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++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to return the count of valid sub-strings int CountSubstring(string 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()
{ string str = "babac" ;
int n = str.length();
cout << CountSubstring(str, n);
return 0;
} |
// 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 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# 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 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. |
<script> // JavaScript 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
var ans = parseInt((n * (n + 1)) / 2);
// Stores recent index of the characters
var a_index = 0;
var b_index = 0;
var c_index = 0;
for ( var 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
var str = "babac" .split( "" );
var n = str.length;
document.write(CountSubstring(str, n));
</script>
|
12
Time Complexity: O(N).
Auxiliary Space: O(1)