Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.
There is a strange printer with the following two special properties:
Given a string s, return the minimum number of turns the printer needed to print it.
Example 1:
Input: s = "aaabbb" Output: 2 Explanation: Print "aaa" first and then print "bbb".
Example 2:
Input: s = "aba" Output: 2 Explanation: Print "aaa" first and then print "b" from the second place of the string, which will cover the existing character 'a'.
Constraints:
1 <= s.length <= 100s consists of lowercase English letters.Problem summary: There is a strange printer with the following two special properties: The printer can only print a sequence of the same character each time. At each turn, the printer can print new characters starting from and ending at any place and will cover the original existing characters. Given a string s, return the minimum number of turns the printer needed to print it.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Dynamic Programming
"aaabbb"
"aba"
remove-boxes)strange-printer-ii)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #664: Strange Printer
class Solution {
public int strangePrinter(String s) {
final int inf = 1 << 30;
int n = s.length();
int[][] f = new int[n][n];
for (var g : f) {
Arrays.fill(g, inf);
}
for (int i = n - 1; i >= 0; --i) {
f[i][i] = 1;
for (int j = i + 1; j < n; ++j) {
if (s.charAt(i) == s.charAt(j)) {
f[i][j] = f[i][j - 1];
} else {
for (int k = i; k < j; ++k) {
f[i][j] = Math.min(f[i][j], f[i][k] + f[k + 1][j]);
}
}
}
}
return f[0][n - 1];
}
}
// Accepted solution for LeetCode #664: Strange Printer
func strangePrinter(s string) int {
n := len(s)
f := make([][]int, n)
for i := range f {
f[i] = make([]int, n)
for j := range f[i] {
f[i][j] = 1 << 30
}
}
for i := n - 1; i >= 0; i-- {
f[i][i] = 1
for j := i + 1; j < n; j++ {
if s[i] == s[j] {
f[i][j] = f[i][j-1]
} else {
for k := i; k < j; k++ {
f[i][j] = min(f[i][j], f[i][k]+f[k+1][j])
}
}
}
}
return f[0][n-1]
}
# Accepted solution for LeetCode #664: Strange Printer
class Solution:
def strangePrinter(self, s: str) -> int:
n = len(s)
f = [[inf] * n for _ in range(n)]
for i in range(n - 1, -1, -1):
f[i][i] = 1
for j in range(i + 1, n):
if s[i] == s[j]:
f[i][j] = f[i][j - 1]
else:
for k in range(i, j):
f[i][j] = min(f[i][j], f[i][k] + f[k + 1][j])
return f[0][-1]
// Accepted solution for LeetCode #664: Strange Printer
/**
* [0664] Strange Printer
*
* There is a strange printer with the following two special properties:
*
* The printer can only print a sequence of the same character each time.
* At each turn, the printer can print new characters starting from and ending at any place and will cover the original existing characters.
*
* Given a string s, return the minimum number of turns the printer needed to print it.
*
* Example 1:
*
* Input: s = "aaabbb"
* Output: 2
* Explanation: Print "aaa" first and then print "bbb".
*
* Example 2:
*
* Input: s = "aba"
* Output: 2
* Explanation: Print "aaa" first and then print "b" from the second place of the string, which will cover the existing character 'a'.
*
*
* Constraints:
*
* 1 <= s.length <= 100
* s consists of lowercase English letters.
*
*/
pub struct Solution {}
// problem: https://leetcode.com/problems/strange-printer/
// discuss: https://leetcode.com/problems/strange-printer/discuss/?currentPage=1&orderBy=most_votes&query=
// submission codes start here
impl Solution {
pub fn strange_printer(s: String) -> i32 {
let s = s.as_bytes();
let n = s.len();
match n {
0 => 0,
n @ _ => {
let mut dp = vec![vec![0; n + 1]; n + 1];
for len in 1..=n {
for i in 0..(n - len + 1) {
let j = i + len - 1;
dp[i][j] = 1 + dp[i + 1][j];
for k in (i + 1)..=j {
if s[k] == s[i] {
dp[i][j] = std::cmp::min(dp[i][j], dp[i][k - 1] + dp[k + 1][j]);
}
}
}
}
dp[0][n - 1]
}
}
}
}
// submission codes end
#[cfg(test)]
mod tests {
use super::*;
#[test]
fn test_0664_example_1() {
let s = "aaabbb".to_string();
let result = 2;
assert_eq!(Solution::strange_printer(s), result);
}
#[test]
fn test_0664_example_2() {
let s = "aba".to_string();
let result = 2;
assert_eq!(Solution::strange_printer(s), result);
}
}
// Accepted solution for LeetCode #664: Strange Printer
function strangePrinter(s: string): number {
const n = s.length;
const f: number[][] = new Array(n).fill(0).map(() => new Array(n).fill(1 << 30));
for (let i = n - 1; i >= 0; --i) {
f[i][i] = 1;
for (let j = i + 1; j < n; ++j) {
if (s[i] === s[j]) {
f[i][j] = f[i][j - 1];
} else {
for (let k = i; k < j; ++k) {
f[i][j] = Math.min(f[i][j], f[i][k] + f[k + 1][j]);
}
}
}
}
return f[0][n - 1];
}
Use this to step through a reusable interview workflow for this problem.
Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.
Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.
Review these before coding to avoid predictable interview regressions.
Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.