日韩久久久精品,亚洲精品久久久久久久久久久,亚洲欧美一区二区三区国产精品 ,一区二区福利

Google Interview - Compute the h-index of a

系統 2845 0

Google:

Compute the h-index of a list of papers, given their citation count. Can you do it in linear time? How about a distributed algorithm for the task?

?

Facebook:

Given: for every paper authored, there is a citation count vector. The h-index is a measure of researcher importance. h-index: The largest number i such that there are i papers each with at least i citations.?

1. Suppose that the citation-vector is sorted, how to efficiently compute the h-index??

2. Suppose that the citation-vector is not sorted, how to efficiently compute the h-index? time complexity? an algorithm with time complexity n?

?

Princeton algorithm:

Given an array of N positive integers, its h-index is the largest integer h such that there are at least h entries in the array greater than or equal to h. Design an algorithm to compute the h-index of an array.?

Hint: median or quicksort-like partitioning and divide-and-conquer.

?

Solution:

- Create an int[] Histogram as big as the maximum number of publications of any particular scientist).?

- If all publication reference counts are stored in another int[] references, then go over this array and, on each publication, if it's reference count is R, then do Histogram[R]++. While doing this, keep the maximum reference count in Max.?

- After building the histogram, do a decreasing loop on int[] Histogram from i=Max, adding Histogram[i] values to int hIndex. When hIndex >= i, return i as the hIndex.?

... As to the distributed part, let several machines build the Histogram of disjoint sets of somebody's publications, and then have one machine add up those histograms and return hIndex as described above.

?

1. binary-search (O(log(n)). If citations[i] >= i then h >= i (if array's in descending order).

2. Here's a O(n) time & space solution in ruby. The trick is you can ignore citation-counts larger than n.

?

If there are 'n' papers in total, this problem can be solved in O(n) with space complexity of O(n). Note that, h-index can be between 0 to n. Say if the h-index is 10, this means, there has to be 10 papers with citation count >= 10. So if we can find out the number of papers with citations >=X for every X (and store it in an array C) where X ranges between 0 to n, then by scanning the count array C from the right to left, we can find the h-index at index i where i == C[i].?

Pseudocode:?
input array A of length n.?

- init array C[0] to C[n] with 0?
- foreach p in A, if p >= n, c[n]++; else c[p] +=1?
- for i=n-1 to 0, c[i]=c[i]+c[i+1]?
- for i=n to 0, if c[i] == i return i

?

      // assume sorted in descending order, O(lgN)
public static int getHIndexFromSorted(int[] citation) {
    int low = 0; int high = citation.length - 1;
    while(low <= high) {
        int idx = (low+high)/2;
        if(citation[idx] >= idx + 1) {
            low = idx + 1;
        } else {
            high = idx - 1;
        }
    }
    return low;
}

// sort the array, O(NlgN)
public static int computeHIndexBySorting(int[] A) {
    Arrays.sort(A);
    int h = 0;
    for (int i = A.length-1; i >= 0; i--) {
        if(A[i] > h) {
            h++;
        } else {
            return h;
        }
    }
    return -1;
}

// no need to sort array, O(N)
public static int computeHIndex(int[] A) {
    int n = A.length;
    int[] s = new int[n+1];
    for(int num : A) {
        num = Math.min(n ,num);
        s[num]++;
    }
    int sum = 0;
    for (int i = s.length-1; i >= 0; i--) {
        sum += s[i];
        if(sum >= i) {
            return i;
        }
    }
    return -1;
}
    

?

?

Reference:

http://en.wikipedia.org/wiki/H-index

http://www.careercup.com/question?id=14585874

http://algs4.cs.princeton.edu/25applications/

http://www.glassdoor.com/Interview/Compute-the-h-index-of-a-list-of-papers-given-their-citation-count-Can-you-do-it-in-linear-time-How-about-a-distributed-QTN_572531.htm

?

Google Interview - Compute the h-index of a list of papers


更多文章、技術交流、商務合作、聯系博主

微信掃碼或搜索:z360901061

微信掃一掃加我為好友

QQ號聯系: 360901061

您的支持是博主寫作最大的動力,如果您喜歡我的文章,感覺我的文章對您有幫助,請用微信掃描下面二維碼支持博主2元、5元、10元、20元等您想捐的金額吧,狠狠點擊下面給點支持吧,站長非常感激您!手機微信長按不能支付解決辦法:請將微信支付二維碼保存到相冊,切換到微信,然后點擊微信右上角掃一掃功能,選擇支付二維碼完成支付。

【本文對您有幫助就好】

您的支持是博主寫作最大的動力,如果您喜歡我的文章,感覺我的文章對您有幫助,請用微信掃描上面二維碼支持博主2元、5元、10元、自定義金額等您想捐的金額吧,站長會非常 感謝您的哦!!!

發表我的評論
最新評論 總共0條評論
主站蜘蛛池模板: 昆山市| 益阳市| 金溪县| 陕西省| 三门峡市| 贞丰县| 正定县| 京山县| 荥阳市| 竹北市| 景泰县| 隆子县| 黔西县| 赤城县| 色达县| 社旗县| 大洼县| 达拉特旗| 九龙坡区| 海城市| 贵定县| 丘北县| 元朗区| 南漳县| 内江市| 油尖旺区| 登封市| 乌鲁木齐县| 孝感市| 桦甸市| 恩施市| 兴业县| 东阳市| 安平县| 蒙城县| 福泉市| 南充市| 梧州市| 怀宁县| 咸丰县| 苏尼特左旗|