Count of lines required to write the given String - GeeksforGeeks

29 Jul, 2021
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: 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 lineBelow is the implementation of the above approach:  
C++
Java
Python3
C#
Javascript
// CPP implementation of the approach#include <bits/stdc++.h>using namespace std; // Function to return the number of lines requiredpair<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 Codeint 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 implementation of the approachclass GFG{ // Function to return the number of lines requiredstatic 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 Codepublic 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 implementation of the approach # Function to return the number of lines requireddef 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 CodeS = "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 answerprint(numberOfLines(S, Widths))
// C# implementation of the approachusing System; class GFG{ // Function to return the number of lines requiredstatic 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 Codepublic 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
<script>// JAVAscript implementation of the approach // Function to return the number of lines requiredfunction 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 Codelet 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 answerlet ans = numberOfLines(S, widths);document.write(ans[0]+ " " + ans[1] +"<br>"); // This code is contributed by rag2127 </script>
(2, 8)
 
sanjeev2552
Rajput-Ji
29AjayKumar
rag2127
Greedy
Pattern Searching
Python
Strings
Strings
Greedy
Pattern Searching
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