Count of lines required to write the given String

Given a string str and an integer array width[] where: 

width[0] = width of character ‘a’ 
width[1] = width of character ‘b’ 
… 
width[25] = width of character ‘z’ 

The task is to find the number of lines it’ll take to write the string str on a paper and the width of the last line upto which it is occupied. 
Note: The width of a line is 10 units.
Examples:

Input: str = “bbbcccdddaa”, 
width[] = {4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1} 
Output: (2, 8) 
“bbbcccddd” will cover first line (9 * 1 = 9 units) 
As ‘a’ has a width of 4 which cannot fit the remaining 1 unit in the first line. 
It’ll have to be written in the second line. 
So, next line will contain “aa” covering 4 * 2 = 8 units. 
We need 1 full line and one line with width 8 units.
Input: str = “abcdefghijklmnopqrstuvwxyz”, 
width[] = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1} 
Output: (3, 6) 
All the characters have the same width of 1. To write all 26 characters, 
We need 2 full lines and one line with width 6 units. 

Approach: We will write each character in the string str one by one. As we write a character, we immediately update (lines, width) that keeps track of how many lines we have used till now and what is the length of the used space in the last line. 
If the width[char] in str fits our current line, we will add it. Otherwise, we will start with a new line
Below is the implementation of the above approach: 

C++

// CPP implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the number of lines required
pair<int, int> numberOfLines(string S, int *widths)
{
    // If string is empty
    if (S.empty())
        return {0, 0};

    // Initialize lines and width
    int lines = 1, width = 0;

    // Iterate through S
    for (auto character : S)
    {
        int w = widths[character - 'a'];
        width += w;

        if (width >= 10)
        {
            lines++;
            width = w;
        }
    }

    // Return lines and width used
    return {lines, width};
}

// Driver Code
int main()
{
    string S = "bbbcccdddaa";
    int widths[] = {4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
                    1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1};

    // Function call to print required answer
    pair<int, int> ans = numberOfLines(S, widths);
    cout << ans.first << " " << ans.second << endl;

    return 0;
}

// This code is contributed by
// sanjeev2552

Java

// JAVA implementation of the approach
class GFG
{

// Function to return the number of lines required
static int[] numberOfLines(String S, int []widths)
{
    // If String is empty
    if (S.isEmpty())
        return new int[]{0, 0};

    // Initialize lines and width
    int lines = 1, width = 0;

    // Iterate through S
    for (char character : S.toCharArray())
    {
        int w = widths[character - 'a'];
        width += w;

        if (width >= 10)
        {
            lines++;
            width = w;
        }
    }

    // Return lines and width used
    return new int[]{lines, width};
}

// Driver Code
public static void main(String[] args)
{
    String S = "bbbcccdddaa";
    int widths[] = {4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
                    1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1};

    // Function call to print required answer
    int []ans = numberOfLines(S, widths);
    System.out.print(ans[0]+ " " + ans[1] +"\n");
}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 implementation of the approach

# Function to return the number of lines required
def numberOfLines(S, widths):

    # If string is empty
    if(S == ""):
        return 0, 0

    # Initialize lines and width
    lines, width = 1, 0

    # Iterate through S
    for c in S:
        w = widths[ord(c) - ord('a')]
        width += w
        if width > 10:
            lines += 1
            width = w

    # Return lines and width used
    return lines, width


# Driver Code
S = "bbbcccdddaa"
Widths = [4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]

# Function call to print required answer
print(numberOfLines(S, Widths))

C#

// C# implementation of the approach
using System;

class GFG
{

// Function to return the number of lines required
static int[] numberOfLines(String S, int []widths)
{
    // If String is empty
    if (S.Length == 0)
        return new int[]{0, 0};

    // Initialize lines and width
    int lines = 1, width = 0;

    // Iterate through S
    foreach (char character in S.ToCharArray())
    {
        int w = widths[character - 'a'];
        width += w;

        if (width >= 10)
        {
            lines++;
            width = w;
        }
    }

    // Return lines and width used
    return new int[]{lines, width};
}

// Driver Code
public static void Main(String[] args)
{
    String S = "bbbcccdddaa";
    int []widths = {4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
                    1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1};

    // Function call to print required answer
    int []ans = numberOfLines(S, widths);
    Console.Write(ans[0]+ " " + ans[1] +"\n");
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>
// JAVAscript implementation of the approach

// Function to return the number of lines required
function numberOfLines(S,widths)
{
    // If String is empty
    if (S.length==0)
        return [0, 0];

    // Initialize lines and width
    let lines = 1, width = 0;

    // Iterate through S
    for (let character of S.split(""))
    {
        let w = widths[character.charCodeAt(0) - 'a'.charCodeAt(0)];
        width += w;

        if (width >= 10)
        {
            lines++;
            width = w;
        }
    }

    // Return lines and width used
    return [lines, width];
}

// Driver Code
let S = "bbbcccdddaa";
let widths = [4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
                1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1];

// Function call to print required answer
let ans = numberOfLines(S, widths);
document.write(ans[0]+ " " + ans[1] +"<br>");

// This code is contributed by rag2127

</script>

Output

2 8

Time Complexity: O(n), where n is the length of the given string.
Auxiliary Space: O(2) ? O(1), no extra space is required, so it is a constant.