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 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 |
(2, 8)
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