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++
// C++ program for above approach#include <iostream>#include <string>using namespace std;// Function to count the number of// substrings consists of 0, 1, and 2int 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 Codeint main(){ string str = "00021"; int count = countSubstrings(str); cout << count; return 0;}// This code is contributed by Kdheeraj. |
Java
// Java program for the above approachimport 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)); }} |
Python3
# Python program for above approach# Function to count the number of# substrings consists of 0, 1, and 2def 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 Codestr = "00021"count = countSubstrings(str)print(count)# This code is contributed by shivanisinghss2110 |
C#
// C# program for the above approachusing 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 |
Javascript
<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> |
Output:
3
Time Complexity: O(N)
Auxiliary Space: O(1)
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