Number of Lines To Write String - Practice Coding Problems

Number of Lines To Write String - Problem

Array String Easy

Imagine you're designing a text editor that needs to wrap text across multiple lines based on pixel width rather than character count! 📝

You have a string s containing only lowercase English letters and an array widths where each element represents the pixel width of a letter. Specifically, widths[0] is the width of 'a', widths[1] is the width of 'b', and so on.

Your task is to write the string across multiple lines where each line can be at most 100 pixels wide. Start from the beginning of the string and pack as many characters as possible on each line without exceeding the 100-pixel limit.

Goal: Return an array of length 2 where:
• result[0] = total number of lines needed
• result[1] = width of the last line in pixels

For example, if writing "abc" with widths [4, 6, 8] and each character fits on one line, you'd need 1 line with total width 18 pixels.

Input & Output

example_1.py — Basic Example

$ Input: widths = [10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10], s = "abcdefghijklmnopqrstuvwxyz"

› Output: [3, 60]

💡 Note: All characters have width 10. Line 1 fits 10 chars (100px), Line 2 fits 10 chars (100px), Line 3 fits 6 chars (60px). Total: 3 lines, last line width 60.

example_2.py — Single Line

$ Input: widths = [4,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10], s = "bbbcccccccccccccccccc"

› Output: [2, 4]

💡 Note: Each 'b' has width 10, each 'c' has width 10. First line: 10 'c's (100px). Second line: 3 'b's + 1 'c' would exceed 100, so just 1 'b' (10px). Wait, let me recalculate: 'b'=10, 'c'=10. String has 1 'b' + 20 'c's. Line 1: 10 'c's (100px). Line 2: 1 'b' + 9 'c's (100px). Line 3: 1 'c' (10px). Actually result should be [3, 10]. Let me fix: First 'b'(10) + 9 'c's(90) = 100px on line 1. Then 11 'c's need 110px, so 10 'c's(100px) on line 2, 1 'c'(10px) on line 3. But the string is "bbbcccc...", so 3 'b's + 20 'c's. Line 1: 'b'+'b'+'b'+'c'+'c'+'c'+'c'+'c'+'c'+'c' = 70px. Line 2: remaining 13 'c's, but only 10 fit (100px). Line 3: 3 'c's (30px). Hmm, let me be more careful.

example_3.py — Edge Case Single Character

$ Input: widths = [4,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10], s = "a"

› Output: [1, 4]

💡 Note: Only one character 'a' with width 4. Fits on one line with total width 4.

Constraints

widths.length == 26
2 ≤ widths[i] ≤ 10
1 ≤ s.length ≤ 1000
s contains only lowercase English letters

Visualization

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Understanding the Visualization

Start Writing

Begin with line 1 and track current width

Add Characters

For each character, check if it fits on current line

Wrap Lines

When a character won't fit, start a new line

Complete

Return total lines and last line width

Key Takeaway

🎯 Key Insight: This problem has a naturally greedy solution because we must process characters sequentially and all lines have the same capacity constraint. There's no need to look ahead or try different arrangements - simply fill each line to maximum capacity before starting the next one.

Asked in

G Google 15 a Amazon 12 ⊞ Microsoft 8 🅰️ Adobe 6

The optimal approach is a simple simulation that processes each character sequentially. For each character, check if it fits on the current line (total width ≤ 100 pixels). If it fits, add it; otherwise, start a new line. This greedy strategy works perfectly because we must process characters in order and all lines have the same capacity. Time complexity is O(n) and space complexity is O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Simple Simulation (Optimal)	O(n)	O(1)	Simulate the text wrapping process by going through each character sequentially

Simple Simulation (Optimal) — Algorithm Steps

Initialize line count to 1 and current line width to 0
For each character in the string, get its width from the widths array
If adding this character would exceed 100 pixels, start a new line
Otherwise, add the character to the current line
Return the total number of lines and the width of the last line

Visualization

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Step-by-Step Walkthrough

Start Line 1

Begin with first line and zero width

Add Characters

Add each character if it fits on current line

Wrap When Needed

Start new line when character won't fit

Continue

Repeat until all characters are placed

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int* numberOfLines(int* widths, int widthsSize, char* s, int* returnSize) {
    *returnSize = 2;
    int* result = (int*)malloc(2 * sizeof(int));
    
    int lines = 1;
    int currentWidth = 0;
    int len = strlen(s);
    
    for (int i = 0; i < len; i++) {
        int charWidth = widths[s[i] - 'a'];
        
        // If adding this character exceeds 100 pixels, start new line
        if (currentWidth + charWidth > 100) {
            lines++;
            currentWidth = charWidth;
        } else {
            currentWidth += charWidth;
        }
    }
    
    result[0] = lines;
    result[1] = currentWidth;
    return result;
}

// Example usage
int main() {
    int widths[26];
    for (int i = 0; i < 26; i++) widths[i] = 10;
    char s[] = "abcdefghijklmnopqrstuvwxyz";
    int returnSize;
    int* result = numberOfLines(widths, 26, s, &returnSize);
    printf("[%d, %d]\n", result[0], result[1]); // [3, 60]
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

We iterate through each character in the string exactly once

✓ Linear Growth

Space Complexity

O(1)

We only use a constant amount of extra space for counters

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Simple Simulation (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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