Given a string str of n lowercase characters, the task is to count the number of substrings of str starting with character X and ending with character Y.
Examples:
Input: str = "abbcaceghcak"
x = 'a', y = 'c'
Output: 5
abbc, abbcac, ac, abbcaceghc, aceghc
Input: str = "neveropen"
x = 'g', y = 'e'
Output: 6
Approach:
- Initialize two counters i.e. tot_count to count the total number of substrings and count_x to count the number of strings that start with X.
- Start traversing the string.
- If the current character is X then increment the count of count_x.
- If the current character is Y, it means a string ends at Y so increment the count of tot_count i.e.
tot_count = tot_count + count_x
- It means that if there exists a Y then it will make a substring with all the X occurs before Y in the string. So, add the count of X to the total count.
- Return total count.
Below is the implementation of above approach:
C++
// C++ implementation to count substrings// starting with character X and ending// with character Y#include <bits/stdc++.h>using namespace std;// function to count substrings starting with// character X and ending with character Yint countSubstr(string str, int n, char x, char y){ // to store total count of // required substrings int tot_count = 0; // to store count of character 'x' // up to the point the string 'str' // has been traversed so far int count_x = 0; // traverse 'str' from left to right for (int i = 0; i < n; i++) { // if true, increment 'count_x' if (str[i] == x) count_x++; // if true accumulate 'count_x' // to 'tot_count' if (str[i] == y) tot_count += count_x; } // required count return tot_count;}// Driver codeint main(){ string str = "abbcaceghcak"; int n = str.size(); char x = 'a', y = 'c'; cout << "Count = " << countSubstr(str, n, x, y); return 0;} |
Java
// Java implementation to count // substrings starting with // character X and ending// with character Yimport java.util.*;import java.lang.*;import java.io.*;class GFG{// function to count substrings // starting with character X and// ending with character Ystatic int countSubstr(String str, int n, char x, char y){ // to store total count of // required substrings int tot_count = 0; // to store count of character // 'x' up to the point the // string 'str' has been // traversed so far int count_x = 0; // traverse 'str' from // left to right for (int i = 0; i < n; i++) { // if true, increment 'count_x' if (str.charAt(i) == x) count_x++; // if true accumulate 'count_x' // to 'tot_count' if (str.charAt(i) == y) tot_count += count_x; } // required count return tot_count;}// Driver codepublic static void main(String args[]){ String str = "abbcaceghcak"; int n = str.length(); char x = 'a', y = 'c'; System.out.print ("Count = " + countSubstr(str, n, x, y));}}// This code is contributed// by Subhadeep |
Python3
# Python 3 implementation to count substrings # starting with character X and ending # with character Y # function to count substrings starting with # character X and ending with character Y def countSubstr(str, n, x, y): # to store total count of # required substrings tot_count = 0 # to store count of character 'x' # up to the point the string 'str' # has been traversed so far count_x = 0 # traverse 'str' from left to right for i in range(n): # if true, increment 'count_x' if str[i] == x: count_x += 1 # if true accumulate 'count_x' # to 'tot_count' if str[i] == y: tot_count += count_x # required count return tot_count# Driver Codestr = 'abbcaceghcak'n = len(str)x, y = 'a', 'c'print('Count =', countSubstr(str, n, x, y))# This code is contributed SamyuktaSHegde |
C#
// C# implementation to count substrings // starting with character X and ending// with character Yusing System;class GFG{// function to count substrings starting // with character X and ending with character Ystatic int countSubstr(string str, int n, char x, char y){ // to store total count of // required substrings int tot_count = 0; // to store count of character 'x' up // to the point the string 'str' has // been traversed so far int count_x = 0; // traverse 'str' from left to right for (int i = 0; i < n; i++) { // if true, increment 'count_x' if (str[i] == x) count_x++; // if true accumulate 'count_x' // to 'tot_count' if (str[i] == y) tot_count += count_x; } // required count return tot_count;}// Driver codepublic static void Main(){ string str = "abbcaceghcak"; int n = str.Length; char x = 'a', y = 'c'; Console.Write("Count = " + countSubstr(str, n, x, y));}}// This code is contributed// by Akanksha Rai |
PHP
<?php// PHP implementation to count substrings// starting with character X and ending// with character Y// function to count substrings starting with// character X and ending with character Yfunction countSubstr($str, $n, $x, $y){ // to store total count of // required substrings $tot_count = 0; // to store count of character 'x' // up to the point the string 'str' // has been traversed so far $count_x = 0; // traverse 'str' from left to right for ($i = 0; $i < $n; $i++) { // if true, increment 'count_x' if ($str[$i] == $x) $count_x++; // if true accumulate 'count_x' // to 'tot_count' if ($str[$i] == $y) $tot_count += $count_x; } // required count return $tot_count;}// Driver code$str = "abbcaceghcak";$n = strlen($str);$x = 'a'; $y = 'c';echo "Count = ". countSubstr($str, $n, $x, $y);// This code is contributed // by Akanksha Rai |
Javascript
<script> // JavaScript implementation to count substrings // starting with character X and ending // with character Y // function to count substrings starting with // character X and ending with character Y function countSubstr(str, n, x, y) { // to store total count of // required substrings var tot_count = 0; // to store count of character 'x' // up to the point the string 'str' // has been traversed so far var count_x = 0; // traverse 'str' from left to right for (var i = 0; i < n; i++) { // if true, increment 'count_x' if (str[i] === x) count_x++; // if true accumulate 'count_x' // to 'tot_count' if (str[i] === y) tot_count += count_x; } // required count return tot_count; } // Driver code var str = "abbcaceghcak"; var n = str.length; var x = "a", y = "c"; document.write("Count = " + countSubstr(str, n, x, y)); </script> |
Output
Count = 5
Complexity Analysis:
- Time Complexity: O(n), to iterate over the array where n is the size of the given string
- Auxiliary Space: O(1),as no extra space is required
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