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Count of sticks required to represent the given string
  • Last Updated : 23 Mar, 2020

Given a string str of uppercase alphabets and numbers, the task is to find the number of matchsticks required to represent it.

Examples:

Input: str = “ABC2”
Output: 22
Explanation:
6 sticks are required to represent A,
7 sticks are required to represent B,
4 sticks are required to represent C,
5 sticks are required to represent 2.
Therefore the total number of matchsticks required is 6 + 7 + 4 + 5 = 22.

Input: str = “GEEKSFORGEEKS”
Output: 66
Explanation:
6 sticks are required to represent G,
5 sticks are required to represent E,
4 sticks are required to represent K,
5 sticks are required to represent S,
4 sticks are required to represent F,
6 sticks are required to represent O,
6 sticks are required to represent R.
Therefore the total number of matchsticks required is 6 + 5 + 5 + 4 + 5 + 4 + 6 + 6 + 6 + 5 + 5 + 4 + 5 = 17.



Approach:

  1. The idea is to store the count of matchstick required to represent a particular alphabet and number as shown in above image.
  2. Traverse the given string str and add the count of matchstick required for each character.

Below is the implementation of the above approach:

C++




// C++ implementation of the
// above approach
#include <bits/stdc++.h>
using namespace std;
  
// stick[] stores the count
// of matchsticks required to
// represent the alphabets
int sticks[] = { 6, 7, 4, 6, 5, 4, 6,
                 5, 2, 4, 4, 3, 6, 6,
                 6, 5, 7, 6, 5, 3, 5,
                 4, 6, 4, 3, 4 };
  
// number[] stores the count
// of matchsticks required to
// represent the numerals
int number[] = { 6, 2, 5, 5, 4, 5, 6,
                 3, 7, 6 };
  
// Function that return the count of
// sticks required to represent
// the given string
int countSticks(string str)
{
    int cnt = 0;
  
    // For every char of the given
    // string
    for (int i = 0; str[i]; i++) {
  
        char ch = str[i];
  
        // Add the count of sticks
        // required to represent the
        // current character
        if (ch >= 'A' && ch <= 'Z') {
            cnt += sticks[ch - 'A'];
        }
        else {
            cnt += number[ch - '0'];
        }
    }
    return cnt;
}
  
// Driver code
int main()
{
    string str = "GEEKSFORGEEKS";
  
    // Function call to find the
    // count of matchsticks
    cout << countSticks(str);
  
    return 0;
}

Java




// Java implementation of the above approach
class GFG {
      
    // stick[] stores the count
    // of matchsticks required to
    // represent the alphabets
    static int sticks[] = { 6, 7, 4, 6, 5, 4, 6,
                     5, 2, 4, 4, 3, 6, 6,
                     6, 5, 7, 6, 5, 3, 5,
                     4, 6, 4, 3, 4 };
      
    // number[] stores the count
    // of matchsticks required to
    // represent the numerals
    static int number[] = { 6, 2, 5, 5, 4, 5, 6,
                     3, 7, 6 };
      
    // Function that return the count of
    // sticks required to represent
    // the given string
    static int countSticks(String str)
    {
        int cnt = 0;
      
        // For every char of the given
        // string
        for (int i = 0; i < str.length(); i++) {
      
            char ch = str.charAt(i);
      
            // Add the count of sticks
            // required to represent the
            // current character
            if (ch >= 'A' && ch <= 'Z') {
                cnt += sticks[ch - 'A'];
            }
            else {
                cnt += number[ch - '0'];
            }
        }
        return cnt;
    }
      
    // Driver code
    public static void main (String[] args) 
    {
        String str = "GEEKSFORGEEKS";
      
        // Function call to find the
        // count of matchsticks
        System.out.println(countSticks(str));
      
    }
}
  
// This code is contributed by AnkitRai01

Python3




# Python3 implementation of the above approach 
  
# stick[] stores the count 
# of matchsticks required to 
# represent the alphabets 
sticks = [ 6, 7, 4, 6, 5, 4, 6
            5, 2, 4, 4, 3, 6, 6
            6, 5, 7, 6, 5, 3, 5
            4, 6, 4, 3, 4 ]; 
  
# number[] stores the count 
# of matchsticks required to 
# represent the numerals 
number = [ 6, 2, 5, 5, 4, 5, 6, 3, 7, 6 ]; 
  
# Function that return the count of 
# sticks required to represent 
# the given string 
def countSticks(string) :
  
    cnt = 0
  
    # For every char of the given 
    # string 
    for i in range(len(string)) :
  
        ch = string[i]; 
  
        # Add the count of sticks 
        # required to represent the 
        # current character 
        if (ch >= 'A' and ch <= 'Z') :
            cnt += sticks[ord(ch) - ord('A')]; 
          
        else :
            cnt += number[ord(ch) - ord('0')]; 
      
    return cnt; 
  
# Driver code 
if __name__ == "__main__"
  
    string = "GEEKSFORGEEKS"
  
    # Function call to find the 
    # count of matchsticks 
    print(countSticks(string)); 
  
# This code is contributed by AnkitRai01

C#




// C# implementation of the above approach
using System;
  
class GFG
{
      
    // stick[] stores the count
    // of matchsticks required to
    // represent the alphabets
    static int []sticks = { 6, 7, 4, 6, 5, 4, 6,
                    5, 2, 4, 4, 3, 6, 6,
                    6, 5, 7, 6, 5, 3, 5,
                    4, 6, 4, 3, 4 };
      
    // number[] stores the count
    // of matchsticks required to
    // represent the numerals
    static int []number = { 6, 2, 5, 5, 4, 5, 6,
                    3, 7, 6 };
      
    // Function that return the count of
    // sticks required to represent
    // the given string
    static int countSticks(string str)
    {
        int cnt = 0;
      
        // For every char of the given
        // string
        for (int i = 0; i < str.Length; i++) 
        {
      
            char ch = str[i];
      
            // Add the count of sticks
            // required to represent the
            // current character
            if (ch >= 'A' && ch <= 'Z')
            {
                cnt += sticks[ch - 'A'];
            }
            else
            {
                cnt += number[ch - '0'];
            }
        }
        return cnt;
    }
      
    // Driver code
    public static void Main() 
    {
        string str = "GEEKSFORGEEKS";
      
        // Function call to find the
        // count of matchsticks
        Console.WriteLine(countSticks(str));
    }
}
  
// This code is contributed by AnkitRai01
Output:
66

Time Complexity: O(N), where N is the length of given string.

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