Count of substrings from given Ternary strings containing characters at least once

Given string str of size N consisting of only 0, 1, and 2, the task is to find the number of substrings that consists of characters 0, 1, and 2 at least once.

Examples: 

Input: str = "0122"
Output: 2
Explanation:
There exists 2 substrings such that the substrings has characters 0, 1, 2 at least once is "012" and "0122". Therefore, the count of substrings is 2.

Input: S = "00021"
Output: 3

Approach: The given problem can be solved using the Sliding Window Technique, the idea is to make the frequency array of the size 3 that contains the occurrence of 0, 1, and 2. Traverse the given string and update the freq array accordingly, if all 3 indexes in the array are greater than zero then count the minimum of them and increment it into variable count. Follow the steps below to solve the problem:

• Initialize an array freq[] of size 3 to store the frequency of all elements in the array.
• Initialize the variable count as 0 to store the answer and i as 0 to maintain the left pointer.
• Iterate over the range [0, N) using the variable j and perform the following tasks:
  • Increase the frequency of the current character str[I] in the array freq[] by 1.
  • Traverse in a while loop till freq[0], freq[1], and freq[2] are greater than 0, decrease the frequency of the character at i-th position by 1 and increase the value of i by 1.
  • Add the value of i to the variable count.
• After performing the above steps, print the value of count as the answer.

Below is the implementation of the above approach.

// C++ program for above approach
#include <iostream>
#include <string>
using namespace std;

// Function to count the number of
// substrings consists of 0, 1, and 2
int countSubstrings(string& str)
{
  
    // Initialize frequency array
    // of size 3
    int freq[3] = { 0 };

    // Stores the resultant count
    int count = 0;
    int i = 0;

    // Traversing string str
    for (int j = 0; j < str.length(); j++) {

        // Update frequency array
        freq[str[j] - '0']++;

        // If all the characters are
        // present counting number of
        // substrings possible
        while (freq[0] > 0 && freq[1] > 0 && freq[2] > 0) {
            freq[str[i++] - '0']--;
        }

        // Update number of substrings
        count += i;
    }

    // Return the number of substrings
    return count;
}

// Driver Code
int main()
{
    string str = "00021";
    int count = countSubstrings(str);
    cout << count;
    return 0;
}

// This code is contributed by Kdheeraj.

// Java program for the above approach
import java.util.*;

class GFG {

    // Function to count the number of
    // substrings consists of 0, 1, and 2
    public static int countSubstrings(String str)
    {
        // Initialize frequency array
        // of size 3
        int[] freq = new int[3];

        // Stores the resultant count
        int count = 0;
        int i = 0;

        // Traversing string str
        for (int j = 0;
             j < str.length(); j++) {

            // Update frequency array
            freq[str.charAt(j) - '0']++;

            // If all the characters are
            // present counting number of
            // substrings possible
            while (freq[0] > 0 && freq[1] > 0
                   && freq[2] > 0) {
                freq[str.charAt(i++) - '0']--;
            }

            // Update number of substrings
            count += i;
        }

        // Return the number of substrings
        return count;
    }

    // Driver Code
    public static void main(String[] args)
    {
        String str = "00021";
        System.out.println(
            countSubstrings(str));
    }
}

# Python program for above approach
# Function to count the number of
# substrings consists of 0, 1, and 2
def countSubstrings(str):

    # Initialize frequency array
    # of size 3
    freq = [ 0 ]*3

    # Stores the resultant count
    count = 0
    i = 0

    # Traversing string str
    for j in range ( 0 ,len(str)):

        # Update frequency array
        freq[ord(str[j]) - ord('0')] += 1

        # If all the characters are
        # present counting number of
        # substrings possible
        while (freq[0] > 0 and freq[1] > 0 and freq[2] > 0):
            i += 1
            freq[ord(str[i]) - ord('0')] -= 1
        
        # Update number of substrings
        count += i

    # Return the number of substrings
    return count

# Driver Code
str = "00021"
count = countSubstrings(str)
print(count)

# This code is contributed by shivanisinghss2110

// C# program for the above approach
using System;

class GFG {

    // Function to count the number of
    // substrings consists of 0, 1, and 2
    public static int countSubstrings(string str)
    {
      
        // Initialize frequency array
        // of size 3
        int[] freq = new int[3];

        // Stores the resultant count
        int count = 0;
        int i = 0;

        // Traversing string str
        for (int j = 0;
             j < str.Length; j++) {

            // Update frequency array
            freq[str[j] - '0']++;

            // If all the characters are
            // present counting number of
            // substrings possible
            while (freq[0] > 0 && freq[1] > 0
                   && freq[2] > 0) {
                freq[str[i++] - '0']--;
            }

            // Update number of substrings
            count += i;
        }

        // Return the number of substrings
        return count;
    }

    // Driver Code
    public static void Main(String[] args)
    {
        string str = "00021";
        Console.Write(countSubstrings(str));
    }
}

// This code is contributed by shivanisinghss2110

 <script>
        // JavaScript Program to implement
        // the above approach

        // Function to count the number of
        // substrings consists of 0, 1, and 2
        function countSubstrings(str) {

            // Initialize frequency array
            // of size 3
            let freq = new Array(3).fill(0)

            // Stores the resultant count
            let count = 0;
            let i = 0;

            // Traversing string str
            for (let j = 0; j < str.length; j++) {

                // Update frequency array
                freq[str.charCodeAt(j) - '0'.charCodeAt(0)]++;

                // If all the characters are
                // present counting number of
                // substrings possible
                while (freq[0] > 0 && freq[1] > 0 && freq[2] > 0) {
                    freq[str.charCodeAt(i++) - '0'.charCodeAt(0)]--;
                }

                // Update number of substrings
                count += i;
            }

            // Return the number of substrings
            return count;
        }

        // Driver Code

        let str = "00021";
        let count = countSubstrings(str);
        document.write(count);


// This code is contributed by Potta Lokesh
    </script>

3

Time Complexity: O(N)
Auxiliary Space: O(1)

kdheeraj

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Article Tags :

• Strings
• Mathematical
• DSA
• sliding-window
• frequency-counting
• substring

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Practice Tags :

• Mathematical
• sliding-window
• Strings