LeetCode #806 — EASY

Number of Lines To Write String

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

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.

Example 1:

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]
Explanation: You can write s as follows:
abcdefghij  // 100 pixels wide
klmnopqrst  // 100 pixels wide
uvwxyz      // 60 pixels wide
There are a total of 3 lines, and the last line is 60 pixels wide.

Example 2:

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]
Explanation: You can write s as follows:
bbbcccdddaa  // 98 pixels wide
a            // 4 pixels wide
There are a total of 2 lines, and the last line is 4 pixels wide.

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[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]
"abcdefghijklmnopqrstuvwxyz"

Example 2

[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]
"bbbcccdddaaa"

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #806: Number of Lines To Write String
class Solution {
    public int[] numberOfLines(int[] widths, String s) {
        int lines = 1, last = 0;
        for (int i = 0; i < s.length(); ++i) {
            int w = widths[s.charAt(i) - 'a'];
            if (last + w <= 100) {
                last += w;
            } else {
                ++lines;
                last = w;
            }
        }
        return new int[] {lines, last};
    }
}

// Accepted solution for LeetCode #806: Number of Lines To Write String
func numberOfLines(widths []int, s string) []int {
	lines, last := 1, 0
	for _, c := range s {
		w := widths[c-'a']
		if last+w <= 100 {
			last += w
		} else {
			lines++
			last = w
		}
	}
	return []int{lines, last}
}

# Accepted solution for LeetCode #806: Number of Lines To Write String
class Solution:
    def numberOfLines(self, widths: List[int], s: str) -> List[int]:
        lines, last = 1, 0
        for w in map(lambda c: widths[ord(c) - ord("a")], s):
            if last + w <= 100:
                last += w
            else:
                lines += 1
                last = w
        return [lines, last]

// Accepted solution for LeetCode #806: Number of Lines To Write String
impl Solution {
    pub fn number_of_lines(widths: Vec<i32>, s: String) -> Vec<i32> {
        let mut lines = 1;
        let mut last = 0;

        for c in s.chars() {
            let idx = ((c as u8) - b'a') as usize;
            let w = widths[idx];
            if last + w <= 100 {
                last += w;
            } else {
                lines += 1;
                last = w;
            }
        }

        vec![lines, last]
    }
}

// Accepted solution for LeetCode #806: Number of Lines To Write String
function numberOfLines(widths: number[], s: string): number[] {
    let [lines, last] = [1, 0];
    for (const c of s) {
        const w = widths[c.charCodeAt(0) - 'a'.charCodeAt(0)];
        if (last + w <= 100) {
            last += w;
        } else {
            ++lines;
            last = w;
        }
    }
    return [lines, last];
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.