Count M-length substrings occurring exactly K times in a string
Last Updated :
27 Jan, 2023
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++
#include <bits/stdc++.h>
using namespace std;
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;
}
int main()
{
string S = "geeksforgeeks";
int M = 2, K = 1;
findCount(S, M, K);
return 0;
}
|
Java
import java.util.*;
class GFG {
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);
}
public static void main(String[] args) {
String S = "geeksforgeeks";
int M = 2, K = 1;
findCount(S, M, K);
}
}
|
Python3
def find_count(s: str, m: int, k: int) -> int:
unmap = {}
for i in range(len(s) - m + 1):
s1 = s[i:i+k]
unmap[s1] = unmap.get(s1, 0) + 1
count = 0
for key, value in unmap.items():
if value == k:
count += 1
return count
S = "geeksforgeeks"
M = 2
K = 1
print(find_count(S, M, K))
|
Javascript
function findCount(S, M, K) {
const unmap = {};
for (let i = 0; i <= S.length - M; i++) {
const s1 = S.substring(i, i + K);
if (!unmap[s1]) unmap[s1] = 0;
unmap[s1]++;
}
let count = 0;
for (const it in unmap) {
if (unmap[it] === K) count++;
}
console.log(count);
}
(function main() {
const S = "geeksforgeeks";
const M = 2;
const K = 1;
findCount(S, M, K);
})();
|
C#
using System;
using System.Linq;
using System.Collections.Generic;
class Program {
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);
}
static void Main(string[] args)
{
string S = "geeksforgeeks";
int M = 2, K = 1;
FindCount(S, M, K);
}
}
|
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++
#include <bits/stdc++.h>
using namespace std;
void computeLPSArray(string pat, int M,
int lps[])
{
int len = 0;
int i = 1;
lps[0] = 0;
while (i < M) {
if (pat[i] == pat[len]) {
len++;
lps[i] = len;
i++;
}
else {
if (len != 0) {
len = lps[len - 1];
}
else {
lps[i] = len;
i++;
}
}
}
}
int KMPSearch(string pat, string txt)
{
int M = pat.length();
int N = txt.length();
int lps[M];
int j = 0;
computeLPSArray(pat, M, lps);
int i = 0;
int res = 0;
int next_i = 0;
while (i < N) {
if (pat[j] == txt[i]) {
j++;
i++;
}
if (j == M) {
j = lps[j - 1];
res++;
if (lps[j] != 0)
i = ++next_i;
j = 0;
}
else if (i < N
&& pat[j] != txt[i]) {
if (j != 0)
j = lps[j - 1];
else
i = i + 1;
}
}
return res;
}
void findCount(string& S, int M, int P)
{
set<string> vec;
int n = S.length();
for (int i = 0; i < n; i++) {
for (int len = 1;
len <= n - i; len++) {
string s = S.substr(i, len);
if (s.length() == M) {
vec.insert(s);
}
}
}
int count = 0;
for (auto it : vec) {
int ans = KMPSearch(it, S);
if (ans == P) {
count++;
}
}
cout << count;
}
int main()
{
string S = "abacaba";
int M = 3, P = 2;
findCount(S, M, P);
return 0;
}
|
Java
import java.io.*;
import java.util.*;
class GFG {
static void computeLPSArray(String pat, int M,
int lps[])
{
int len = 0;
int i = 1;
lps[0] = 0;
while (i < M) {
if (pat.charAt(i) == pat.charAt(len)) {
len++;
lps[i] = len;
i++;
}
else {
if (len != 0) {
len = lps[len - 1];
}
else {
lps[i] = len;
i++;
}
}
}
}
static int KMPSearch(String pat, String txt)
{
int M = pat.length();
int N = txt.length();
int lps[] = new int[M];
int j = 0;
computeLPSArray(pat, M, lps);
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) {
j = lps[j - 1];
res++;
if (lps[j] != 0)
i = ++next_i;
j = 0;
}
else if (i < N
&& pat.charAt(j) != txt.charAt(i)) {
if (j != 0)
j = lps[j - 1];
else
i = i + 1;
}
}
return res;
}
static void findCount(String S, int M, int P)
{
TreeSet<String> vec = new TreeSet<>();
int n = S.length();
for (int i = 0; i < n; i++) {
for (int len = 1; len <= n - i; len++) {
String s = S.substring(i, i + len);
if (s.length() == M) {
vec.add(s);
}
}
}
int count = 0;
for (String it : vec) {
int ans = KMPSearch(it, S);
if (ans == P) {
count++;
}
}
System.out.println(count);
}
public static void main(String[] args)
{
String S = "abacaba";
int M = 3, P = 2;
findCount(S, M, P);
}
}
|
Python3
def computeLPSArray(pat, M, lps):
len1 = 0
i = 1
lps[0] = 0
while (i < M):
if (pat[i] == pat[len1]):
len1 += 1
lps[i] = len1
i += 1
else:
if (len1 != 0):
len1 = lps[len1 - 1]
else:
lps[i] = len1
i += 1
def KMPSearch(pat, txt):
M = len(pat)
N = len(txt)
lps = [0 for i in range(M)]
j = 0
computeLPSArray(pat, M, lps)
i = 0
res = 0
next_i = 0
while (i < N):
if (pat[j] == txt[i]):
j += 1
i += 1
if (j == M):
j = lps[j - 1]
res += 1
if (lps[j] != 0):
next_i += 1
i = next_i
j = 0
elif (i < N and pat[j] != txt[i]):
if (j != 0):
j = lps[j - 1]
else:
i = i + 1
return res
def findCount(S, M, P):
vec = set()
n = len(S)
for i in range(n):
for len1 in range(n - i + 1):
s = S[i:len1]
count = 1
for it in vec:
ans = KMPSearch(it, S)
if (ans == P):
count += 1
print(count)
if __name__ == '__main__':
S = "abacaba"
M = 3
P = 2
findCount(S, M, P)
|
C#
using System;
using System.Collections.Generic;
class GFG
{
static void computeLPSArray(string pat, int M, int[] lps)
{
int len = 0;
int i = 1;
lps[0] = 0;
while (i < M)
{
if (pat[i] == pat[len])
{
len++;
lps[i] = len;
i++;
}
else {
if (len != 0) {
len = lps[len - 1];
}
else {
lps[i] = len;
i++;
}
}
}
}
static int KMPSearch(string pat, string txt)
{
int M = pat.Length;
int N = txt.Length;
int[] lps = new int[M];
int j = 0;
computeLPSArray(pat, M, lps);
int i = 0;
int res = 0;
int next_i = 0;
while (i < N) {
if (pat[j] == txt[i]) {
j++;
i++;
}
if (j == M) {
j = lps[j - 1];
res++;
if (lps[j] != 0)
i = ++next_i;
j = 0;
}
else if (i < N
&& pat[j] != txt[i]) {
if (j != 0)
j = lps[j - 1];
else
i = i + 1;
}
}
return res;
}
static void findCount(string S, int M, int P)
{
HashSet<string> vec = new HashSet<string>();
int n = S.Length;
for (int i = 0; i < n; i++) {
for (int len = 1;
len <= n - i; len++) {
string s = S.Substring(i, len);
if (s.Length == M) {
vec.Add(s);
}
}
}
int count = 0;
foreach(string it in vec) {
int ans = KMPSearch(it, S);
if (ans == P) {
count++;
}
}
Console.WriteLine(count);
}
static void Main() {
string S = "abacaba";
int M = 3, P = 2;
findCount(S, M, P);
}
}
|
Javascript
<script>
function computeLPSArray(pat, M, lps)
{
var len = 0;
var i = 1;
lps[0] = 0;
while (i < M) {
if (pat[i] == pat[len]) {
len++;
lps[i] = len;
i++;
}
else {
if (len != 0) {
len = lps[len - 1];
}
else {
lps[i] = len;
i++;
}
}
}
}
function KMPSearch(pat, txt)
{
var M = pat.length;
var N = txt.length;
var lps = new Array(M);
var j = 0;
computeLPSArray(pat, M, lps);
var i = 0;
var res = 0;
var next_i = 0;
while (i < N) {
if (pat[j] == txt[i]) {
j++;
i++;
}
if (j == M) {
j = lps[j - 1];
res++;
if (lps[j] != 0)
i = ++next_i;
j = 0;
}
else if (i < N
&& pat[j] != txt[i]) {
if (j != 0)
j = lps[j - 1];
else
i = i + 1;
}
}
return res;
}
function findCount( S, M, P)
{
var vec = new Set();
var n = S.length;
for (var i = 0; i < n; i++) {
for (var len = 1;
len <= n - i; len++) {
var s = S.substring(i, len);
if (s.length == M) {
vec.add(s);
}
}
}
var count = 0;
for (const it of vec){
var ans = KMPSearch(it, S);
if (ans == P) {
count++;
}
}
document.write( count);
}
var S = "abacaba";
var M = 3, P = 2;
findCount(S, M, P);
</script>
|
Time Complexity: O((N*M) + (N2 – M2))
Auxiliary Space: O(N – M)
Similar Reads
Smallest String consisting of a String S exactly K times as a Substring
Given a string S of length N and integer K, find the smallest length string which contains the string S as a sub string exactly K times. Examples: Input: S = "abba", K = 3 Output: abbabbabba Explanation: The string "abba" occurs K times in the string abbabbabba, i.e. {abbabbabba, abbabbabba, abbabba
6 min read
Count the number of vowels occurring in all the substrings of given string
Given a string of length N of lowercase characters containing 0 or more vowels, the task is to find the count of vowels that occurred in all the substrings of the given string. Examples: Input: str = "abc" Output: 3The given string "abc" contains only one vowel = 'a' Substrings of "abc" are = {"a",
12 min read
Count subarrays having exactly K elements occurring at least twice
Given an array arr[] consisting of N integers and a positive integer K, the task is to count the number of subarrays having exactly K elements occurring at least twice. Examples: Input: arr[] = {1, 1, 1, 2, 2}, K = 1Output: 7Explanation: The subarrays having exactly 1 element occurring at least twic
11 min read
Count of Distinct Substrings occurring consecutively in a given String
Given a string str, the task is to find the number of distinct substrings that are placed consecutively in the given string. Examples: Input: str = "geeksgeeksforgeeks" Output: 2 Explanation: geeksgeeksforgeeks -> {"geeks"} geeksgeeksforgeeks -> {"e"} Only one consecutive occurrence of "e" is
14 min read
Count substrings with each character occurring at most k times
Given a string S. Count number of substrings in which each character occurs at most k times. Assume that the string consists of only lowercase English alphabets. Examples: Input : S = ab k = 1 Output : 3 All the substrings a, b, ab have individual character count less than 1. Input : S = aaabb k = 2
15+ min read
Count of K length substrings containing exactly X vowels
Given string str of N characters containing both uppercase and lowercase English alphabets, the task is to find the count of substrings of size K containing exactly X vowels. Examples: Input: str = "GeeksForGeeks", K = 2, X = 2Output: 2Explanation: The given string only contains 2 substrings of size
7 min read
Convert to a string that is repetition of a substring of k length
Given a string, find if it is possible to convert it to a string that is the repetition of a substring with k characters. To convert, we can replace one substring of length k starting at index i (zero-based indexing) such that i is divisible by K, with k characters. Examples: Input: str = "bdac", k
7 min read
Count of strings whose prefix match with the given string to a given length k
Given an array of strings arr[] and given some queries where each query consists of a string str and an integer k. The task is to find the count of strings in arr[] whose prefix of length k matches with the k length prefix of str.Examples: Input: arr[] = {"abba", "abbb", "abbc", "abbd", "abaa", "abc
15+ min read
Count ways to split string into K Substrings starting with even digit and min length M
Given a string s of length n consisting of digits from '0' to '9'. The task is to determine the total number of possible ways to partition the string into k substrings such that : Each substring has a minimum length of m (m >= 2).Substring must start with an even digit number and end with an odd
15+ min read
Print all words occurring in a sentence exactly K times
Given a string S consisting of lowercase alphabets and an integer K, the task is to print all the words that occur K times in the string S. Examples: Input: S = "banana is in yellow and sun flower is also in yellow", K = 2Output: "is" "yellow" "in"Explanation: The words "is", "yellow" and "in" occur
13 min read