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 = “neveropen”
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 sameint 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 functionint main(){ string s = "neveropen"; 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 equalimport 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 samestatic 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 functionpublic static void main(String args[]){ String s = "neveropen"; 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 samedef 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 Codeif __name__ == '__main__': s = "neveropen" 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 equalusing System;class GFG{// Function to return the number of ways // of removing a sub-string from s such// that all the remaining characters are samestatic 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 Codepublic static void Main(){ string s = "neveropen"; 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 $s such// that all the remaining characters are samefunction no_of_ways($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 = "neveropen";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 = "neveropen"; 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
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