Number of ways to remove a sub-string from S such that all remaining characters are same

Given a string str consisting only of lowercase English alphabets, the task is to count the number of ways to remove exactly one sub-string from str such that all remaining characters are same. 

Note: There are at least two different characters in str and we can remove the whole string as well.

Examples:

Input: str = “abaa” 
Output: 6 
We can remove the following sub-strings to satisfy the given condition: 
str[0:1], str[0:2], str[0:3], str[1:1], str[1:2] and str[1:3]

Input: str = “geeksforgeeks” 
Output: 3 
We remove complete string. 
We remove all except first. 
We remove all except last 

Approach:

• Store the length of prefix and suffix of same characters from the string in variables count_left and count_right.
• It is obvious that this prefix and suffix wouldn’t overlap, since there are at least two different characters in str.
• Now there are 2 cases: 
  • When str[0] = str[n – 1] then every character of the prefix (including the character just after the prefix ends) will act as the starting character of the sub-string and every character of the suffix (including the character just before the suffix starts) will act as the ending character of the sub-string. So, total valid sub-strings will be count = (count_left + 1) * (count_right + 1).
  • When str[0] != str[n – 1]. then count = count_left + count_right + 1.

Below is the implementation of the above approach: 

C++

// C++ program to count number of ways of
// removing a substring from a string such
// that all remaining characters are equal
#include <bits/stdc++.h>
using namespace std;

// Function to return the number of ways
// of removing a sub-string from s such
// that all the remaining characters are same
int no_of_ways(string s)
{
    int n = s.length();

    // To store the count of prefix and suffix
    int count_left = 0, count_right = 0;

    // Loop to count prefix
    for (int i = 0; i < n; ++i) {
        if (s[i] == s[0]) {
            ++count_left;
        }
        else
            break;
    }

    // Loop to count suffix
    for (int i = n - 1; i >= 0; --i) {
        if (s[i] == s[n - 1]) {
            ++count_right;
        }
        else
            break;
    }

    // First and last characters of the
    // string are same
    if (s[0] == s[n - 1])
        return ((count_left + 1) * (count_right + 1));

    // Otherwise
    else
        return (count_left + count_right + 1);
}

// Driver function
int main()
{
    string s = "geeksforgeeks";
    cout << no_of_ways(s);

    return 0;
}

Java

// Java program to count number of ways of
// removing a substring from a string such
// that all remaining characters are equal
import java.util.*;

class solution
{

// Function to return the number of ways
// of removing a sub-string from s such
// that all the remaining characters are same
static int no_of_ways(String s)
{
    int n = s.length();

    // To store the count of prefix and suffix
    int count_left = 0, count_right = 0;

    // Loop to count prefix
    for (int i = 0; i < n; ++i) {
        if (s.charAt(i) == s.charAt(0)) {
            ++count_left;
        }
        else
            break;
    }

    // Loop to count suffix
    for (int i = n - 1; i >= 0; --i) {
        if (s.charAt(i) == s.charAt(n - 1)) {
            ++count_right;
        }
        else
            break;
    }

    // First and last characters of the
    // string are same
    if (s.charAt(0) == s.charAt(n - 1))
        return ((count_left + 1) * (count_right + 1));

    // Otherwise
    else
        return (count_left + count_right + 1);
}

// Driver function
public static void main(String args[])
{
    String s = "geeksforgeeks";
    System.out.println(no_of_ways(s));

}
}

Python3

# Python 3 program to count number of
# ways of removing a substring from a
# string such that all remaining
# characters are equal

# Function to return the number of
# ways of removing a sub-string from
# s such that all the remaining
# characters are same
def no_of_ways(s):
    n = len(s)

    # To store the count of
    # prefix and suffix
    count_left = 0
    count_right = 0

    # Loop to count prefix
    for i in range(0, n, 1):
        if (s[i] == s[0]):
            count_left += 1
        else:
            break

    # Loop to count suffix
    i = n - 1
    while(i >= 0):
        if (s[i] == s[n - 1]):
            count_right += 1
        else:
            break

        i -= 1

    # First and last characters of
    # the string are same
    if (s[0] == s[n - 1]):
        return ((count_left + 1) *
                (count_right + 1))

    # Otherwise
    else:
        return (count_left + count_right + 1)

# Driver Code
if __name__ == '__main__':
    s = "geeksforgeeks"
    print(no_of_ways(s))

# This code is contributed by
# Sahil_shelengia

C#

// C# program to count number of ways of
// removing a substring from a string such
// that all remaining characters are equal
using System;

class GFG
{

// Function to return the number of ways
// of removing a sub-string from s such
// that all the remaining characters are same
static int no_of_ways(string s)
{
    int n = s.Length;

    // To store the count of prefix and suffix
    int count_left = 0, count_right = 0;

    // Loop to count prefix
    for (int i = 0; i < n; ++i)
    {
        if (s[i] == s[0])
        {
            ++count_left;
        }
        else
            break;
    }

    // Loop to count suffix
    for (int i = n - 1; i >= 0; --i)
    {
        if (s[i] == s[n - 1])
        {
            ++count_right;
        }
        else
            break;
    }

    // First and last characters of the
    // string are same
    if (s[0] == s[n - 1])
        return ((count_left + 1) *
                (count_right + 1));

    // Otherwise
    else
        return (count_left + count_right + 1);
}

// Driver Code
public static void Main()
{
    string s = "geeksforgeeks";
    Console.WriteLine(no_of_ways(s));
}
}

// This code is contributed
// by Akanksha Rai

PHP

<?php
// PHP program to count number of ways of
// removing a substring from a string such
// that all remaining characters are equal

// Function to return the number of ways
// of removing a sub-string from $s such
// that all the remaining characters are same
function no_of_ways($s)
{
    $n = strlen($s);

    // To store the count of prefix and suffix
    $count_left = 0;
    $count_right = 0;

    // Loop to count prefix
    for ($i = 0; $i < $n; ++$i)
    {
        if ($s[$i] == $s[0])
        {
            ++$count_left;
        }
        else
            break;
    }

    // Loop to count suffix
    for ($i = $n - 1; $i >= 0; --$i)
    {
        if ($s[$i] == $s[$n - 1])
        {
            ++$count_right;
        }
        else
            break;
    }

    // First and last characters of the
    // string are same
    if ($s[0] == $s[$n - 1])
        return (($count_left + 1) *
                ($count_right + 1));

    // Otherwise
    else
        return ($count_left + $count_right + 1);
}

// Driver Code
$s = "geeksforgeeks";
echo no_of_ways($s);

// This code is contributed by ihritik
?>

Javascript

<script>

    // JavaScript program to count number of ways of
    // removing a substring from a string such
    // that all remaining characters are equal

    // Function to return the number of ways
    // of removing a sub-string from s such
    // that all the remaining characters are same
    function no_of_ways(s)
    {
        let n = s.length;

        // To store the count of prefix and suffix
        let count_left = 0, count_right = 0;

        // Loop to count prefix
        for (let i = 0; i < n; ++i)
        {
            if (s[i] == s[0])
            {
                ++count_left;
            }
            else
                break;
        }

        // Loop to count suffix
        for (let i = n - 1; i >= 0; --i)
        {
            if (s[i] == s[n - 1])
            {
                ++count_right;
            }
            else
                break;
        }

        // First and last characters of the
        // string are same
        if (s[0] == s[n - 1])
            return ((count_left + 1) *
                    (count_right + 1));

        // Otherwise
        else
            return (count_left + count_right + 1);
    }

    let s = "geeksforgeeks";
    document.write(no_of_ways(s));

</script>

Output

3

Complexity Analysis:

• Time Complexity: O(n), where n is the size of the given string
• Auxiliary Space: O(1), as no extra space is required