Count strings that end with the given pattern
Last Updated :
29 Nov, 2022
Given a pattern pat and a string array sArr[], the task is to count the number of strings from the array that ends with the given pattern.
Examples:
Input: pat = "ks", sArr[] = {"geeks", "geeksforgeeks", "games", "unit"}
Output: 2
Only string "geeks" and "geeksforgeeks" end with the pattern "ks".
Input: pat = "abc", sArr[] = {"abcd", "abcc", "aaa", "bbb"}
Output: 0
Approach:
- Initialize count = 0 and start traversing the given string array.
- For every string str, initialize strLen = len(str) and patLen = len(pattern).
- If patLen > strLen then skips to the next string as the current string cannot end with the given pattern.
- Else match the string with the pattern starting from the end. If the string matches the pattern then update count = count + 1.
- Print the count in the end.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function that return true if str
// ends with pat
bool endsWith(string str, string pat)
{
int patLen = pat.length();
int strLen = str.length();
// Pattern is larger in length than
// the string
if (patLen > strLen)
return false;
// We match starting from the end while
// patLen is greater than or equal to 0.
patLen--;
strLen--;
while (patLen >= 0) {
// If at any index str doesn't match
// with pattern
if (pat[patLen] != str[strLen])
return false;
patLen--;
strLen--;
}
// If str ends with the given pattern
return true;
}
// Function to return the count of required
// strings
int countOfStrings(string pat, int n,
string sArr[])
{
int count = 0;
for (int i = 0; i < n; i++)
// If current string ends with
// the given pattern
if (endsWith(sArr[i], pat))
count++;
return count;
}
// Driver code
int main()
{
string pat = "ks";
int n = 4;
string sArr[] = { "geeks", "geeksforgeeks", "games", "unit" };
cout << countOfStrings(pat, n, sArr);
return 0;
}
// Java implementation of the approach
class GfG
{
// Function that return true
// if str ends with pat
static boolean endsWith(String str, String pat)
{
int patLen = pat.length();
int strLen = str.length();
// Pattern is larger in length
// than the string
if (patLen > strLen)
return false;
// We match starting from the end while
// patLen is greater than or equal to 0.
patLen--;
strLen--;
while (patLen >= 0)
{
// If at any index str doesn't match
// with pattern
if (pat.charAt(patLen) != str.charAt(strLen))
return false;
patLen--;
strLen--;
}
// If str ends with the given pattern
return true;
}
// Function to return the
// count of required strings
static int countOfStrings(String pat, int n,
String sArr[])
{
int count = 0;
for (int i = 0; i < n; i++)
{
// If current string ends with
// the given pattern
if (endsWith(sArr[i], pat))
count++;
}
return count;
}
// Driver code
public static void main(String []args)
{
String pat = "ks";
int n = 4;
String sArr[] = { "geeks", "geeksforgeeks", "games", "unit" };
System.out.println(countOfStrings(pat, n, sArr));
}
}
// This code is contributed by Rituraj Jain
# Python3 implementation of the approach
# Function that return true if str1
# ends with pat
def endsWith(str1, pat):
patLen = len(pat)
str1Len = len(str1)
# Pattern is larger in length
# than the string
if (patLen > str1Len):
return False
# We match starting from the end while
# patLen is greater than or equal to 0.
patLen -= 1
str1Len -= 1
while (patLen >= 0):
# If at any index str1 doesn't match
# with pattern
if (pat[patLen] != str1[str1Len]):
return False
patLen -= 1
str1Len -= 1
# If str1 ends with the given pattern
return True
# Function to return the count of
# required strings
def countOfstrings(pat, n, sArr):
count = 0
for i in range(n):
# If current string ends with
# the given pattern
if (endsWith(sArr[i], pat) == True):
count += 1
return count
# Driver code
pat = "ks"
n = 4
sArr= [ "geeks", "geeksforgeeks",
"games", "unit"]
print(countOfstrings(pat, n, sArr))
# This code is contributed by
# Mohit kumar 29
// C# implementation of the approach
using System;
class GFG
{
// Function that return true if str
// ends with pat
static bool endsWith(string str, string pat)
{
int patLen = pat.Length;
int strLen = str.Length;
// Pattern is larger in length than
// the string
if (patLen > strLen)
return false;
// We match starting from the end while
// patLen is greater than or equal to 0.
patLen--;
strLen--;
while (patLen >= 0)
{
// If at any index str doesn't match
// with pattern
if (pat[patLen] != str[strLen])
return false;
patLen--;
strLen--;
}
// If str ends with the given pattern
return true;
}
// Function to return the count of required
// strings
static int countOfStrings(string pat, int n,
string[] sArr)
{
int count = 0;
for (int i = 0; i < n; i++)
// If current string ends with
// the given pattern
if (endsWith(sArr[i], pat))
count++;
return count;
}
// Driver code
public static void Main()
{
string pat = "ks";
int n = 4;
string[] sArr = { "geeks", "geeksforgeeks",
"games", "unit" };
Console.WriteLine(countOfStrings(pat, n, sArr));
}
}
// This code is contributed by Akanksha Rai
<?php
// PHP implementation of the approach
// Function that return true if str
// ends with pat
function endsWith($str, $pat)
{
$patLen = strlen($pat);
$strLen = strlen($str);
// Pattern is larger in length than
// the string
if ($patLen > $strLen)
return false;
// We match starting from the end while
// patLen is greater than or equal to 0.
$patLen--;
$strLen--;
while ($patLen >= 0)
{
// If at any index str doesn't match
// with pattern
if ($pat[$patLen] != $str[$strLen])
return false;
$patLen--;
$strLen--;
}
// If str ends with the given pattern
return true;
}
// Function to return the count of required
// strings
function countOfStrings($pat, $n, $sArr)
{
$count = 0;
for ($i = 0; $i < $n; $i++)
// If current string ends with
// the given pattern
if (endsWith($sArr[$i], $pat))
$count++;
return $count;
}
// Driver code
$pat = "ks";
$n = 4;
$sArr = array("geeks", "geeksforgeeks",
"games", "unit");
echo countOfStrings($pat, $n, $sArr);
// This code is contributed by mits
?>
<script>
// JavaScript implementation of the approach
// Function that return true
// if str ends with pat
function endsWith(str,pat)
{
let patLen = pat.length;
let strLen = str.length;
// Pattern is larger in length
// than the string
if (patLen > strLen)
return false;
// We match starting from the end while
// patLen is greater than or equal to 0.
patLen--;
strLen--;
while (patLen >= 0)
{
// If at any index str doesn't match
// with pattern
if (pat[patLen] != str[strLen])
return false;
patLen--;
strLen--;
}
// If str ends with the given pattern
return true;
}
// Function to return the
// count of required strings
function countOfStrings(pat,n,sArr)
{
let count = 0;
for (let i = 0; i < n; i++)
{
// If current string ends with
// the given pattern
if (endsWith(sArr[i], pat))
count++;
}
return count;
}
// Driver code
let pat = "ks";
let n = 4;
let sArr=[ "geeks", "geeksforgeeks", "games", "unit"];
document.write(countOfStrings(pat, n, sArr));
// This code is contributed by unknown2108
</script>
Time Complexity: O(m * n), where m is the length of pattern string and n is the size of the string array.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
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