不要混淆K c的含义（大写字母做变量名很容易出错）。
代码
int solution(vector<int> &A) {
int N = A.size();
vector<int> npeaks(N+1, 0);//npeaks[i]代表第i个元素（不包括）之前peak的数量
int maxD = 0;//最远两peak间坐标差D
int last_peak = -1;//处理第一个peak到起始的距离
for(int i = 1; i < N-1; i++){
if(A[i]>A[i+1] && A[i]>A[i-1]){
npeaks[i+1] = npeaks[i] + 1;
maxD = max(maxD, i - last_peak);
last_peak = i;
}
else{
npeaks[i+1] = npeaks[i];
}
}
maxD = max(maxD, N - last_peak);//处理最后一个peak到末端的距离
npeaks[N] = npeaks[N-1];
if(npeaks[N] < 1) return 0;
if(maxD > N/2) return 1;
for(int K = maxD/2; K <= maxD; K++){//slice长度
if(N%K == 0){
bool isvalid = true;
int c = N/K;//slice数量
for(int i = 1; i <= c; i++){
if(npeaks[i*K] - npeaks[(i-1)*K] < 1){
isvalid = false;
break;
}
}
if(isvalid) return c;
}
}
cout<<"fail"<<endl;
for(int K = maxD+1;;K++) {
if(N%K == 0) return N/K;//不能return K 呀
}
}
【题解】【数组】【Prime and composite numbers】【Codility】Peaks