Open In App
Related Articles

Count M-length substrings occurring exactly K times in a string

Improve Article
Improve
Save Article
Save
Like Article
Like

Given a string S of length N and two integers M and K, the task is to count the number of substrings of length M occurring exactly K times in the string S

Examples:

Input: S = “abacaba”, M = 3, K = 2
Output: 1
Explanation: All distinct substrings of length 3 are “aba”, “bac”, “aca”, “cab”.
Out of all these substrings, only “aba” occurs twice in the string S.
Therefore, the count is 1.

Input: S = “geeksforgeeks”, M = 2, K = 1
Output: 4
Explanation:
All distinct substrings of length 2 are “ge”, “ee”, “ek”, “ks”, “sf”, “fo”, “or”, “rg”.
Out of all these strings, “sf”, “fo”, “or”, “rg” occurs once in the string S.
Therefore, the count is 4.

Naive Approach: The simplest approach is to generate all substrings of length M and store the frequency of each substring in the string S in a Map. Now, traverse the Map and if the frequency is equal to K, then increment count by 1. After completing the above steps, print count as the result. 

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find count of substrings
// of length M occurring exactly P times
// in the string, S
void findCount(string& S, int M, int K)
{
 
    unordered_map<string, int> unmap;
 
    for (int i = 0; i <= S.size() - M; i++) {
        string s1 = S.substr(i, K);
        unmap[s1]++;
    }
 
    int count = 0;
 
    for (auto it : unmap) {
        if (it.second == K)
            count++;
    }
 
    cout << count;
}
 
// Driver Code
int main()
{
    string S = "geeksforgeeks";
    int M = 2, K = 1;
 
    // Function Call
    findCount(S, M, K);
 
    return 0;
}


Java




// Java code to implement the approach
 
import java.util.*;
 
class GFG {
    // Function to find count of substrings
    // of length M occurring exactly P times
    // in the string, S
    static void findCount(String S, int M, int K) {
        HashMap<String, Integer> unmap = new HashMap<String, Integer>();
 
        for (int i = 0; i <= S.length() - M; i++) {
            String s1 = S.substring(i, i + K);
            if (unmap.containsKey(s1)) {
                unmap.put(s1, unmap.get(s1) + 1);
            }
            else {
                unmap.put(s1, 1);
            }
        }
 
        int count = 0;
 
        for (Map.Entry<String, Integer> it : unmap.entrySet()) {
            if (it.getValue() == K)
                count++;
        }
 
        System.out.println(count);
    }
 
    // Driver Code
    public static void main(String[] args) {
        String S = "geeksforgeeks";
        int M = 2, K = 1;
 
        // Function Call
        findCount(S, M, K);
    }
}
 
 
// This code is contributed by phasing17


Python3




# Python3 code to implement the approach
 
#  Function to find count of substrings
# of length M occurring exactly P times
# in the string, S
def find_count(s: str, m: int, k: int) -> int:
    # create an empty dictionary
    unmap = {}
 
    # iterate through the string, s, with a sliding window of size m
    for i in range(len(s) - m + 1):
        # extract the substring of size k
        s1 = s[i:i+k]
        # add the substring to the dictionary, or increment its count if it already exists
        unmap[s1] = unmap.get(s1, 0) + 1
 
    # initialize a count variable
    count = 0
    # iterate through the dictionary
    for key, value in unmap.items():
        # if the value (count of substring) is equal to k, increment the count variable
        if value == k:
            count += 1
    # return the count
    return count
 
# Driver code
S = "geeksforgeeks"
M = 2
K = 1
 
print(find_count(S, M, K))
 
# This code is contributed by phasing17.


Javascript




// JS program to implement the approach
 
// Function to find count of substrings
// of length M occurring exactly P times
// in the string, S
function findCount(S, M, K) {
     
  // Initializing a map to store counts
  const unmap = {};
 
  // Iterating over all but the last M indices
  for (let i = 0; i <= S.length - M; i++) {
    // Extracting the substring of length K starting from that index
    const s1 = S.substring(i, i + K);
     
    // Updating unmap
    if (!unmap[s1]) unmap[s1] = 0;
    unmap[s1]++;
  }
 
  let count = 0;
 
  for (const it in unmap) {
    if (unmap[it] === K) count++;
  }
 
  console.log(count);
}
 
// Driver Code
(function main() {
  const S = "geeksforgeeks";
  const M = 2;
  const K = 1;
 
  // Function Call
  findCount(S, M, K);
})();
 
 
// This code is contributed by phasing17


C#




using System;
using System.Linq;
using System.Collections.Generic;
 
class Program {
    // Function to find count of substrings
    // of length M occurring exactly P times
    // in the string, S
    static void FindCount(string S, int M, int K)
    {
        Dictionary<string, int> unmap
            = new Dictionary<string, int>();
 
        for (int i = 0; i <= S.Length - M; i++) {
            string s1 = S.Substring(i, K);
            if (unmap.ContainsKey(s1)) {
                unmap[s1]++;
            }
            else {
                unmap.Add(s1, 1);
            }
        }
 
        int count = 0;
 
        foreach(KeyValuePair<string, int> it in unmap)
        {
            if (it.Value == K)
                count++;
        }
 
        Console.WriteLine(count);
    }
 
    // Driver Code
    static void Main(string[] args)
    {
        string S = "geeksforgeeks";
        int M = 2, K = 1;
 
        // Function Call
        FindCount(S, M, K);
    }
}


Output

4

Time Complexity: O(N*M), where N and M are the length of the given string and the length of the substring needed respectively.
Auxiliary Space: O(N)

Efficient Approach: The above approach can be optimized by using the KMP algorithm for finding the frequency of a substring in the string. Follow the steps to solve the problem:

  • Initialize a variable, say count as 0, to store the number of the required substring.
  • Generate all substrings of length M from the string S and insert them in an array, say arr[].
  • Traverse the array arr[] and for each string in the array, calculate its frequency in the string S using KMP algorithm.
  • If the frequency of the string is equal to P, then increment the count by 1.
  • After completing the above steps, print the value of count as the resultant count of substrings.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to compute the LPS array
void computeLPSArray(string pat, int M,
                     int lps[])
{
    // Length of the previous
    // longest prefix suffix
    int len = 0;
    int i = 1;
    lps[0] = 0;
 
    // Iterate from [1, M - 1] to find lps[i]
    while (i < M) {
 
        // If the characters match
        if (pat[i] == pat[len]) {
 
            len++;
            lps[i] = len;
            i++;
        }
 
        // If pat[i] != pat[len]
        else {
 
            // If length is non-zero
            if (len != 0) {
                len = lps[len - 1];
 
                // Also, note that i is
                // not incremented here
            }
 
            // Otherwise
            else {
                lps[i] = len;
                i++;
            }
        }
    }
}
 
// Function to find the frequency of
// pat in the string txt
int KMPSearch(string pat, string txt)
{
    // Stores length of both strings
    int M = pat.length();
    int N = txt.length();
 
    // Initialize lps[] to store the
    // longest prefix suffix values
    // for the string pattern
    int lps[M];
 
    // Store the index for pat[]
    int j = 0;
 
    // Preprocess the pattern
    // (calculate lps[] array)
    computeLPSArray(pat, M, lps);
 
    // Store the index for txt[]
    int i = 0;
    int res = 0;
    int next_i = 0;
 
    while (i < N) {
        if (pat[j] == txt[i]) {
            j++;
            i++;
        }
        if (j == M) {
 
            // If pattern is found the
            // first time, iterate again
            // to check for more patterns
            j = lps[j - 1];
            res++;
 
            // Start i to check for more
            // than once occurrence
            // of pattern, 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 the required frequency
    return res;
}
 
// Function to find count of substrings
// of length M occurring exactly P times
// in the string, S
void findCount(string& S, int M, int P)
{
 
    // Store all substrings of length M
    set<string> vec;
 
    // Store the size of the string, S
    int n = S.length();
 
    // Pick starting point
    for (int i = 0; i < n; i++) {
 
        // Pick ending point
        for (int len = 1;
             len <= n - i; len++) {
 
            // If the substring is of
            // length M, insert it in vec
            string s = S.substr(i, len);
            if (s.length() == M) {
                vec.insert(s);
            }
        }
    }
 
    // Initialise count as 0 to store
    // the required count of substrings
    int count = 0;
 
    // Iterate through the set of
    // substrings
    for (auto it : vec) {
 
        // Store its frequency
        int ans = KMPSearch(it, S);
 
        // If frequency is equal to P
        if (ans == P) {
 
            // Increment count by 1
            count++;
        }
    }
 
    // Print the answer
    cout << count;
}
 
// Driver Code
int main()
{
    string S = "abacaba";
    int M = 3, P = 2;
 
    // Function Call
    findCount(S, M, P);
 
    return 0;
}


Java




// Java Program to implement
// the above approach
 
import java.io.*;
import java.util.*;
 
class GFG {
 
    // Function to compute the LPS array
    static void computeLPSArray(String pat, int M,
                                int lps[])
    {
        // Length of the previous
        // longest prefix suffix
        int len = 0;
        int i = 1;
        lps[0] = 0;
 
        // Iterate from [1, M - 1] to find lps[i]
        while (i < M) {
 
            // If the characters match
            if (pat.charAt(i) == pat.charAt(len)) {
 
                len++;
                lps[i] = len;
                i++;
            }
 
            // If pat[i] != pat[len]
            else {
 
                // If length is non-zero
                if (len != 0) {
                    len = lps[len - 1];
 
                    // Also, note that i is
                    // not incremented here
                }
 
                // Otherwise
                else {
                    lps[i] = len;
                    i++;
                }
            }
        }
    }
 
    // Function to find the frequency of
    // pat in the string txt
    static int KMPSearch(String pat, String txt)
    {
        // Stores length of both strings
        int M = pat.length();
        int N = txt.length();
 
        // Initialize lps[] to store the
        // longest prefix suffix values
        // for the string pattern
        int lps[] = new int[M];
 
        // Store the index for pat[]
        int j = 0;
 
        // Preprocess the pattern
        // (calculate lps[] array)
        computeLPSArray(pat, M, lps);
 
        // Store the index for txt[]
        int i = 0;
        int res = 0;
        int next_i = 0;
 
        while (i < N) {
            if (pat.charAt(j) == txt.charAt(i)) {
                j++;
                i++;
            }
            if (j == M) {
 
                // If pattern is found the
                // first time, iterate again
                // to check for more patterns
                j = lps[j - 1];
                res++;
 
                // Start i to check for more
                // than once occurrence
                // of pattern, 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 the required frequency
        return res;
    }
 
    // Function to find count of substrings
    // of length M occurring exactly P times
    // in the string, S
    static void findCount(String S, int M, int P)
    {
 
        // Store all substrings of length M
        // set<string> vec;
        TreeSet<String> vec = new TreeSet<>();
 
        // Store the size of the string, S
        int n = S.length();
 
        // Pick starting point
        for (int i = 0; i < n; i++) {
 
            // Pick ending point
            for (int len = 1; len <= n - i; len++) {
 
                // If the substring is of
                // length M, insert it in vec
                String s = S.substring(i, i + len);
                if (s.length() == M) {
                    vec.add(s);
                }
            }
        }
 
        // Initialise count as 0 to store
        // the required count of substrings
        int count = 0;
 
        // Iterate through the set of
        // substrings
        for (String it : vec) {
 
            // Store its frequency
            int ans = KMPSearch(it, S);
 
            // If frequency is equal to P
            if (ans == P) {
 
                // Increment count by 1
                count++;
            }
        }
 
        // Print the answer
        System.out.println(count);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
 
        String S = "abacaba";
        int M = 3, P = 2;
 
        // Function Call
        findCount(S, M, P);
    }
}
 
// This code is contributed by kingash.


Python3




# Python 3 program for the above approach
 
# Function to compute the LPS array
def computeLPSArray(pat, M, lps):
   
    # Length of the previous
    # longest prefix suffix
    len1 = 0
    i = 1
    lps[0] = 0
 
    # Iterate from [1, M - 1] to find lps[i]
    while (i < M):
       
        # If the characters match
        if (pat[i] == pat[len1]):
            len1 += 1
            lps[i] = len1
            i += 1
 
        # If pat[i] != pat[len]
        else:
            # If length is non-zero
            if (len1 != 0):
                len1 = lps[len1 - 1]
 
                # Also, note that i is
                # not incremented here
 
            # Otherwise
            else:
                lps[i] = len1
                i += 1
 
# Function to find the frequency of
# pat in the string txt
def KMPSearch(pat, txt):
   
    # Stores length of both strings
    M = len(pat)
    N = len(txt)
 
    # Initialize lps[] to store the
    # longest prefix suffix values
    # for the string pattern
    lps = [0 for i in range(M)]
 
    # Store the index for pat[]
    j = 0
 
    # Preprocess the pattern
    # (calculate lps[] array)
    computeLPSArray(pat, M, lps)
 
    # Store the index for txt[]
    i = 0
    res = 0
    next_i = 0
 
    while (i < N):
        if (pat[j] == txt[i]):
            j += 1
            i += 1
        if (j == M):
           
            # If pattern is found the
            # first time, iterate again
            # to check for more patterns
            j = lps[j - 1]
            res += 1
 
            # Start i to check for more
            # than once occurrence
            # of pattern, reset i to
            # previous start + 1
            if (lps[j] != 0):
                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 the required frequency
    return res
 
# Function to find count of substrings
# of length M occurring exactly P times
# in the string, S
def findCount(S, M, P):
   
    # Store all substrings of length M
    vec = set()
 
    # Store the size of the string, S
    n = len(S)
 
    # Pick starting point
    for i in range(n):
       
        # Pick ending point
        for len1 in range(n - i + 1):
           
            # If the substring is of
            # length M, insert it in vec
            s = S[i:len1]
             
          #  if (len1(s) == M):
           #     vec.add(s)
 
    # Initialise count as 0 to store
    # the required count of substrings
    count = 1
 
    # Iterate through the set of
    # substrings
    for it in vec:
       
        # Store its frequency
        ans = KMPSearch(it, S)
 
        # If frequency is equal to P
        if (ans == P):
           
            # Increment count by 1
            count += 1
 
    # Print the answer
    print(count)
 
# Driver Code
if __name__ == '__main__':
    S = "abacaba"
    M = 3
    P = 2
 
    # Function Call
    findCount(S, M, P)
     
    # This code is contributed by ipg2016107.


C#




// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
 
  // Function to compute the LPS array
  static void computeLPSArray(string pat, int M, int[] lps)
  {
 
    // Length of the previous
    // longest prefix suffix
    int len = 0;
    int i = 1;
    lps[0] = 0;
 
    // Iterate from [1, M - 1] to find lps[i]
    while (i < M)
    {
 
      // If the characters match
      if (pat[i] == pat[len])
      {
        len++;
        lps[i] = len;
        i++;
      }
 
      // If pat[i] != pat[len]
      else {
 
        // If length is non-zero
        if (len != 0) {
          len = lps[len - 1];
 
          // Also, note that i is
          // not incremented here
        }
 
        // Otherwise
        else {
          lps[i] = len;
          i++;
        }
      }
    }
  }
 
  // Function to find the frequency of
  // pat in the string txt
  static int KMPSearch(string pat, string txt)
  {
 
    // Stores length of both strings
    int M = pat.Length;
    int N = txt.Length;
 
    // Initialize lps[] to store the
    // longest prefix suffix values
    // for the string pattern
    int[] lps = new int[M];
 
    // Store the index for pat[]
    int j = 0;
 
    // Preprocess the pattern
    // (calculate lps[] array)
    computeLPSArray(pat, M, lps);
 
    // Store the index for txt[]
    int i = 0;
    int res = 0;
    int next_i = 0;
 
    while (i < N) {
      if (pat[j] == txt[i]) {
        j++;
        i++;
      }
      if (j == M) {
 
        // If pattern is found the
        // first time, iterate again
        // to check for more patterns
        j = lps[j - 1];
        res++;
 
        // Start i to check for more
        // than once occurrence
        // of pattern, 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 the required frequency
    return res;
  }
 
  // Function to find count of substrings
  // of length M occurring exactly P times
  // in the string, S
  static void findCount(string S, int M, int P)
  {
 
    // Store all substrings of length M
    HashSet<string> vec = new HashSet<string>();
 
    // Store the size of the string, S
    int n = S.Length;
 
    // Pick starting point
    for (int i = 0; i < n; i++) {
 
      // Pick ending point
      for (int len = 1;
           len <= n - i; len++) {
 
        // If the substring is of
        // length M, insert it in vec
        string s = S.Substring(i, len);
        if (s.Length == M) {
          vec.Add(s);
        }
      }
    }
 
    // Initialise count as 0 to store
    // the required count of substrings
    int count = 0;
 
    // Iterate through the set of
    // substrings
    foreach(string it in vec) {
 
      // Store its frequency
      int ans = KMPSearch(it, S);
 
      // If frequency is equal to P
      if (ans == P) {
 
        // Increment count by 1
        count++;
      }
    }
 
    // Print the answer
    Console.WriteLine(count);
  }
 
  // Driver code
  static void Main() {
    string S = "abacaba";
    int M = 3, P = 2;
 
    // Function Call
    findCount(S, M, P);
  }
}
 
// This code is contributed by divyeshrabadiya07.


Javascript




<script>
 
//Javascript implementation of the approach
 
// Function to compute the LPS array
function computeLPSArray(pat, M, lps)
{
    // Length of the previous
    // longest prefix suffix
    var len = 0;
    var i = 1;
    lps[0] = 0;
 
    // Iterate from [1, M - 1] to find lps[i]
    while (i < M) {
 
        // If the characters match
        if (pat[i] == pat[len]) {
 
            len++;
            lps[i] = len;
            i++;
        }
 
        // If pat[i] != pat[len]
        else {
 
            // If length is non-zero
            if (len != 0) {
                len = lps[len - 1];
 
                // Also, note that i is
                // not incremented here
            }
 
            // Otherwise
            else {
                lps[i] = len;
                i++;
            }
        }
    }
}
 
// Function to find the frequency of
// pat in the string txt
function KMPSearch(pat, txt)
{
    // Stores length of both strings
    var M = pat.length;
    var N = txt.length;
 
    // Initialize lps[] to store the
    // longest prefix suffix values
    // for the string pattern
    var lps = new Array(M);
 
    // Store the index for pat[]
    var j = 0;
 
    // Preprocess the pattern
    // (calculate lps[] array)
    computeLPSArray(pat, M, lps);
 
    // Store the index for txt[]
    var i = 0;
    var res = 0;
    var next_i = 0;
 
    while (i < N) {
        if (pat[j] == txt[i]) {
            j++;
            i++;
        }
        if (j == M) {
 
            // If pattern is found the
            // first time, iterate again
            // to check for more patterns
            j = lps[j - 1];
            res++;
 
            // Start i to check for more
            // than once occurrence
            // of pattern, 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 the required frequency
    return res;
}
 
// Function to find count of substrings
// of length M occurring exactly P times
// in the string, S
function findCount( S, M, P)
{
 
    // Store all substrings of length M
    var vec = new Set();
 
    // Store the size of the string, S
    var n = S.length;
 
    // Pick starting point
    for (var i = 0; i < n; i++) {
 
        // Pick ending point
        for (var len = 1;
             len <= n - i; len++) {
 
            // If the substring is of
            // length M, insert it in vec
            var s = S.substring(i, len);
            if (s.length == M) {
                vec.add(s);
            }
        }
    }
 
    // Initialise count as 0 to store
    // the required count of substrings
    var count = 0;
 
    // Iterate through the set of
    // substrings
    for (const it of vec){
 
        // Store its frequency
        var ans = KMPSearch(it, S);
 
        // If frequency is equal to P
        if (ans == P) {
 
            // Increment count by 1
            count++;
        }
    }
 
    // Print the answer
    document.write( count);
}
 
 
var S = "abacaba";
var M = 3, P = 2;
// Function Call
findCount(S, M, P);
 
// This code is contributed by SoumikMondal
 
</script>


Output: 

1

 

Time Complexity: O((N*M) + (N2 – M2))
Auxiliary Space: O(N – M)


Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!

Last Updated : 27 Jan, 2023
Like Article
Save Article
Previous
Next
Similar Reads
Complete Tutorials