Count of matchsticks required to represent the given number

Given a large integer as a string str, the task is find the number of matchsticks required to represent it.

Examples:

Input: str = “56”
Output: 11
5 sticks are required to represent 5 and
6 sticks are required to represent 6.



Input: str = “548712458645878”
Output: 74

Approach: Store the count of match sticks required to represent every digit from 0 to 9 in an array sticks[]. Now traverse the given string digit by digit and add the count of sticks required for the current digit.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// stick[i] stores the count of sticks
// required to represent the digit i
const int sticks[] = { 6, 2, 5, 5, 4, 5,
                       6, 3, 7, 6 };
  
// Function to return the count of
// matchsticks required to represent
// the given number
int countSticks(string str, int n)
{
    int cnt = 0;
  
    // For every digit of the given number
    for (int i = 0; i < n; i++) {
  
        // Add the count of sticks required
        // to represent the current digit
        cnt += (sticks[str[i] - '0']);
    }
  
    return cnt;
}
  
// Driver code
int main()
{
    string str = "56";
    int n = str.length();
  
    cout << countSticks(str, n);
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.util.*;
  
class GFG 
{
  
// stick[i] stores the count of sticks
// required to represent the digit i
static int sticks[] = { 6, 2, 5, 5, 4, 5,
                        6, 3, 7, 6 };
  
// Function to return the count of
// matchsticks required to represent
// the given number
static int countSticks(String str, int n)
{
    int cnt = 0;
  
    // For every digit of the given number
    for (int i = 0; i < n; i++)
    {
  
        // Add the count of sticks required
        // to represent the current digit
        cnt += (sticks[str.charAt(i) - '0']);
    }
    return cnt;
}
  
// Driver code
public static void main(String []args) 
{
    String str = "56";
    int n = str.length();
  
    System.out.println(countSticks(str, n));
}
}
  
// This code is contributed by 29AjayKumar

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Python3

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# Python3 implementation of the approach 
  
# stick[i] stores the count of sticks 
# required to represent the digit i 
sticks = [ 6, 2, 5, 5, 4, 5
           6, 3, 7, 6 ]; 
  
# Function to return the count of 
# matchsticks required to represent 
# the given number 
def countSticks(string, n) :
  
    cnt = 0
  
    # For every digit of the given number 
    for i in range(n) :
  
        # Add the count of sticks required 
        # to represent the current digit 
        cnt += (sticks[ord(string[i]) - ord('0')]); 
  
    return cnt; 
  
# Driver code 
if __name__ == "__main__"
  
    string = "56"
    n = len(string); 
  
    print(countSticks(string, n)); 
  
# This code is contributed by AnkitRai01

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C#

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// C# implementation of the approach
using System;
                      
class GFG 
{
  
// stick[i] stores the count of sticks
// required to represent the digit i
static int []sticks = { 6, 2, 5, 5, 4, 5,
                        6, 3, 7, 6 };
  
// Function to return the count of
// matchsticks required to represent
// the given number
static int countSticks(String str, int n)
{
    int cnt = 0;
  
    // For every digit of the given number
    for (int i = 0; i < n; i++)
    {
  
        // Add the count of sticks required
        // to represent the current digit
        cnt += (sticks[str[i] - '0']);
    }
    return cnt;
}
  
// Driver code
public static void Main(String []args) 
{
    String str = "56";
    int n = str.Length;
  
    Console.WriteLine(countSticks(str, n));
}
}
  
// This code is contributed by 29AjayKumar

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Output:

11

Time Complexity: O(n)



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Improved By : AnkitRai01, 29AjayKumar