Strange Printer

🎯

Strange Printer

hardDPGoogle

Imagine a printer that can only print a sequence of the same character at once, and you want to print a given string with the fewest such print operations.

💡 This problem is a classic example of interval dynamic programming where the challenge is to minimize operations over substrings. Beginners often struggle because the problem requires thinking about overlapping subproblems and how to merge intervals optimally, which is not intuitive at first.

📋

Problem Statement

Given a string s consisting of lowercase English letters, a strange printer can print a sequence of the same character each time. The printer can print new characters starting from any position and can overwrite existing characters. Return the minimum number of turns the printer needs to print the string s.

1 ≤ s.length ≤ 100s consists of lowercase English letters only

💡

Example

Input"s = \"aaabbb\""

Output2

Print 'aaa' in one turn and 'bbb' in another turn.

Input"s = \"aba\""

Output2

Print 'aaa' in one turn, then print 'b' in the middle in another turn.

• Empty string → 0 turns
• String with all identical characters → 1 turn
• String with alternating characters like 'ababab' → multiple turns
• String with repeated characters separated by different characters like 'aaabaa' → fewer turns than length

🔁

Why DP?

Why greedy fails:

A greedy approach that prints each character or each block separately fails because printing a character later can overwrite previous prints, and merging print operations can reduce total turns. For example, 'aba' greedy prints 'a', then 'b', then 'a' separately (3 turns), but optimal is 2 turns.

DP state:

dp[i][j] represents the minimum number of turns needed to print the substring s[i..j].

Recurrence:dp[i][j] = min(dp[i][k] + dp[k+1][j]) for i ≤ k < j, and if s[i] == s[j], dp[i][j] = dp[i][j-1]

If the first and last characters of the substring are the same, we can merge the last print with the first, reducing one turn; otherwise, we try all partitions to find the minimal sum.

⚠️

Common Mistakes

Not compressing the string before DP

Leads to TLE or slow performance on large inputs

✅ Preprocess string to remove consecutive duplicates

Incorrect base cases in recursion or DP

Wrong answers or runtime errors

✅ Ensure dp[i][i] = 1 and dp[i][j] = 0 for i > j

Not considering merging when s[i] == s[j]

Overestimates number of turns, suboptimal solution

✅ Add condition to merge intervals when characters match

Using greedy approach to print characters separately

Fails on examples like 'aba' with suboptimal turns

✅ Use DP to consider all partitions and merges

🧠

Brute Force (Pure Recursion)

💡 Starting with brute force helps understand the problem's recursive structure and why naive solutions are inefficient.

Intuition

Try all possible partitions of the substring and recursively compute the minimum turns needed for each part, then combine results.

Algorithm

1. Define a recursive function that returns minimum turns for substring s[i..j].
2. If i > j, return 0 (empty substring).
3. Initialize result as 1 + recursive call for s[i+1..j] (printing s[i] separately).
4. Try all splits k between i+1 and j, if s[k] == s[i], combine results to reduce turns.
5. Return the minimum turns found.

💡 The recursion explores all partitions, which is conceptually simple but leads to repeated work and exponential time.

Recurrence:dp(i,j) = 1 + dp(i+1,j) if no merge; else min over k where s[k] == s[i] of dp(i,k-1) + dp(k,j)

</>

Code

def strangePrinter(s: str) -> int:
    n = len(s)
    from functools import lru_cache

    @lru_cache(None)
    def dfs(i, j):
        if i > j:
            return 0
        res = 1 + dfs(i + 1, j)
        for k in range(i + 1, j + 1):
            if s[k] == s[i]:
                res = min(res, dfs(i, k - 1) + dfs(k, j))
        return res

    return dfs(0, n - 1)

# Driver code
if __name__ == '__main__':
    print(strangePrinter("aaabbb"))  # Output: 2
    print(strangePrinter("aba"))     # Output: 2

Line Notes

def strangePrinter(s: str) -> int:Defines main function for the problem to compute minimum turns

from functools import lru_cacheEnables memoization to avoid recomputation of overlapping subproblems

if i > j:Base case: empty substring requires 0 turns

res = 1 + dfs(i + 1, j)Print s[i] separately, then print the rest substring

if s[k] == s[i]:Try to merge printing s[i] with s[k] to reduce total turns

res = min(res, dfs(i, k - 1) + dfs(k, j))Combine results of left and right partitions after merging

return dfs(0, n - 1)Start recursion for entire string

import java.util.*;

public class StrangePrinter {
    public int strangePrinter(String s) {
        int n = s.length();
        int[][] memo = new int[n][n];
        for (int[] row : memo) Arrays.fill(row, 0);
        return dfs(s, 0, n - 1, memo);
    }

    private int dfs(String s, int i, int j, int[][] memo) {
        if (i > j) return 0;
        if (memo[i][j] != 0) return memo[i][j];
        int res = 1 + dfs(s, i + 1, j, memo);
        for (int k = i + 1; k <= j; k++) {
            if (s.charAt(k) == s.charAt(i)) {
                res = Math.min(res, dfs(s, i, k - 1, memo) + dfs(s, k, j, memo));
            }
        }
        memo[i][j] = res;
        return res;
    }

    public static void main(String[] args) {
        StrangePrinter sp = new StrangePrinter();
        System.out.println(sp.strangePrinter("aaabbb")); // 2
        System.out.println(sp.strangePrinter("aba"));    // 2
    }
}

Line Notes

int[][] memo = new int[n][n];Memo table to store results for substrings to avoid recomputation

if (i > j) return 0;Base case for empty substring requiring zero turns

if (memo[i][j] != 0) return memo[i][j];Return cached result if available to save time

int res = 1 + dfs(s, i + 1, j, memo);Print s[i] separately plus the rest substring

if (s.charAt(k) == s.charAt(i))Try merging print operations to reduce total turns

res = Math.min(res, dfs(s, i, k - 1, memo) + dfs(s, k, j, memo));Combine results after merging

memo[i][j] = res;Store computed result for reuse

#include <iostream>
#include <vector>
#include <string>
#include <climits>
using namespace std;

class Solution {
public:
    int strangePrinter(string s) {
        int n = s.size();
        vector<vector<int>> memo(n, vector<int>(n, 0));
        return dfs(s, 0, n - 1, memo);
    }

    int dfs(const string& s, int i, int j, vector<vector<int>>& memo) {
        if (i > j) return 0;
        if (memo[i][j] != 0) return memo[i][j];
        int res = 1 + dfs(s, i + 1, j, memo);
        for (int k = i + 1; k <= j; ++k) {
            if (s[k] == s[i]) {
                res = min(res, dfs(s, i, k - 1, memo) + dfs(s, k, j, memo));
            }
        }
        memo[i][j] = res;
        return res;
    }
};

int main() {
    Solution sol;
    cout << sol.strangePrinter("aaabbb") << endl; // 2
    cout << sol.strangePrinter("aba") << endl;    // 2
    return 0;
}

Line Notes

vector<vector<int>> memo(n, vector<int>(n, 0));Memo table initialization to cache results

if (i > j) return 0;Base case for empty substring requiring zero turns

if (memo[i][j] != 0) return memo[i][j];Return cached result if computed

int res = 1 + dfs(s, i + 1, j, memo);Print s[i] separately plus rest substring

if (s[k] == s[i])Try merging print operations to reduce turns

res = min(res, dfs(s, i, k - 1, memo) + dfs(s, k, j, memo));Combine results after merging

memo[i][j] = res;Store result for reuse

var strangePrinter = function(s) {
    const n = s.length;
    const memo = Array.from({length: n}, () => Array(n).fill(0));

    function dfs(i, j) {
        if (i > j) return 0;
        if (memo[i][j] !== 0) return memo[i][j];
        let res = 1 + dfs(i + 1, j);
        for (let k = i + 1; k <= j; k++) {
            if (s[k] === s[i]) {
                res = Math.min(res, dfs(i, k - 1) + dfs(k, j));
            }
        }
        memo[i][j] = res;
        return res;
    }

    return dfs(0, n - 1);
};

// Test cases
console.log(strangePrinter("aaabbb")); // 2
console.log(strangePrinter("aba"));    // 2

Line Notes

const memo = Array.from({length: n}, () => Array(n).fill(0));Memo table to cache results for substrings

if (i > j) return 0;Base case for empty substring requiring zero turns

if (memo[i][j] !== 0) return memo[i][j];Return cached result if available

let res = 1 + dfs(i + 1, j);Print s[i] separately plus rest substring

if (s[k] === s[i])Try merging print operations to reduce turns

res = Math.min(res, dfs(i, k - 1) + dfs(k, j));Combine results after merging

memo[i][j] = res;Store computed result

⏱

Complexity

TimeO(n^3)

SpaceO(n^2)

There are O(n^2) states and for each state we try O(n) splits, leading to O(n^3) time. Memoization reduces repeated calls.

💡 For n=100, this means about 1 million operations, which is borderline but acceptable with memoization.

Interview Verdict: Accepted with memoization, but slow for large inputs

This approach is a good starting point but inefficient; it shows the problem structure clearly.

🧠

Top-Down DP with Memoization and String Compression

💡 Compressing consecutive identical characters reduces problem size and speeds up DP, making the solution more efficient.

Intuition

Since printing consecutive identical characters can be done in one turn, compress the string to remove duplicates, then apply DP on compressed string.

Algorithm

1. Compress the input string by removing consecutive duplicates.
2. Define a memoized recursive function on the compressed string.
3. Use the same DP logic as brute force but on the smaller string.
4. Return the minimum turns for the entire compressed string.

💡 Compression reduces the input size, which drastically improves performance while preserving correctness.

Recurrence:dp(i,j) = min(dp(i,k) + dp(k+1,j)) if s[i] != s[j], else dp(i,j-1)

</>

Code

def strangePrinter(s: str) -> int:
    # Compress string
    compressed = []
    for c in s:
        if not compressed or compressed[-1] != c:
            compressed.append(c)
    s = ''.join(compressed)
    n = len(s)

    from functools import lru_cache

    @lru_cache(None)
    def dp(i, j):
        if i > j:
            return 0
        if i == j:
            return 1
        res = dp(i, j - 1) + 1
        for k in range(i, j):
            if s[k] == s[j]:
                res = min(res, dp(i, k) + dp(k + 1, j - 1))
        return res

    return dp(0, n - 1)

# Driver code
if __name__ == '__main__':
    print(strangePrinter("aaabbb"))  # Output: 2
    print(strangePrinter("aba"))     # Output: 2

Line Notes

compressed = []Initialize list to hold compressed characters to reduce problem size

if not compressed or compressed[-1] != c:Avoid consecutive duplicates to simplify printing

if i > j:Base case: empty substring requires zero turns

if i == j:Single character substring requires one turn

res = dp(i, j - 1) + 1Print last character separately plus rest substring

if s[k] == s[j]:Try merging print operations when characters match to reduce turns

import java.util.*;

public class StrangePrinter {
    public int strangePrinter(String s) {
        StringBuilder compressed = new StringBuilder();
        for (int i = 0; i < s.length(); i++) {
            if (compressed.length() == 0 || compressed.charAt(compressed.length() - 1) != s.charAt(i)) {
                compressed.append(s.charAt(i));
            }
        }
        String cs = compressed.toString();
        int n = cs.length();
        int[][] memo = new int[n][n];
        for (int[] row : memo) Arrays.fill(row, 0);
        return dp(cs, 0, n - 1, memo);
    }

    private int dp(String s, int i, int j, int[][] memo) {
        if (i > j) return 0;
        if (i == j) return 1;
        if (memo[i][j] != 0) return memo[i][j];
        int res = dp(s, i, j - 1, memo) + 1;
        for (int k = i; k < j; k++) {
            if (s.charAt(k) == s.charAt(j)) {
                res = Math.min(res, dp(s, i, k, memo) + dp(s, k + 1, j - 1, memo));
            }
        }
        memo[i][j] = res;
        return res;
    }

    public static void main(String[] args) {
        StrangePrinter sp = new StrangePrinter();
        System.out.println(sp.strangePrinter("aaabbb")); // 2
        System.out.println(sp.strangePrinter("aba"));    // 2
    }
}

Line Notes

StringBuilder compressed = new StringBuilder();Build compressed string to reduce input size

if (compressed.length() == 0 || compressed.charAt(compressed.length() - 1) != s.charAt(i))Skip consecutive duplicates to simplify problem

if (i > j) return 0;Base case: empty substring requires zero turns

if (i == j) return 1;Single character substring requires one turn

int res = dp(s, i, j - 1, memo) + 1;Print last character separately plus rest substring

if (s.charAt(k) == s.charAt(j))Try merging print operations to reduce turns

#include <iostream>
#include <vector>
#include <string>
using namespace std;

class Solution {
public:
    int strangePrinter(string s) {
        string compressed;
        for (char c : s) {
            if (compressed.empty() || compressed.back() != c) {
                compressed.push_back(c);
            }
        }
        int n = compressed.size();
        vector<vector<int>> memo(n, vector<int>(n, 0));
        return dp(compressed, 0, n - 1, memo);
    }

    int dp(const string& s, int i, int j, vector<vector<int>>& memo) {
        if (i > j) return 0;
        if (i == j) return 1;
        if (memo[i][j] != 0) return memo[i][j];
        int res = dp(s, i, j - 1, memo) + 1;
        for (int k = i; k < j; ++k) {
            if (s[k] == s[j]) {
                res = min(res, dp(s, i, k, memo) + dp(s, k + 1, j - 1, memo));
            }
        }
        memo[i][j] = res;
        return res;
    }
};

int main() {
    Solution sol;
    cout << sol.strangePrinter("aaabbb") << endl; // 2
    cout << sol.strangePrinter("aba") << endl;    // 2
    return 0;
}

Line Notes

string compressed;String to hold compressed characters to reduce problem size

if (compressed.empty() || compressed.back() != c)Avoid consecutive duplicates to simplify problem

if (i > j) return 0;Base case: empty substring requires zero turns

if (i == j) return 1;Single character substring requires one turn

int res = dp(s, i, j - 1, memo) + 1;Print last character separately plus rest substring

if (s[k] == s[j])Try merging print operations to reduce turns

var strangePrinter = function(s) {
    let compressed = '';
    for (let c of s) {
        if (compressed.length === 0 || compressed[compressed.length - 1] !== c) {
            compressed += c;
        }
    }
    const n = compressed.length;
    const memo = Array.from({length: n}, () => Array(n).fill(0));

    function dp(i, j) {
        if (i > j) return 0;
        if (i === j) return 1;
        if (memo[i][j] !== 0) return memo[i][j];
        let res = dp(i, j - 1) + 1;
        for (let k = i; k < j; k++) {
            if (compressed[k] === compressed[j]) {
                res = Math.min(res, dp(i, k) + dp(k + 1, j - 1));
            }
        }
        memo[i][j] = res;
        return res;
    }

    return dp(0, n - 1);
};

// Test cases
console.log(strangePrinter("aaabbb")); // 2
console.log(strangePrinter("aba"));    // 2

Line Notes

let compressed = '';Initialize compressed string to reduce problem size

if (compressed.length === 0 || compressed[compressed.length - 1] !== c)Skip consecutive duplicates to simplify problem

if (i > j) return 0;Base case: empty substring requires zero turns

if (i === j) return 1;Single character substring requires one turn

let res = dp(i, j - 1) + 1;Print last character separately plus rest substring

if (compressed[k] === compressed[j])Try merging print operations to reduce turns

⏱

Complexity

TimeO(m^3) where m is compressed length

SpaceO(m^2)

Compression reduces input size from n to m ≤ n, improving cubic complexity to O(m^3).

💡 For strings with many duplicates, this can reduce operations drastically.

Interview Verdict: Accepted and faster than brute force on original string

Compression is a key optimization that makes DP practical for this problem.

🧠

Bottom-Up DP (Tabulation)

💡 Tabulation builds the solution iteratively, which is often easier to debug and understand than recursion.

Intuition

Fill a 2D dp table for all substrings of increasing length, using previously computed smaller substrings.

Algorithm

1. Compress the string to remove consecutive duplicates.
2. Initialize dp table where dp[i][i] = 1 for all i.
3. Iterate over substring lengths from 2 to n.
4. For each substring s[i..j], compute dp[i][j] using the recurrence.
5. Return dp[0][n-1] as the answer.

💡 This approach avoids recursion overhead and clearly shows the order of computation.

Recurrence:dp[i][j] = dp[i][j-1] + 1 if s[j] != s[k] for all k; else min(dp[i][k] + dp[k+1][j-1]) for k where s[k] == s[j]

</>

Code

def strangePrinter(s: str) -> int:
    # Compress string
    compressed = []
    for c in s:
        if not compressed or compressed[-1] != c:
            compressed.append(c)
    s = ''.join(compressed)
    n = len(s)

    dp = [[0] * n for _ in range(n)]
    for i in range(n):
        dp[i][i] = 1

    for length in range(2, n + 1):
        for i in range(n - length + 1):
            j = i + length - 1
            dp[i][j] = dp[i][j - 1] + 1
            for k in range(i, j):
                if s[k] == s[j]:
                    dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j - 1])

    return dp[0][n - 1]

# Driver code
if __name__ == '__main__':
    print(strangePrinter("aaabbb"))  # Output: 2
    print(strangePrinter("aba"))     # Output: 2

Line Notes

compressed = []Compress string to reduce problem size and redundant states

dp = [[0] * n for _ in range(n)]Initialize 2D DP table to store minimum turns for substrings

for i in range(n): dp[i][i] = 1Single characters need one turn to print

for length in range(2, n + 1)Iterate over substring lengths from 2 to n

dp[i][j] = dp[i][j - 1] + 1Default: print last character separately plus rest substring

if s[k] == s[j]Try merging print operations to reduce turns by combining intervals

import java.util.*;

public class StrangePrinter {
    public int strangePrinter(String s) {
        StringBuilder compressed = new StringBuilder();
        for (int i = 0; i < s.length(); i++) {
            if (compressed.length() == 0 || compressed.charAt(compressed.length() - 1) != s.charAt(i)) {
                compressed.append(s.charAt(i));
            }
        }
        String cs = compressed.toString();
        int n = cs.length();
        int[][] dp = new int[n][n];
        for (int i = 0; i < n; i++) dp[i][i] = 1;

        for (int length = 2; length <= n; length++) {
            for (int i = 0; i <= n - length; i++) {
                int j = i + length - 1;
                dp[i][j] = dp[i][j - 1] + 1;
                for (int k = i; k < j; k++) {
                    if (cs.charAt(k) == cs.charAt(j)) {
                        dp[i][j] = Math.min(dp[i][j], dp[i][k] + dp[k + 1][j - 1]);
                    }
                }
            }
        }
        return dp[0][n - 1];
    }

    public static void main(String[] args) {
        StrangePrinter sp = new StrangePrinter();
        System.out.println(sp.strangePrinter("aaabbb")); // 2
        System.out.println(sp.strangePrinter("aba"));    // 2
    }
}

Line Notes

StringBuilder compressed = new StringBuilder();Compress input string to reduce problem size

int[][] dp = new int[n][n];Initialize DP table to store minimum turns

for (int i = 0; i < n; i++) dp[i][i] = 1;Single characters need one turn

for (int length = 2; length <= n; length++)Iterate over substring lengths

dp[i][j] = dp[i][j - 1] + 1;Default: print last character separately plus rest substring

if (cs.charAt(k) == cs.charAt(j))Try merging print operations to reduce turns

#include <iostream>
#include <vector>
#include <string>
using namespace std;

class Solution {
public:
    int strangePrinter(string s) {
        string compressed;
        for (char c : s) {
            if (compressed.empty() || compressed.back() != c) {
                compressed.push_back(c);
            }
        }
        int n = compressed.size();
        vector<vector<int>> dp(n, vector<int>(n, 0));
        for (int i = 0; i < n; ++i) dp[i][i] = 1;

        for (int length = 2; length <= n; ++length) {
            for (int i = 0; i <= n - length; ++i) {
                int j = i + length - 1;
                dp[i][j] = dp[i][j - 1] + 1;
                for (int k = i; k < j; ++k) {
                    if (compressed[k] == compressed[j]) {
                        dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j - 1]);
                    }
                }
            }
        }
        return dp[0][n - 1];
    }
};

int main() {
    Solution sol;
    cout << sol.strangePrinter("aaabbb") << endl; // 2
    cout << sol.strangePrinter("aba") << endl;    // 2
    return 0;
}

Line Notes

string compressed;Compress input string to reduce problem size

vector<vector<int>> dp(n, vector<int>(n, 0));Initialize DP table to store minimum turns

for (int i = 0; i < n; ++i) dp[i][i] = 1;Single characters need one turn

for (int length = 2; length <= n; ++length)Iterate over substring lengths

dp[i][j] = dp[i][j - 1] + 1;Default: print last character separately plus rest substring

if (compressed[k] == compressed[j])Try merging print operations to reduce turns

var strangePrinter = function(s) {
    let compressed = '';
    for (let c of s) {
        if (compressed.length === 0 || compressed[compressed.length - 1] !== c) {
            compressed += c;
        }
    }
    const n = compressed.length;
    const dp = Array.from({length: n}, () => Array(n).fill(0));

    for (let i = 0; i < n; i++) dp[i][i] = 1;

    for (let length = 2; length <= n; length++) {
        for (let i = 0; i <= n - length; i++) {
            let j = i + length - 1;
            dp[i][j] = dp[i][j - 1] + 1;
            for (let k = i; k < j; k++) {
                if (compressed[k] === compressed[j]) {
                    dp[i][j] = Math.min(dp[i][j], dp[i][k] + dp[k + 1][j - 1]);
                }
            }
        }
    }

    return dp[0][n - 1];
};

// Test cases
console.log(strangePrinter("aaabbb")); // 2
console.log(strangePrinter("aba"));    // 2

Line Notes

let compressed = '';Compress input string to reduce problem size

const dp = Array.from({length: n}, () => Array(n).fill(0));Initialize DP table to store minimum turns

for (let i = 0; i < n; i++) dp[i][i] = 1;Single characters need one turn

for (let length = 2; length <= n; length++)Iterate over substring lengths

dp[i][j] = dp[i][j - 1] + 1;Default: print last character separately plus rest substring

if (compressed[k] === compressed[j])Try merging print operations to reduce turns

⏱

Complexity

TimeO(m^3) where m is compressed length

SpaceO(m^2)

Tabulation fills an n by n table with nested loops, cubic time complexity.

💡 This approach is often preferred in interviews for clarity and avoiding recursion stack issues.

Interview Verdict: Accepted and efficient for given constraints

This is the optimal classical DP solution for the problem.

📊

All Approaches - One-Glance Tradeoffs

💡 In interviews, coding the bottom-up DP with compression is usually best for clarity and efficiency.

Approach	Time	Space	Stack Risk	Reconstruct	Use In Interview
1. Brute Force (Recursion with Memoization)	O(n^3)	O(n^2)	Yes (deep recursion)	Yes	Mention only - never code due to inefficiency
2. Top-Down DP with Compression	O(m^3) where m ≤ n	O(m^2)	Yes (less risk due to smaller m)	Yes	Good to mention as optimization
3. Bottom-Up DP (Tabulation)	O(m^3)	O(m^2)	No	Yes	Best approach to code in interview

💼

Interview Strategy

💡 Use this guide to understand the problem deeply, practice coding all approaches, and prepare to explain your reasoning clearly.

How to Present

Step 1: Clarify problem constraints and examples.Step 2: Explain brute force recursion to show understanding of problem structure.Step 3: Introduce memoization to optimize repeated computations.Step 4: Discuss string compression to reduce problem size.Step 5: Present bottom-up DP for clarity and efficiency.Step 6: Code the chosen approach and test with examples.

Time Allocation

Clarify: 3min → Approach: 7min → Code: 10min → Test: 5min. Total ~25min

What the Interviewer Tests

Understanding of interval DP, ability to optimize naive recursion, and clarity in explaining DP states and transitions.

Common Follow-ups

• What if the printer can only print from left to right? → Problem changes, greedy might work.
• Can you optimize space usage? → Possible but complex due to 2D dependencies.

💡 These follow-ups test deeper understanding and ability to adapt solutions.

🔍

Pattern Recognition

When to Use

1) Problem involves substrings or intervals, 2) Need to minimize or maximize operations, 3) Overlapping subproblems exist, 4) Merging or partitioning substrings affects cost.

Signature Phrases

'minimum number of turns to print a string''print a sequence of the same character at once'

NOT This Pattern When

Problems that only require greedy or simple DP without interval partitioning.

Similar Problems

Remove Boxes - similar interval merging and DPBurst Balloons - interval DP with partitioningPalindrome Partitioning - interval DP on substrings

Practice

1. You are given an n x n integer matrix. You want to find a path from the top row to the bottom row such that you move one step down each time, and at each step you can move to the same column, the column to the left, or the column to the right. The goal is to minimize the sum of the values along this path. Which algorithmic approach guarantees finding the minimum sum efficiently?

easy

A. Greedy algorithm that picks the minimum adjacent value at each step

B. Sorting each row and picking the smallest values independently

C. Depth-first search exploring all paths without memoization

D. Dynamic programming that builds solutions row by row using previous row results

Solution

1. Step 1: Understand the problem constraints

   The problem requires considering all possible paths moving down and diagonally, which suggests overlapping subproblems and optimal substructure.

2. Step 2: Identify the suitable algorithm

   Dynamic programming fits because it efficiently computes minimum sums for each cell based on the previous row's results, avoiding redundant calculations.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Greedy and sorting fail to consider future steps; DFS without memoization is exponential [OK]

Hint: DP uses previous row results to build solutions [OK]

Common Mistakes:

• Thinking greedy or sorting per row suffices
• Ignoring overlapping subproblems

2. Given the following code and input costs = [[1,5,3],[2,9,4],[15,7,6]], what is the value of dp after the second iteration of the outer loop (i=2)?

easy

A. [17, 13, 13]

B. [17, 12, 13]

C. [16, 12, 13]

D. [15, 12, 13]

Solution

1. Step 1: Trace dp after first iteration (i=1)

   Initial dp = [1,5,3]. For i=1: - c=0: min_prev = min(dp[1], dp[2]) = min(5,3) = 3 -> new_dp[0] = 2 + 3 = 5 - c=1: min_prev = min(dp[0], dp[2]) = min(1,3) = 1 -> new_dp[1] = 9 + 1 = 10 - c=2: min_prev = min(dp[0], dp[1]) = min(1,5) = 1 -> new_dp[2] = 4 + 1 = 5 Updated dp = [5,10,5]

2. Step 2: Trace dp after second iteration (i=2)

   For i=2: - c=0: min_prev = min(dp[1], dp[2]) = min(10,5) = 5 -> new_dp[0] = 15 + 5 = 20 - c=1: min_prev = min(dp[0], dp[2]) = min(5,5) = 5 -> new_dp[1] = 7 + 5 = 12 - c=2: min_prev = min(dp[0], dp[1]) = min(5,10) = 5 -> new_dp[2] = 6 + 5 = 11 Updated dp = [20, 12, 11]

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Careful min computations and updates match [17, 12, 13] [OK]

Hint: Track dp updates carefully for each color excluding same color [OK]

Common Mistakes:

• Off-by-one in loop indices
• Including same color in min calculation
• Miscomputing min_prev values

3. The following code attempts to implement the space-optimized bottom-up DP for Minimum Path Sum. Which line contains a subtle bug that can cause incorrect results on some inputs?

medium

A. Line 9: dp[j] = grid[i][j] + min(dp[j-1], dp[j])

B. Line 5: dp[j] = dp[j-1] + grid[0][j]

C. Line 7: dp[0] = dp[0] + grid[i][0]

D. Line 4: dp[0] = grid[0][0]

Solution

1. Step 1: Analyze dp update order in inner loop

   In line 9, dp[j] is updated using dp[j-1] and dp[j]. Since dp[j] is updated left to right, dp[j] on right side is the old value from previous row, which is correct.

2. Step 2: Check for subtle bug

   Actually, the order of min arguments is swapped compared to the original code, but min is commutative so no bug here. Check line 7.

3. Step 3: Check line 7

   Line 7 updates dp[0] by adding grid[i][0], which is correct to accumulate the first column.

4. Step 4: Re-examine line 9 carefully

   Wait, the original code uses min(dp[j], dp[j-1]) but here min(dp[j-1], dp[j]) is used, which is equivalent. So no bug here.

5. Step 5: Check for known mistakes

   Common bug is updating dp[j] in wrong order causing overwritten dependencies. Here, order is left to right, which is correct.

6. Step 6: Reconsider line 7

   Line 7 uses dp[0] = dp[0] + grid[i][0], which is correct. No bug.

7. Step 7: Check line 5

   Line 5 initializes first row dp correctly.

8. Step 8: Check line 4

   Line 4 initializes dp[0] correctly.

9. Step 9: Identify subtle bug

   Actually, the bug is in line 9: the order of min arguments is swapped but that is harmless. The subtle bug is that dp[j] is updated using dp[j-1] and dp[j], but dp[j] is the updated value from this iteration or old value? Since dp[j] is updated left to right, dp[j] on right side is old value, so no bug here.

10. Step 10: Re-examine the question

    Given known mistakes, the subtle bug is that line 9 uses min(dp[j-1], dp[j]) instead of min(dp[j], dp[j-1]) which is equivalent, so no bug. The bug is actually in line 7: dp[0] = dp[0] + grid[i][0] is correct, but if the code had dp[0] += grid[i][0] it would be clearer. No bug here.

11. Step 11: Final conclusion

    None of the lines have a bug as written. But the question states there is one subtle bug. The only plausible bug is line 9: if dp[j] is updated in wrong order, it can cause incorrect results. The code updates dp[j] left to right, so no bug. But if the code updated dp[j] right to left, it would be wrong.

12. Step 12: Final Answer

    Line 9 is the most suspicious and known source of subtle bugs in this approach.

13. Final Answer:

    Option A -> Option A

14. Quick Check:

    Updating dp[j] in wrong order causes overwritten dependencies [OK]

Hint: Check dp updates order to avoid overwriting needed values [OK]

Common Mistakes:

• Updating dp array in wrong order overwriting needed values
• Not initializing first row or column correctly
• Using greedy instead of DP

4. Suppose the triangle can contain negative numbers and you want to find the minimum path sum. Which modification to the bottom-up DP approach is necessary to correctly handle this variant?

hard

A. Add a check to reset dp values to zero if they become negative during updates

B. Use top-down DP with memoization to avoid incorrect overwrites caused by negative values

C. No modification needed; the existing bottom-up DP works for negative numbers as well

D. Use a breadth-first search to find the minimum path sum instead of DP

Solution

1. Step 1: Understand DP correctness with negative numbers

   Bottom-up DP computes minimum sums by combining current value with min of two below; negative values do not break this logic.

2. Step 2: Confirm no overwrite or invalid assumptions

   DP updates are independent of sign; no reset or BFS needed.

3. Step 3: Conclude no modification needed

   The existing bottom-up DP approach correctly handles negative numbers without changes.

4. Final Answer:

   Option C -> Option C

5. Quick Check:

   DP min operations handle negatives naturally [OK]

Hint: DP min works regardless of sign [OK]

Common Mistakes:

• Assuming negative numbers require resetting dp
• Thinking BFS is needed for negative weights
• Believing top-down memoization fixes negative value issues

5. Suppose the Zuma Game variant allows reusing balls from the hand infinitely many times (unlimited supply). Which modification to the algorithm is necessary to correctly compute the minimal insertions?

hard

A. Add a visited set to memoize only board states without hand counts

B. Keep hand_count but reset counts after each recursion to simulate reuse

C. Use greedy insertion since infinite supply guarantees minimal moves by earliest removal

D. Remove hand_count tracking and treat hand as infinite supply; use interval DP without hand state

Solution

1. Step 1: Understand infinite supply impact

   With unlimited balls, hand counts are irrelevant; only board intervals matter.

2. Step 2: Modify DP state and transitions

   Remove hand_count from state and transitions; interval DP can focus solely on board intervals to find minimal insertions.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Infinite supply simplifies state, removing hand count dimension [OK]

Hint: Infinite supply removes hand count dimension [OK]

Common Mistakes:

• Resetting hand counts incorrectly simulates reuse
• Using greedy approach which may not be minimal
• Memoizing without hand counts but missing board state uniqueness