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 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
// 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
# 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#
// 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 |
Javascript
<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
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.
