Question

我试图解决String Function Calculation problem from Hackerrank。在这个问题中，我们给出了一个字符串作为输入，并要求在输入字符串的所有子串中打印一个代表以下函数最大值的数字：

f（s，t）=子串的次数＆＃39;出现在字符串＆＃39; t＆＃39; *子串的长度＆＃39;

我提交了以下答案：

import Data.List

main :: IO()
main = do
    stringInput <- getLine
    print $ solution stringInput

solution :: String -> Int
solution input = maximum $ map sum $ map (map length) $ group $ sort $ substrings input

substrings :: String -> [String]
substrings s = tail . inits =<< tails s

这个想法是：

获取s的所有子字符串。 let s = "aaaaaa"; substrings s = ["a","aa","aaa","aaaa","aaaaa","aaaaaa","a","aa","aaa","aaaa","aaaaa","a","aa","aaa","aaaa","a","aa","aaa","a","aa","a"]
排序。 ["a","a","a","a","a","a","aa","aa","aa","aa","aa","aaa","aaa","aaa","aaa","aaaa" ,"aaaa","aaaa","aaaaa","aaaaa","aaaaaa"]
将它分组。 [["a","a","a","a","a","a"],["aa","aa","aa","aa","aa"],["aaa","aaa","aaa","aaa"],["aaaa","aaaa","aaaa"],["aaaaa","aaaaa"],["aaaaaa"]]
获取每个子字符串的各个长度。 [[1,1,1,1,1,1],[2,2,2,2,2],[3,3,3,3],[4,4,4],[5,5],[6]]
对结果列表求和。 [6,10,12,12,10,6]。
获得最大值。 12。

这通过了初步测试。但是，当我提交它时，它会失败所有其他测试，运行错误＆＃39;。

测试案例编号。 2，第一个失败，运行1.47秒并输入以下内容：

"aacbbabbabbbbbaaaaaaabbbbcacacbcabaccaabbbcaaabbccccbbbcbccccbbcaabaaabcbaacbcbaccaaaccbccbcaacbaccbaacbbabbabbbbbaaaaaaabbbbcacacbcabaccaabbbcaaabbccccbbbcbccccbbcaabaaabcbaacbcbaccaaaccbccbcaacbaccbaacbbabbabbbbbaaaaaaabbbbcacacbcabaccaabbbcaaabbccccbbbcbccccbbcaabaaabcbaacbcbaccaaaccbccbcaacbaccbaacbbabbabbbbbaaaaaaabbbbcacacbcabaccaabbbcaaabbccccbbbcbccccbbcaabaaabcbaacbcbaccaaaccbccbcaacbaccbaacbbabbabbbbbaaaaaaabbbbcacacbcabaccaabbbcaaabbccccbbbcbccccbbcaabaaabcbaacbcbaccaaaccbccbcaacbaccb"

你能帮我弄清楚我做错了什么，或者发生了什么事？

Answer 1

这不起作用。排序变得非常昂贵（记忆明智）作为中间体此列表中的产品仍保留在内存中，这些产品很大。这是一个记忆
错误

更好的方法是使用后缀数组O(n log²n) 然后创建
使用Kasai's Algorithm in O(n)的最长前缀数组（LCP）然后使用LCP数组进行问题的其余部分。

计算LCP[i] and LCP[i+1] 如果它们相等，则意味着有两个相等的子字符串以这种方式继续。

Answer 2

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;

#define nb nexta
#define head height
#define rank b

const int maxn = 100010;
char s[maxn];
int n, id[maxn], height[maxn], b[maxn], nexta[maxn];

bool cmp(const int& i, const int& j)
{
    return s[i] < s[j];
}

void SuffixSort()
{
    int i, j, k, h;
    for (i = 0; i < n; i++) id[i] = i;
    sort(id, id + n, cmp);
    for (i = 0; i < n; i++)
    {
        if (i == 0 || s[id[i]] != s[id[i - 1]])
            b[id[i]] = i;
        else b[id[i]] = b[id[i - 1]];
    }
    for (h = 1; h < n; h <<= 1)
    {
        for (i = 0; i < n; i++)
            head[i] = nexta[i] = -1;
        for (i = n - 1; i >= 0; i--)
        {
            if (id[i])
            {
                j = id[i] - h;
                if (j < 0) j += n;
                nexta[j] = head[b[j]];
                head[b[j]] = j;
            }
        }
        j = n - h;
        nexta[j] = head[b[j]];
        head[b[j]] = j;
        for (i = k = 0; i < n; i++)
            if (head[i] >= 0)
                for (j = head[i]; j >= 0; j = nexta[j])
                    id[k++] = j;
        for (i = 0; i < n; i++)
            if (i>0 && id[i] + h < n&&id[i - 1] + h < n&&b[id[i]] == b[id[i - 1]] && b[id[i] + h] == b[id[i - 1] + h])
                nb[id[i]] = nb[id[i - 1]];
            else
                nb[id[i]] = i;
        for (i = 0; i < n; i++)
            b[i] = nb[i];
    }
}

void GetHeight()
{
    int i, j, h; height[0] = 0;
    for (i = 0; i < n; i++)
        rank[id[i]] = i;
    for (h = 0, i = 0; i < n; i++)
    {
        if (rank[i] > 0)
        {
            j = id[rank[i] - 1];
            while (s[i + h] == s[j + h])++h;
            height[rank[i]] = h;
            if (h>0) --h;
        }
    }
}

int st[maxn], top;

int main()
{
    cin >> s;
    n = strlen(s);
    top = 0;
    SuffixSort();
    GetHeight();
    height[n] = 0;
    int best = n;
    st[top++] = 0;
    for (int i = 1; i < n + 1; i++)
    {
        //cout << height[i] << " ";
        while (top != 0 && height[i] < height[st[top - 1]])
        {
            int val = height[st[top - 1]];
            top--;
            best = max(best, val * (top == 0 ? i : i - st[top - 1]));
        }

        if (top == 0 || height[i] >= height[st[top - 1]])
            st[top++] = i;
    }
    cout << best << endl;
    return 0;
}

Hackerrank字符串函数计算

f（s，t）=子串的次数＆＃39;出现在字符串＆＃39; t＆＃39; *子串的长度＆＃39;

2 个答案: