given a string s and a number k, find the number of substrings of length k such that each character in the substring is present only once.

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Source: https://www.geeksforgeeks.org/number-substrings-count-character-k/ Given a string and an integer k, find number of substrings in which all the different characters occurs exactly k times.

Number of substrings with count of each character as k

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Source: https://www.geeksforgeeks.org/number-substrings-count-character-k/

Given a string and an integer k, find number of substrings in which all the different characters occurs exactly k times.

Looking for a solution in O(n), using two pointers/sliding window approach. I'm able to find only longest substrings satisfying this criteria but not substrings within that long substring.

For ex: ababbaba, k = 2

My solution finds abab, ababba etc, but not bb within ababba.

Can someone help me with the logic?

stringdata-structuressubstring

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edited Jun 12, 2020 at 18:13

asked Jun 12, 2020 at 17:55

NewDev 511 1 gold badge 2 2 silver badges 7 7 bronze badges

Also, would like to know if this can be done in O(n) or need O(n2) algorithm –

NewDev

Jun 12, 2020 at 18:07

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7 Answers

If you could edit your question to include your solution code, I'd be happy to help you with that.

For now I'm sharing my solution code (in java) which runs in O(n2). I've added enough comments to make the code self explanatory. Nonetheless the logic for the solution is as follows:

As you correctly pointed out, the problem can be solved using sliding window approach (with variable window size). The solution below considers all possible sub-strings, using nested for loops for setting start and end indices. For each sub-string, we check if every element in the sub-string occurs exactly k times.

To avoid recalculating the count for every sub-string, we maintain the count in a map, and keep putting new elements in the map as we increment the end index (slide the window). This ensures that our solution runs in O(n2) and not O(n3).

To further improve efficiency, we only check the count of individual elements if the sub-string's size matches our requirement. e.g. for n unique elements (keys in the map), the size of required sub-string would be n*k. If the sub-string's size doesn't match this value, there's no need to check how many times the individual characters occur.

import java.util.*; /**

* Java program to count the number of perfect substrings in a given string. A

* substring is considered perfect if all the elements within the substring

* occur exactly k number of times.

* * @author Codextor */

public class PerfectSubstring {

public static void main(String[] args) {

String s = "aabbcc";

int k = 2;

System.out.println(perfectSubstring(s, k));

s = "aabccc"; k = 2;

System.out.println(perfectSubstring(s, k));

} /**

* Returns the number of perfect substrings in the given string for the

* specified value of k

* @param s The string to check for perfect substrings

* @param k The number of times every element should occur within the substring

* @return int The number of perfect substrings

public static int perfectSubstring(String s, int k) {

int finalCount = 0; /*

* Set the initial starting index for the subarray as 0, and increment it with

* every iteration, till the last index of the string is reached.

for (int start = 0; start < s.length(); start++) {

* Use a HashMap to store the count of every character in the subarray. We'll

* start with an empty map everytime we update the starting index

Map

frequencyMap = new HashMap<>();

* Set the initial ending index for the subarray equal to the starting index and

* increment it with every iteration, till the last index of the string is

* reached. */

for (int end = start; end < s.length(); end++) {

* Get the count of the character at end index and increase it by 1. If the

* character is not present in the map, use 0 as the default count

char c = s.charAt(end);

int count = frequencyMap.getOrDefault(c, 0);

frequencyMap.put(c, count + 1);

* Check if the length of the subarray equals the desired length. The desired

* length is the number of unique characters we've seen so far (size of the map)

* multilied by k (the number of times each character should occur). If the

* length is as per requiremets, check if each element occurs exactly k times

if (frequencyMap.size() * k == (end - start + 1)) {

if (check(frequencyMap, k)) {

finalCount++; } } } } return finalCount; } /**

* Returns true if every value in the map is equal to k

* @param map The map whose values are to be checked

* @param k The required value for keys in the map

* @return true if every value in the map is equal to k

public static boolean check(Map

map, int k) {

* Iterate through all the values (frequency of each character), comparing them

* with k */

for (Integer i : map.values()) {

if (i != k) { return false; } } return true; } } Share Improve this answer

स्रोत : stackoverflow.com

How many substrings can be formed from a character string of length n?

Answer (1 of 9): Let us solve this problem in segments. let the string be 1234. The substring of lenght 1 is 1, 2, 3, and 4. The substring of lenght 2 is 12, 23, 34. The substring of length 3 is 123, 234. and at last substring of lenght 4 is 1234. so it is simply 4 + 3 + 2 + 1. Or you can say n*(...

How many substrings can be formed from a character string of length n?

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Leticia Lorena Rodriguez

Studied Computer Science at University of Buenos Aires5y

Originally Answered: how many substrings can be formed from a character string of length n ?

To solve this problem, we can first think in: “how many substrings of length 1 can be formed from a character string of length n?”

It’s a simple answer. If we take step of 1 character, we can from n substrings from the original string (we are going to call it s).

For example, if s = “abcde”, s with length n = 5, we can form these substrings of length 1 : “a”, “b”, “c”, “d”, “e” (5). So, we formed n substrings.

But in the original question, we want to know how many substrings of any length. This answer is not enough.

Let see how many substrings of length 2 can be formed. We are going to take 1 char

Number of substrings with count of each character as k

A Computer Science portal for geeks. It contains well written, well thought and well explained computer science and programming articles, quizzes and practice/competitive programming/company interview Questions.

Number of substrings with count of each character as k

Difficulty Level : Easy

Last Updated : 12 Dec, 2022

Read Discuss(8) Courses Practice Video

Given a string and an integer k, find the number of substrings in which all the different characters occur exactly k times.

Examples:Input : s = "aabbcc"

k = 2

Output : 6

The substrings are aa, bb, cc,

aabb, bbcc and aabbcc.

Input : s = "aabccc"

k = 2 Output : 3

There are three substrings aa,

cc and cc

Recommended: Please try your approach on first, before moving on to the solution.

Naive Approach: The idea is to traverse through all substrings. We fix a starting point, traverse through all substrings starting with the picked point, we keep incrementing frequencies of all characters. If all frequencies become k, we increment the result. If the count of any frequency becomes more than k, we break and change starting point.

Below is the implementation of the above approach:

C++

// C++ program to count number of substrings

// with counts of distinct characters as k.

#include

using namespace std;

const int MAX_CHAR = 26;

// Returns true if all values

// in freq[] are either 0 or k.

bool check(int freq[], int k)

{

for (int i = 0; i < MAX_CHAR; i++)

if (freq[i] && freq[i] != k)

return false; return true; }

// Returns count of substrings where frequency

// of every present character is k

int substrings(string s, int k)

{

int res = 0; // Initialize result

// Pick a starting point

for (int i = 0; s[i]; i++) {

// Initialize all frequencies as 0

// for this starting point

int freq[MAX_CHAR] = { 0 };

// One by one pick ending points

for (int j = i; s[j]; j++) {

// Increment frequency of current char

int index = s[j] - 'a';

freq[index]++;

// If frequency becomes more than

// k, we can't have more substrings

// starting with i

if (freq[index] > k)

break;

// If frequency becomes k, then check

// other frequencies as well.

else if (freq[index] == k &&

check(freq, k) == true)

res++; } } return res; } // Driver code int main() {

string s = "aabbcc";

int k = 2;

cout << substrings(s, k) << endl;

s = "aabbc"; k = 2;

cout << substrings(s, k) << endl;

}

Java

// Java program to count number of substrings

// with counts of distinct characters as k.

import java.io.*; class GFG {

static int MAX_CHAR = 26;

// Returns true if all values

// in freq[] are either 0 or k.

static boolean check(int freq[], int k)

{

for (int i = 0; i < MAX_CHAR; i++)

if (freq[i] !=0 && freq[i] != k)

return false; return true; }

// Returns count of substrings where frequency

// of every present character is k

static int substrings(String s, int k)

{

int res = 0; // Initialize result

// Pick a starting point

for (int i = 0; i< s.length(); i++)

{

// Initialize all frequencies as 0

// for this starting point

int freq[] = new int[MAX_CHAR];

// One by one pick ending points

for (int j = i; j

{

// Increment frequency of current char

int index = s.charAt(j) - 'a';

freq[index]++;

// If frequency becomes more than

// k, we can't have more substrings

// starting with i

if (freq[index] > k)

break;

// If frequency becomes k, then check

// other frequencies as well.

else if (freq[index] == k &&

check(freq, k) == true)

res++; } } return res; } // Driver code

public static void main(String[] args)

{

String s = "aabbcc";

int k = 2;

System.out.println(substrings(s, k));

s = "aabbc"; k = 2;

System.out.println(substrings(s, k));

} }

// This code has been contributed by 29AjayKumar

Python3

# Python3 program to count number of substrings

# with counts of distinct characters as k.

MAX_CHAR = 26

# Returns true if all values

# in freq[] are either 0 or k.

def check(freq, k):

for i in range(0, MAX_CHAR):

if(freq[i] and freq[i] != k):

return False return True

# Returns count of substrings where

# frequency of every present character is k

def substrings(s, k):

res = 0 # Initialize result

# Pick a starting point

for i in range(0, len(s)):

# Initialize all frequencies as 0

# for this starting point

freq = [0] * MAX_CHAR

# One by one pick ending points

for j in range(i, len(s)):

# Increment frequency of current char

index = ord(s[j]) - ord('a')

freq[index] += 1

# If frequency becomes more than

# k, we can't have more substrings

# starting with i

if(freq[index] > k):

break

# If frequency becomes k, then check

# other frequencies as well

elif(freq[index] == k and

check(freq, k) == True):

res += 1 return res # Driver Code

if __name__ == "__main__":

स्रोत : www.geeksforgeeks.org

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Mohammed 1 month ago

Guys, does anyone know the answer?

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given a string s and a number k, find the number of substrings of length k such that each character in the substring is present only once.

Mohammed

string

Number of substrings with count of each character as k

7 Answers

How many substrings can be formed from a character string of length n?

Number of substrings with count of each character as k

Number of substrings with count of each character as k

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