Skip to content
Related Articles

Related Articles

Count of sub-strings that do not contain all the characters from the set {‘a’, ‘b’, ‘c’} at the same time
  • Difficulty Level : Hard
  • Last Updated : 20 Nov, 2019

Given a string str consisting only of the characters ‘a’, ‘b’ and ‘c’, find the number of sub-strings that do not contain all the three characters at the same time.

Examples:

Input: str = “abc”
Output: 5
The possible substrings are “a”, “b”, “c”, “ab” and “bc”

Input: str = “babac”
Output: 12

Approach: The idea is to use three variables a_index, b_index and c_index to store the latest occurrence of the characters a, b and c and another variable ans to store the number of substrings that don’t have at least one of a, b or c and initialize its value to the total number of substrings with length n i.e, n*(n+1)/2, where n is the length of the string.



Now simply traverse the string from the beginning. At each point update the value of variables to the latest value whenever encountered.Since we are indexing with 0, so we are updating the index as i+1 at each step.

Also at each step, you need to find the index of minimum occurrence of the remaining of the two characters not encountered in the current step.

Then simply deduct this index from the variable ans. This is because once you found the index of the minimum of the remaining characters you are sure that any more substrings formed by moving downwards in index will also contain all the three characters.

Hence the total numbers of all such substrings(ones containing all a, b, and c) formed at this step is the found index.

Below is the implementation of the above approach:

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the count of valid sub-strings
int CountSubstring(char str[], int n)
{
  
    // Variable ans to store all the possible substrings
    // Initialize its value as total number of substrings
    // that can be formed from the given string
    int ans = (n * (n + 1)) / 2;
  
    // Stores recent index of the characters
    int a_index = 0;
    int b_index = 0;
    int c_index = 0;
  
    for (int i = 0; i < n; i++) {
  
        // If character is a update a's index
        // and the variable ans
        if (str[i] == 'a') {
            a_index = i + 1;
            ans -= min(b_index, c_index);
        }
  
        // If character is b update b's index
        // and the variable ans
        else if (str[i] == 'b') {
            b_index = i + 1;
            ans -= min(a_index, c_index);
        }
  
        // If character is c update c's index
        // and the variable ans
        else {
            c_index = i + 1;
            ans -= min(a_index, b_index);
        }
    }
  
    return ans;
}
  
// Driver code
int main()
{
    char str[] = "babac";
    int n = strlen(str);
  
    cout << CountSubstring(str, n);
  
    return 0;
}

Java




// Java implementation of the approach
class GFG 
{
  
    // Function to return the count of valid sub-strings
    static int CountSubstring(char str[], int n) 
    {
  
        // Variable ans to store all the possible substrings
        // Initialize its value as total number of substrings
        // that can be formed from the given string
        int ans = (n * (n + 1)) / 2;
  
        // Stores recent index of the characters
        int a_index = 0;
        int b_index = 0;
        int c_index = 0;
  
        for (int i = 0; i < n; i++) 
        {
  
            // If character is a update a's index
            // and the variable ans
            if (str[i] == 'a'
            {
                a_index = i + 1;
                ans -= Math.min(b_index, c_index);
            
            // If character is b update b's index
            // and the variable ans
            else if (str[i] == 'b')
            {
                b_index = i + 1;
                ans -= Math.min(a_index, c_index);
            
            // If character is c update c's index
            // and the variable ans
            else 
            {
                c_index = i + 1;
                ans -= Math.min(a_index, b_index);
            }
        }
  
        return ans;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        char str[] = "babac".toCharArray();
        int n = str.length;
  
        System.out.println(CountSubstring(str, n));
    }
}
  
// This code contributed by Rajput-Ji

Python3




# Python3 implementation of the approach
  
# Function to return the count of 
# valid sub-Strings
def CountSubString(Str, n):
  
    # Variable ans to store all the possible subStrings
    # Initialize its value as total number of subStrings
    # that can be formed from the given String
    ans = (n * (n + 1)) // 2
  
    # Stores recent index of the characters
    a_index = 0
    b_index = 0
    c_index = 0
  
    for i in range(n):
  
        # If character is a update a's index
        # and the variable ans
        if (Str[i] == 'a'):
            a_index = i + 1
            ans -= min(b_index, c_index)
  
        # If character is b update b's index
        # and the variable ans
        elif (Str[i] == 'b'):
            b_index = i + 1
            ans -= min(a_index, c_index)
          
        # If character is c update c's index
        # and the variable ans
        else:
            c_index = i + 1
            ans -= min(a_index, b_index)
  
    return ans
  
# Driver code
Str = "babac"
n = len(Str)
  
print(CountSubString(Str, n))
  
# This code is contributed by mohit kumar

C#




// C# implementation of the approach
using System;
  
class GFG 
{
  
// Function to return the count of
// valid sub-strings
static int CountSubstring(char []str, int n) 
{
  
    // Variable ans to store all the possible substrings
    // Initialize its value as total number of substrings
    // that can be formed from the given string
    int ans = (n * (n + 1)) / 2;
  
    // Stores recent index of the characters
    int a_index = 0;
    int b_index = 0;
    int c_index = 0;
  
    for (int i = 0; i < n; i++) 
    {
  
        // If character is a update a's index
        // and the variable ans
        if (str[i] == 'a'
        {
            a_index = i + 1;
            ans -= Math.Min(b_index, c_index);
        
        // If character is b update b's index
        // and the variable ans
        else if (str[i] == 'b')
        {
            b_index = i + 1;
            ans -= Math.Min(a_index, c_index);
        
        // If character is c update c's index
        // and the variable ans
        else
        {
            c_index = i + 1;
            ans -= Math.Min(a_index, b_index);
        }
    }
  
    return ans;
}
  
// Driver code
public static void Main()
{
    char []str = "babac".ToCharArray();
    int n = str.Length;
  
    Console.WriteLine(CountSubstring(str, n));
}
}
  
// This code contributed 
// by Akanksha Rai

PHP




<?php
    // PHP implementation of the approach
    // Function to return the count of valid sub-strings
    function CountSubstring($str,$n
      
    {
  
        // Variable ans to store all the possible substrings
        // Initialize its value as total number of substrings
        // that can be formed from the given string
        $ans = ($n * ($n + 1)) / 2;
  
        // Stores recent index of the characters
        $a_index = 0;
        $b_index = 0;
        $c_index = 0;
  
        for ($i = 0; $i < $n; $i++) 
    {
  
  
            // If character is a update a's index
            // and the variable ans
            if ($str[$i] == 'a'
            {
                $a_index = $i + 1;
                $ans -= min($b_index, $c_index);
            
            // If character is b update b's index
            // and the variable ans
            else if ($str[$i] == 'b')
            {
                $b_index = $i + 1;
                $ans -= min($a_index, $c_index);
            
            // If character is c update c's index
            // and the variable ans
            else
            {
                $c_index = $i + 1;
                $ans -= min($a_index, $b_index);
            }
        }
  
        return $ans;
    }
  
    // Driver code
    {
        $str = str_split("babac");
        $n = sizeof($str);
  
        echo(CountSubstring($str, $n));
    }
  
  
// This code contributed by Code_Mech.
Output:
12

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer Geeks Classes Live and Geeks Classes Live USA

My Personal Notes arrow_drop_up
Recommended Articles
Page :