Number of Lines To Write String - Practice Coding Problems

Number of Lines To Write String - Problem

Array String Easy

You are given a string s of lowercase English letters and an array widths denoting how many pixels wide each lowercase English letter is.

Specifically, widths[0] is the width of 'a', widths[1] is the width of 'b', and so on.

You are trying to write s across several lines, where each line is no longer than 100 pixels.

Starting at the beginning of s, write as many letters on the first line such that the total width does not exceed 100 pixels. Then, from where you stopped in s, continue writing as many letters as you can on the second line. Continue this process until you have written all of s.

Return an array result of length 2 where:

result[0] is the total number of lines
result[1] is the width of the last line in pixels

Input & Output

Example 1 — Basic Case

$ 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. First line fits 10 chars (100px), second line fits 10 chars (100px), third line has 6 chars (60px). Result: 3 lines, last line 60px.

Example 2 — Different Widths

$ 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 = "bbbcccdddaaa"

› Output: [2,4]

💡 Note: b=10px, c=10px, d=10px, a=4px. First line: bbbcccddda (94px). Adding another 'a' would make 98px ≤ 100px, so 'aa' fits. Result: 2 lines, last line 4px.

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

Input

String and character widths array

Process

Track line width, start new line when exceeding 100px

Output

Return [total lines, last line width]

Key Takeaway

🎯 Key Insight: Simulate the writing process by tracking current line width and creating new lines when the next character would exceed the 100-pixel limit

Asked in

G Google 25 f Facebook 20

The key insight is to simulate the writing process by tracking the current line width and starting a new line whenever adding the next character would exceed 100 pixels. Best approach is linear scan in O(n) time, O(1) space.

Common Approaches

Approach	Time	Space	Notes
✓ Linear Scan	O(n)	O(1)	Iterate through string, track line width, start new line when needed
Single Pass with Early Tracking	O(n)	O(1)	Same linear approach but with optimized character width lookup

Linear Scan — Algorithm Steps

Step 1: Initialize line count to 1 and current line width to 0
Step 2: For each character, get its width from the widths array
Step 3: If adding this character exceeds 100 pixels, start new line
Step 4: Otherwise, add character width to current line

Visualization

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Step-by-Step Walkthrough

Initialize

Start with 1 line, 0 current width

Process Characters

Add character width, create new line if exceeds 100

Result

Return [total lines, last line width]

Code -

solution.c — C

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

void solution(int* widths, char* s, int* result) {
    int lines = 1;
    int currentWidth = 0;
    int len = strlen(s);
    
    for (int i = 0; i < len; i++) {
        int charWidth = widths[s[i] - 'a'];
        if (currentWidth + charWidth > 100) {
            lines++;
            currentWidth = charWidth;
        } else {
            currentWidth += charWidth;
        }
    }
    
    result[0] = lines;
    result[1] = currentWidth;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    // Parse widths array
    int widths[26];
    char* token = strtok(line + 1, ",]");
    int i = 0;
    while (token != NULL && i < 26) {
        widths[i++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    char s[1000];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0; // Remove newline
    
    int result[2];
    solution(widths, s, result);
    printf("[%d,%d]\n", result[0], result[1]);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through string of length n

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra variables

✓ Linear Space

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Ln 1, Col 1

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Linear Scan — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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