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++ 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 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 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. |
// 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 |
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);
}
} |
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++ 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 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. |
# 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# 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. |
<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> |
1
Time Complexity: O((N*M) + (N2 – M2))
Auxiliary Space: O(N – M)