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 need to find the cheapest cost to travel from a source city to a destination city with at most K stops, given a list of flights with costs. Which algorithmic approach guarantees finding the optimal solution efficiently under these constraints?

easy

A. Greedy algorithm using Dijkstra's shortest path without modification

B. Topological sort followed by single pass relaxation of edges

C. Simple depth-first search exploring all paths without pruning

D. Dynamic programming using a bottom-up approach iterating over stops

Solution

1. Step 1: Understand the problem constraints

   The problem requires finding the cheapest flight with at most K stops, which limits path length and requires considering multiple paths.

2. Step 2: Identify suitable algorithm

   Greedy Dijkstra fails because it doesn't limit stops; DFS is exponential; topological sort requires DAG which flights graph may not be. Bottom-up DP iterates over stops and relaxes edges, guaranteeing optimal cost within K stops.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Bottom-up DP with stops limit ensures optimal solution [OK]

Hint: DP with stops limit ensures optimal cost [OK]

Common Mistakes:

• Using Dijkstra without stop limit
• Trying DFS without pruning
• Assuming DAG for topological sort

2. You are given a 2D binary matrix filled with 0's and 1's. The task is to find the largest square containing only 1's and return its area. Which algorithmic approach guarantees an optimal solution with polynomial time complexity?

easy

A. Greedy approach that expands squares from each cell until a zero is found

B. Sorting rows and columns to find the largest continuous block of 1's

C. Depth-First Search exploring all possible squares recursively without memoization

D. Dynamic Programming that builds solutions for smaller squares to find larger ones

Solution

1. Step 1: Understand problem constraints

   The problem requires finding the largest square of 1's, which depends on overlapping subproblems of smaller squares.

2. Step 2: Recognize DP pattern

   Dynamic Programming efficiently computes maximal squares by using previously computed smaller squares, avoiding redundant checks.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   DP uses subproblem solutions to build larger squares [OK]

Hint: Largest square -> DP using smaller squares [OK]

Common Mistakes:

• Thinking greedy expansion always works
• Confusing DFS with DP
• Assuming sorting helps find squares

3. You need to find the number of distinct ways to move from the top-left corner to the bottom-right corner of an m x n grid, moving only down or right at each step. Which algorithmic approach guarantees an efficient and optimal solution for this problem?

easy

A. Pure brute force recursion exploring all possible paths without memoization

B. Greedy algorithm that always moves towards the direction with fewer remaining steps

C. Dynamic Programming that builds solutions from smaller subproblems using a grid-based state representation

D. Divide and conquer by splitting the grid into quadrants and combining results

Solution

1. Step 1: Understand problem constraints

   The problem requires counting all unique paths with only right and down moves, which naturally forms overlapping subproblems.

2. Step 2: Identify suitable approach

   Dynamic Programming efficiently solves overlapping subproblems by storing intermediate results, unlike greedy or pure recursion which are either incorrect or inefficient.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   DP uses subproblem reuse -> optimal and efficient [OK]

Hint: Counting paths with overlapping subproblems -> DP [OK]

Common Mistakes:

• Thinking greedy can find all paths
• Using pure recursion without memoization

4. What is the time complexity of the bottom-up dynamic programming solution for the Dungeon Game problem on an m x n grid, and why might some candidates mistakenly think it is higher?

medium

A. O(2^{m+n}) because all paths are explored recursively

B. O(m + n) since only one path is considered at a time

C. O(m * n * max(|dungeon[i][j]|)) due to health value range affecting computations

D. O(m * n) because each cell is computed once using constant time operations

Solution

1. Step 1: Identify loops

   The bottom-up DP uses nested loops over m rows and n columns, each cell computed once.

2. Step 2: Analyze per-cell work

   Each dp[i][j] calculation is O(1), involving min and max operations.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   DP table size m*n and constant work per cell [OK]

Hint: DP fills m*n table once, no recursion overhead [OK]

Common Mistakes:

• Confusing brute force recursion with DP
• Thinking health values affect complexity

5. Suppose the problem is modified so that you can reuse the same color for adjacent houses, but you want to minimize the total cost. Which modification to the DP solution is correct?

hard

A. Use brute force recursion without memoization since constraints are relaxed.

B. Keep excluding the previous color but add a penalty cost for same color adjacency.

C. Remove the condition excluding the previous color when computing min_prev; just take min(dp) for all colors.

D. Use a greedy approach picking the cheapest color for each house independently.

Solution

1. Step 1: Understand constraint relaxation

   Allowing same color for adjacent houses removes the need to exclude previous color in DP transitions.

2. Step 2: Modify DP accordingly

   Now min_prev is simply min(dp) over all colors, no exclusion needed. This simplifies the DP and still finds minimum cost.

3. Step 3: Evaluate other options

   Adding penalty is unnecessary. Brute force is inefficient. Greedy fails because costs vary per house and color.

4. Final Answer:

   Option C -> Option C

5. Quick Check:

   Relaxed constraints allow including all colors in min calculation [OK]

Hint: Relaxed constraints mean no exclusion of previous color in DP [OK]

Common Mistakes:

• Still excluding previous color unnecessarily
• Adding penalty complicates solution
• Using brute force or greedy incorrectly