Frequency of a Substring in a String

Last Updated : 18 Apr, 2023

Given an input string and a pattern, the task is to find the frequency of occurrences of the string pattern in a given string.

Examples:

Input: pattern = "man", string = "dhimanman"
Output: 2

Input: pattern = "nn", string = "Banana"
Output: 0

Input: pattern = "aa", string = "aaaaa"
Output : 4

Recommended: Please solve it on "PRACTICE" first, before moving on to the solution.

Approach:

A simple solution is to match characters one by one. And whenever we see a complete match, increment count. For this, we can use Naive pattern searching.

Below is the implementation of the above approach.

C++

// Simple C++ program to count occurrences
// of pat in txt.
#include <bits/stdc++.h>
using namespace std;

int countFreq(string& pat, string& txt)
{
    int M = pat.length();
    int N = txt.length();
    int res = 0;

    /* A loop to slide pat[] one by one */
    for (int i = 0; i <= N - M; i++) {
        /* For current index i, check for
           pattern match */
        int j;
        for (j = 0; j < M; j++)
            if (txt[i + j] != pat[j])
                break;

        // if pat[0...M-1] = txt[i, i+1, ...i+M-1]
        if (j == M) {
            res++;
        }
    }
    return res;
}

/* Driver program to test above function */
int main()
{
    string txt = "dhimanman";
    string pat = "man";
    cout << countFreq(pat, txt);
    return 0;
}

Java

// Simple Java program to count occurrences
// of pat in txt.

class GFG {

    static int countFreq(String pat, String txt)
    {
        int M = pat.length();
        int N = txt.length();
        int res = 0;

        /* A loop to slide pat[] one by one */
        for (int i = 0; i <= N - M; i++) {
            /* For current index i, check for
        pattern match */
            int j;
            for (j = 0; j < M; j++) {
                if (txt.charAt(i + j) != pat.charAt(j)) {
                    break;
                }
            }

            // if pat[0...M-1] = txt[i, i+1, ...i+M-1]
            if (j == M) {
                res++;
                j = 0;
            }
        }
        return res;
    }

    /* Driver program to test above function */
    static public void main(String[] args)
    {
        String txt = "dhimanman";
        String pat = "man";
        System.out.println(countFreq(pat, txt));
    }
}

// This code is contributed by 29AjayKumar

Python3

# Simple python program to count
# occurrences of pat in txt.


def countFreq(pat, txt):
    M = len(pat)
    N = len(txt)
    res = 0

    # A loop to slide pat[] one by one
    for i in range(N - M + 1):

        # For current index i, check
        # for pattern match
        j = 0
        while j < M:
            if (txt[i + j] != pat[j]):
                break
            j += 1

        if (j == M):
            res += 1
            j = 0
    return res


# Driver Code
if __name__ == '__main__':
    txt = "dhimanman"
    pat = "man"
    print(countFreq(pat, txt))

# This code is contributed
# by PrinciRaj1992

// Simple C# program to count occurrences
// of pat in txt.
using System;
public class GFG {

    static int countFreq(String pat, String txt)
    {
        int M = pat.Length;
        int N = txt.Length;
        int res = 0;

        /* A loop to slide pat[] one by one */
        for (int i = 0; i <= N - M; i++) {
            /* For current index i, check for
        pattern match */
            int j;
            for (j = 0; j < M; j++) {
                if (txt[i + j] != pat[j]) {
                    break;
                }
            }

            // if pat[0...M-1] = txt[i, i+1, ...i+M-1]
            if (j == M) {
                res++;
                j = 0;
            }
        }
        return res;
    }

    /* Driver program to test above function */
    static public void Main()
    {
        String txt = "dhimanman";
        String pat = "man";
        Console.Write(countFreq(pat, txt));
    }
}

// This code is contributed by 29AjayKumar

PHP

<?php
// Simple PHP program to count occurrences
// of pat in txt.

function countFreq($pat, $txt)
{
    $M = strlen($pat);
    $N = strlen($txt);
    $res = 0;

    /* A loop to slide pat[] one by one */
    for ($i = 0; $i <= $N - $M; $i++)
    { 
        /* For current index i, check for 
        pattern match */
        for ($j = 0; $j < $M; $j++)
            if ($txt[$i+$j] != $pat[$j])
                break;

        // if pat[0...M-1] = txt[i, i+1, ...i+M-1]
        if ($j == $M) 
        {
            $res++;
            $j = 0;
        }
    }
    return $res;
}

// Driver Code
$txt = "dhimanman";
$pat = "man";
echo countFreq($pat, $txt);

// This code is contributed 
// by Akanksha Rai

JavaScript

<script>

// JavaScript program to count occurrences
// of pat in txt.
let mod = 100000007;
  
function countFreq(pat, txt)
{       
    let M = pat.length;       
    let N = txt.length;       
    let res = 0;

    // A loop to slide pat[] one by one 
    for(let i = 0; i <= N - M; i++) 
    {
        
        // For current index i, check for
        // pattern match 
        let j;           
        for(j = 0; j < M; j++)
        {
            if (txt[i + j] != pat[j]) 
            {
                break;
            }
        }

        // If pat[0...M-1] = txt[i, i+1, ...i+M-1]
        if (j == M)
        {               
            res++;               
            j = 0;               
        }           
    }       
    return res;       
}

// Driver Code
let txt = "dhimanman";       
let pat = "man";  

document.write(countFreq(pat, txt)); 

// This code is contributed by code_hunt

</script>

Output

Time Complexity: O(M * N)

Auxiliary Space: O(1)

Efficient Approach:

An efficient solution is to use KMP algorithm.

Below is the implementation of the above approach.

C++

// C++ program to count occurrences
// of pattern in a text.
#include <iostream>
using namespace std;

void computeLPSArray(string pat, int M, int lps[])
{

    // Length of the previous longest
    // prefix suffix
    int len = 0;
    int i = 1;
    lps[0] = 0; // lps[0] is always 0

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

            // This is tricky. Consider the example.
            // AAACAAAA and i = 7. The idea is similar
            // to search step.
            if (len != 0) {
                len = lps[len - 1];

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

int KMPSearch(string pat, string txt)
{
    int M = pat.length();
    int N = txt.length();

    // Create lps[] that will hold the longest
    // prefix suffix values for pattern
    int lps[M];
    int j = 0; // index for pat[]

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

    int i = 0; // index for txt[]
    int res = 0;
    int next_i = 0;

    while (i < N) {
        if (pat[j] == txt[i]) {
            j++;
            i++;
        }
        if (j == M) {

            // When we find pattern first time,
            // we iterate again to check if there
            // exists more pattern
            j = lps[j - 1];
            res++;
        }

        // Mismatch after j matches
        else if (i < N && pat[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 res;
}

// Driver code
int main()
{
    string txt = "geeksforgeeks";
    string pat = "eeks";
    int ans = KMPSearch(pat, txt);

    cout << ans;

    return 0;
}

// This code is contributed by akhilsaini

Java

// Java program to count occurrences of pattern
// in a text.

class KMP_String_Matching {
    int KMPSearch(String pat, String txt)
    {
        int M = pat.length();
        int N = txt.length();

        // create lps[] that will hold the longest
        // prefix suffix values for pattern
        int lps[] = new int[M];
        int j = 0; // index for pat[]

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

        int i = 0; // index for txt[]
        int res = 0;
        int next_i = 0;

        while (i < N) {
            if (pat.charAt(j) == txt.charAt(i)) {
                j++;
                i++;
            }
            if (j == M) {
                // When we find pattern first time,
                // we iterate again to check if there
                // exists more pattern
                j = lps[j - 1];
                res++;

                // We start i to check for more than once
                // appearance of pattern, we will reset i
                // to previous start+1
                if (lps[j] != 0)
                    i = ++next_i;
                j = 0;
            }

            // mismatch after j matches
            else if (i < N
                     && pat.charAt(j) != txt.charAt(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 res;
    }

    void computeLPSArray(String pat, int M, int lps[])
    {
        // length of the previous longest prefix suffix
        int len = 0;
        int i = 1;
        lps[0] = 0; // lps[0] is always 0

        // the loop calculates lps[i] for i = 1 to M-1
        while (i < M) {
            if (pat.charAt(i) == pat.charAt(len)) {
                len++;
                lps[i] = len;
                i++;
            }
            else // (pat[i] != pat[len])
            {
                // This is tricky. Consider the example.
                // AAACAAAA and i = 7. The idea is similar
                // to search step.
                if (len != 0) {
                    len = lps[len - 1];

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

    // Driver program to test above function
    public static void main(String args[])
    {
        String txt = "geeksforgeeks";
        String pat = "eeks";
        int ans
            = new KMP_String_Matching().KMPSearch(pat, txt);
        System.out.println(ans);
    }
}

Python3

# Python3 program to count occurrences of
# pattern in a text.


def KMPSearch(pat, txt):

    M = len(pat)
    N = len(txt)

    # Create lps[] that will hold the longest
    # prefix suffix values for pattern
    lps = [None] * M
    j = 0  # index for pat[]

    # Preprocess the pattern (calculate lps[]
    # array)
    computeLPSArray(pat, M, lps)

    i = 0  # index for txt[]
    res = 0
    next_i = 0

    while (i < N):
        if pat[j] == txt[i]:
            j = j + 1
            i = i + 1
        if j == M:

            # When we find pattern first time,
            # we iterate again to check if there
            # exists more pattern
            j = lps[j - 1]
            res = res + 1

            # We start i to check for more than once
            # appearance of pattern, we will reset i
            # to previous start+1
            if lps[j] != 0:
                next_i = next_i + 1
                i = next_i
                j = 0

        # Mismatch after j matches
        elif ((i < N) and (pat[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 res


def computeLPSArray(pat, M, lps):

    # Length of the previous longest
    # prefix suffix
    len = 0
    i = 1
    lps[0] = 0  # lps[0] is always 0

    # The loop calculates lps[i] for
    # i = 1 to M-1
    while (i < M):
        if pat[i] == pat[len]:
            len = len + 1
            lps[i] = len
            i = i + 1

        else:  # (pat[i] != pat[len])

            # This is tricky. Consider the example.
            # AAACAAAA and i = 7. The idea is similar
            # to search step.
            if len != 0:
                len = lps[len - 1]

                # Also, note that we do not increment
                # i here

            else:  # if (len == 0)
                lps[i] = len
                i = i + 1


# Driver code
if __name__ == "__main__":

    txt = "geeksforgeeks"
    pat = "eeks"
    ans = KMPSearch(pat, txt)

    print(ans)

# This code is contributed by akhilsaini

// C# program to count occurrences of pattern
// in a text.
using System;

public class KMP_String_Matching {
    int KMPSearch(String pat, String txt)
    {
        int M = pat.Length;
        int N = txt.Length;

        // create lps[] that will hold the longest
        // prefix suffix values for pattern
        int[] lps = new int[M];
        int j = 0; // index for pat[]

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

        int i = 0; // index for txt[]
        int res = 0;
        int next_i = 0;

        while (i < N) {
            if (pat[j] == txt[i]) {
                j++;
                i++;
            }
            if (j == M) {
                // When we find pattern first time,
                // we iterate again to check if there
                // exists more pattern
                j = lps[j - 1];
                res++;

                // We start i to check for more than once
                // appearance of pattern, we will reset i
                // to previous start+1
                if (lps[j] != 0)
                    i = ++next_i;
                j = 0;
            }

            // mismatch after j matches
            else if (i < N && pat[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 res;
    }

    void computeLPSArray(String pat, int M, int[] lps)
    {
        // length of the previous longest prefix suffix
        int len = 0;
        int i = 1;
        lps[0] = 0; // lps[0] is always 0

        // the loop calculates lps[i] for i = 1 to M-1
        while (i < M) {
            if (pat[i] == pat[len]) {
                len++;
                lps[i] = len;
                i++;
            }
            else // (pat[i] != pat[len])
            {
                // This is tricky. Consider the example.
                // AAACAAAA and i = 7. The idea is similar
                // to search step.
                if (len != 0) {
                    len = lps[len - 1];

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

    // Driver code
    public static void Main(String[] args)
    {
        String txt = "geeksforgeeks";
        String pat = "eeks";
        int ans
            = new KMP_String_Matching().KMPSearch(pat, txt);
        Console.WriteLine(ans);
    }
}

// This code is contributed by Princi Singh

JavaScript

<script>

    // JavaScript program to count occurrences
    // of pattern in a text.


    function computeLPSArray(pat,M,lps)
    {

        // Length of the previous longest
        // prefix suffix
        let len = 0;
        let i = 1;
        lps[0] = 0; // lps[0] is always 0

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

                // This is tricky. Consider the example.
                // AAACAAAA and i = 7. The idea is similar
                // to search step.
                if (len != 0)
                {
                    len = lps[len - 1];

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

    function KMPSearch(pat,txt)
    {
        let M = pat.length;
        let N = txt.length;

        // Create lps[] that will hold the longest
        // prefix suffix values for pattern
        let lps = new Array(M);
        lps.fill(0);
        let j = 0; // index for pat[]

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

        let i = 0; // index for txt[]
        let res = 0;
        let next_i = 0;

        while (i < N)
        {
            if (pat[j] == txt[i])
            {
                j++;
                i++;
            }
            if (j == M)
            {

                // When we find pattern first time,
                // we iterate again to check if there
                // exists more pattern
                j = lps[j - 1];
                res++;
                // We start i to check for more than once
                // appearance of pattern, we will reset i
                // to previous start+1
                if (lps[j]!=0)
                    i = ++next_i;
                j = 0;
            }

            // Mismatch after j matches
            else if (i < N && pat[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 res;
    }

    // Driver code

    let txt = "geeksforgeeks";
    let pat = "eeks";
    let ans = KMPSearch(pat, txt);
    
    document.write(ans);
    
</script>

Output

Time Complexity: O(M + N)

Auxiliary Space: O(M) As an array of size M is used to store the longest prefix suffix values for the pattern.

Frequency of a Substring in a String

Dhiman Mayank

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Practice Tags :

Frequency of a Substring in a String

Recommended: Please solve it on "PRACTICE" first, before moving on to the solution.

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