Problem Name: Substring Frequency

LightOj-1255

Discussion:

In this problem you are given two non-empty strings A and B, both contain lower case English alphabets.
You have to find the number of times B occurs as a substring of A.
Note : Count the number of times of B that occurs as a substring of A.

Solution Idea:

In naive approach,
It checks for all character of the A to the B. This algorithm is helpful for smaller texts.
It does not need any pre-processing phases. We can find substring by checking once for the string.
Thetime complexity of Naive Approach is O(m*n).The m is the size of B and n is the size of the A.
But we can reduce this time complexity is O(n+m) using by KMP(Knuth Morris Pratt) Pattern Searching If you don't know how KMP works then check out these sites.

• KMP Algorithm
• [KMP Algorithm -by Abdul Bari] Beautiful explanation of how the KMP string matching algorithm will work.

CPP Code:

#include <bits/stdc++.h>
using namespace std;

// Fills lps[] for given patttern pat[0..M-1]
void computeLPSArray(char *pattern, int M, int *lps)
{
    // length of the previous longest prefix suffix
    int len = 0;

    lps[0] = 0;    // lps[0] is always 0

    // the loop calculates lps[i] for i = 1 to M-1
    int i = 1;
    while (i < M)
    {
        if (pattern[i] == pattern[len])
        {
            len++;
            lps[i] = len;
            i++;
        }
        else    // (pat[i] != pat[len])
        {
            if (len != 0)
            {
                len = lps[len - 1];

                // Also, note that we do not increment
                // i here
            }
            else    // if (len == 0)
            {
                lps[i] = 0;
                i++;
            }
        }
    }
}

int KMPSearch(char *pattern, char *txt)
{
    int M = strlen(pattern);
    int N = strlen(txt);

    // create lps[] that will hold the longest prefix suffix
    // values for pattern
    int lps[M];

    // Preprocess the pattern (calculate lps[] array)
    computeLPSArray(pattern, M, lps);

    int i = 0;    // index for txt[]
    int j = 0;    // index for pat[]
    int total = 0;
    while (i < N)
    {
        if (pattern[j] == txt[i])
        {
            j++;
            i++;
        }

        if (j == M)
        {
            total++;
            j = lps[j - 1];
        }

        // mismatch after j matches
        else if (i < N && pattern[j] != txt[i])
        {
            // Do not match lps[0..lps[j-1]] characters,
            // they will match anyway
            if (j != 0)
                j = lps[j - 1];
            else
                i = i + 1;
        }
    }

    return total;
}

int main()
{
    int test;
    cin >> test;
    char txt[1000001], pattern[1000001];
    for (int t = 1; t <= test; t++)
    {
        cin >> txt >> pattern;
        int total = KMPSearch(pattern, txt);
        cout << "Case " << t << ": " << total << '\n';

    }
}

Happy Coding!!

Written by:Md. Rasel Meya

Tutorial source