Find the maximum number of flags that can be set on mountain peaks.
A non-empty zero-indexed array A consisting of N integers is given.
A peak is an array element which is larger than its neighbours. More precisely, it is an index P such that 0 < P < N − 1 and A[P − 1] < A[P] > A[P + 1].
For example, the following array A:
A[0] = 1
A[1] = 5
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 four peaks: elements 1, 3, 5 and 10.
You are going on a trip to a range of mountains whose relative heights are represented by array A, as shown in a figure below. You have to choose how many flags you should take with you. The goal is to set the maximum number of flags on the peaks, according to certain rules.

Flags can only be set on peaks. What’s more, if you take K flags, then the distance between any two flags should be greater than or equal to K. The distance between indices P and Q is the absolute value |P − Q|.
For example, given the mountain range represented by array A, above, with N = 12, if you take:
two flags, you can set them on peaks 1 and 5;
three flags, you can set them on peaks 1, 5 and 10;
four flags, you can set only three flags, on peaks 1, 5 and 10.
You can therefore set a maximum of three flags in this case.
Write a function:
class Solution { public int solution(int[] A); }
that, given a non-empty zero-indexed array A of N integers, returns the maximum number of flags that can be set on the peaks of the array.
For example, the following array A:
A[0] = 1
A[1] = 5
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..200,000];
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);
expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).
i個flag將整個空間分爲i-1塊，每塊的間隔最小爲i，並且兩端必須是peak。i <= Math.min(length,(int)Math.sqrt(A.length)+1);
預先將所有peak找出形成一個遞增數組，採用窮舉法依次減小i值，返回第一個滿足條件的i

import java.util.*;
class Solution {
    public int solution(int[] A) {
        // write your code in Java SE 8
        if(A.length<3) return 0;
        int[] peaks = new int[A.length];
        int length = 0;
        for(int i=1; i<A.length-1; i++){
            // find peak
            if(A[i-1]<A[i] && A[i]>A[i+1]){
                peaks[length++] = i;
                i++; //the next data can`t be peak;
            }
        }
        peaks[length] = Integer.MAX_VALUE;
        int max = Math.min(length,(int)Math.sqrt(A.length)+1);

        for(int i=max; i>0; i--){
            int left = 0;
            int block=0;
            for(block=0; block<i-1; block++){
                int right = Arrays.binarySearch(peaks,left,length,peaks[left]+i);
                System.out.println(i+"- "+block + right);
                if(right>=0){
                    left = right;     
                }
                else{
                    left = Math.abs(right+1);
                    if(left >= length) break;
                }
                //left peaks are less than left block 
                //or left items are less than left block* each block size
                // if((length-1)-left < i-block || (A.length-1)-peaks[left] < (i-block)*i) 
                // break;
            }
            if(block == i-1){
                return i;
            }
        }  
        return 0;
    }
}