Peaks

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)
    auto N = A.size();
    vector<int> peaks;
    for (size_t i = 1; i < N - 1; ++i) {
        if (A[i - 1] < A[i] && A[i] > A[i + 1]) {
            peaks.push_back(i);
        }
    }
    
    int len = peaks.size();
    if (len <= 2) {
        return len;
    }
    
    // from the "biggest possible number" to smaller number
    for(int numBlocks = N; numBlocks >= 1; numBlocks--){
                
        if (N % numBlocks == 0){ // it is divisible   
        
            int blockSize = N / numBlocks; 
            int ithOkBlock = 0;  // the ith block has peak(s)

            // test all the peaks 
            // if a peak is found in the ith block
            // then, go to the (i+1)th block
            for(int peaksIndex : peaks){
                if( peaksIndex / blockSize == ithOkBlock){ // peak in the ith block
                    ithOkBlock++; // go to check (i+1)th block
                }
            }   
            
            // ithOkBlock: the number of blocks having peak(s) 
            // if all the blocks have peak(s)
            // then, return the number of blocks
            // note: we test from the biggest possible number
            // so, once we find it, we can just return it
            // (no need to check the smaller possible numbers)
            if (ithOkBlock == numBlocks){
                return numBlocks;
            }
        }
    }   
    
    return 0; 
}