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Count of Palindrome Strings in given Array of strings

  • Last Updated : 27 Jan, 2022

Given an array of strings arr[] of size N where each string consists only of lowercase English letter. The task is to return the count of all palindromic string in the array.

Examples:

Input: arr[] = {“abc”,”car”,”ada”,”racecar”,”cool”}
Output: 2
Explanation: “ada” and “racecar” are the two palindrome strings.

Input: arr[] = {“def”,”aba”}
Output: 1
Explanation: “aba” is the only palindrome string.

 

Approach: The solution is based on greedy approach. Check every string of an array if it is palindrome or not and also keep track of the count. Follow the steps below to solve the problem:

  • Initialize a count variable ans as 0.
  • Iterate over the range [0, N) using the variable i and if arr[i] is a palindrome, then increment the value of ans.
  • After performing the above steps, print the value of ans as the answer.

Below is the implementation of the above approach.

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if given string
// is Palindrome or not
bool isPalindrome(string& s)
{
    // Copy string s char into string a
    string a = s;
    reverse(s.begin(), s.end());
 
    // Check if two string are equal or not
    return s == a;
}
 
// Function to return count
// of Palindrome string
int PalindromicStrings(string arr[], int N)
{
    int ans = 0;
 
    // Loop to find palindrome string
    for (int i = 0; i < N; i++) {
 
        // Checking if given string is
        // palindrome or not
        if (isPalindrome(arr[i])) {
 
            // Update answer variable
            ans++;
        }
    }
    return ans;
}
 
// Driver Code
int main()
{
 
    string arr[]
        = { "abc", "car", "ada",
           "racecar", "cool" };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Print required answer
    cout << PalindromicStrings(arr, N);
 
    return 0;
}

Java




// java program for the above approach
class GFG
{
 
  // Function to check if given String
  // is Palindrome or not
  static boolean isPalindrome(String str)
  {
 
    // Start from leftmost and rightmost corners of str
    int l = 0;
    int h = str.length() - 1;
 
    // Keep comparing characters while they are same
    while (h > l)
    {
      if (str.charAt(l++) != str.charAt(h--))
      {
        return false;
      }
    }
    return true;
  }
 
  // Function to return all Palindrome String
  static int PalindromicStrings(String []arr,
                                int N)
  {
 
    int ans = 0;
 
    // Loop to find palindrome String
    for (int i = 0; i < N; i++) {
 
      // Checking if given String is
      // palindrome or not
      if (isPalindrome(arr[i])) {
 
        // Update answer variable
        ans++;
      }
    }
    return ans;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
 
    String []arr
      = { "abc", "car", "ada", "racecar", "cool" };
    int N = arr.length;
 
    System.out.print(PalindromicStrings(arr, N));
 
  }
}
 
// This code is contributed by Rajput-Ji

Python3




# Python program for the above approach
 
# Function to check if given String
# is Palindrome or not
def isPalindrome(str):
   
    # Start from leftmost and rightmost corners of str
    l = 0;
    h = len(str) - 1;
 
    # Keep comparing characters while they are same
    while (h > l):
        if (str[l] != str[h]):
            return False;
        l += 1;
        h -= 1;
 
    return True;
 
# Function to return all Palindrome String
def PalindromicStrings(arr, N):
    ans = 0;
 
    # Loop to find palindrome String
    for i in range(N):
 
        # Checking if given String is
        # palindrome or not
        if (isPalindrome(arr[i])):
           
            # Update answer variable
            ans += 1;
 
    return ans;
 
# Driver Code
if __name__ == '__main__':
    arr = ["abc", "car", "ada", "racecar", "cool"];
    N = len(arr);
 
    print(PalindromicStrings(arr, N));
     
# This code is contributed by 29AjayKumar

C#




// C# program for the above approach
using System;
using System.Collections;
class GFG
{
 
  // Function to check if given string
  // is Palindrome or not
  static bool isPalindrome(string str)
  {
 
    // Start from leftmost and rightmost corners of str
    int l = 0;
    int h = str.Length - 1;
 
    // Keep comparing characters while they are same
    while (h > l)
    {
      if (str[l++] != str[h--])
      {
        return false;
      }
    }
    return true;
  }
 
  // Function to return all Palindrome string
  static int PalindromicStrings(string []arr,
                                int N)
  {
 
    int ans = 0;
 
    // Loop to find palindrome string
    for (int i = 0; i < N; i++) {
 
      // Checking if given string is
      // palindrome or not
      if (isPalindrome(arr[i])) {
 
        // Update answer variable
        ans++;
      }
    }
    return ans;
  }
 
  // Driver Code
  public static void Main()
  {
 
    string []arr
      = { "abc", "car", "ada", "racecar", "cool" };
    int N = arr.Length;
 
    Console.Write(PalindromicStrings(arr, N));
 
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript




<script>
// Javascript program for the above approach
 
// Function to check if given string
// is Palindrome or not
function isPalindrome(s)
{
 
    // Copy string s char into string a
    let a = s;
    s = s.split("").reverse().join("");
 
    // Check if two string are equal or not
    return s == a;
}
 
// Function to return count
// of Palindrome string
function PalindromicStrings(arr, N) {
    let ans = 0;
 
    // Loop to find palindrome string
    for (let i = 0; i < N; i++) {
 
        // Checking if given string is
        // palindrome or not
        if (isPalindrome(arr[i])) {
 
            // Update answer variable
            ans++;
        }
    }
    return ans;
}
 
// Driver Code
let arr = ["abc", "car", "ada", "racecar", "cool"];
let N = arr.length;
 
// Print required answer
document.write(PalindromicStrings(arr, N));
 
// This code is contributed by saurabh_jaiswal.
</script>

 
 

Output
2

 

Time Complexity: O(N * W) where W is the average length of the strings
Auxiliary Space: O(1)

 


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