Palindrome Partitioning DP Minimum Cuts in DSA Rust - Time & Space Complexity

Time Complexity: Palindrome Partitioning DP Minimum Cuts

O(n³)

Understanding Time Complexity

We want to know how the time needed to find the minimum cuts for palindrome partitions grows as the input string gets longer.

How does the number of steps change when the string size increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn min_cut(s: &str) -> usize {
    let n = s.len();
    let s_chars: Vec = s.chars().collect();
    let mut dp = vec![0usize; n + 1];
    let mut palindrome = vec![vec![false; n]; n];

    for i in 0..=n {
        dp[i] = n - i;
    }

    for i in (0..n).rev() {
        for j in i..n {
            if s_chars[i] == s_chars[j] && (j - i < 2 || palindrome[i + 1][j - 1]) {
                palindrome[i][j] = true;
                dp[i] = ::std::cmp::min(dp[i], 1usize + dp[j + 1]);
            }
        }
    }
    dp[0] - 1
}

This code finds the minimum number of cuts needed to split a string into palindrome parts using dynamic programming.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops checking all substring pairs (i, j) to find palindromes.
• How many times: Outer loop runs n times, inner loop runs up to n times, so roughly n x n = n² times.

How Execution Grows With Input

As the string length grows, the number of substring checks grows roughly with the square of the length.

Input Size (n)	Approx. Operations
10	About 100 checks
100	About 10,000 checks
1000	About 1,000,000 checks

Pattern observation: Doubling the input size roughly quadruples the work because of the nested loops.

Final Time Complexity

Time Complexity: O(n³)

This means the time needed grows roughly with the cube of the string length, so longer strings take much more time.

Common Mistake

[X] Wrong: "The algorithm runs in linear time because it just scans the string twice."

[OK] Correct: The nested loops check all possible substrings, which means many repeated checks, making it grow much faster than linear.

Interview Connect

Understanding this time complexity helps you explain how dynamic programming improves over naive methods and shows your grasp of nested loops and substring checks.

Self-Check

"What if we used memoization with recursion instead of bottom-up DP? How would the time complexity change?"

Practice

1. What does the "Palindrome Partitioning DP Minimum Cuts" problem aim to find for a given string s?

easy

A. The minimum number of cuts needed to partition s into palindromic substrings

B. The longest palindromic substring in s

C. The total number of palindromic substrings in s

D. The maximum number of cuts possible in s

Solution

1. Step 1: Understand the problem goal

   The problem asks for the fewest cuts needed so that every piece after cutting is a palindrome.

2. Step 2: Clarify what minimum cuts mean

   Minimum cuts means the smallest number of splits to make all parts palindromes.

3. Final Answer:

   The minimum number of cuts needed to partition s into palindromic substrings -> Option A

4. Quick Check:

   Minimum cuts = minimum splits for palindrome parts [OK]

Hint: Minimum cuts split string into palindrome parts [OK]

Common Mistakes:

• Confusing minimum cuts with longest palindrome
• Thinking about total palindrome count instead of cuts
• Assuming maximum cuts is the goal

2. Which of the following Rust code snippets correctly initializes a 2D vector dp of size n x n with false values for palindrome checks?

easy

A. let dp = vec![vec![true; n]; n];

B. let dp = vec![false; n * n];

C. let dp = vec![vec![false; n * n]; 1];

D. let dp = vec![vec![false; n]; n];

Solution

1. Step 1: Understand 2D vector initialization in Rust

   To create an n by n 2D vector filled with false, use vec![vec![false; n]; n].

2. Step 2: Check each option

   let dp = vec![vec![false; n]; n]; correctly creates a 2D vector with n rows and n columns filled with false.

3. Final Answer:

   let dp = vec![vec![false; n]; n]; -> Option D

4. Quick Check:

   2D vector with false = vec![vec![false; n]; n] [OK]

Hint: Use vec![vec![false; n]; n] for 2D boolean vector [OK]

Common Mistakes:

• Using 1D vector instead of 2D
• Filling with true instead of false
• Incorrect dimensions in inner or outer vec

3. Given the Rust function below for minimum cuts, what is the output for input "aab"?

fn min_cut(s: &str) -> i32 {
    let n = s.len();
    let s_bytes = s.as_bytes();
    let mut dp = vec![0; n + 1];
    let mut palindrome = vec![vec![false; n]; n];
    for i in 0..=n {
        dp[i] = i as i32 - 1;
    }
    for i in 0..n {
        for j in i..n {
            if s_bytes[i] == s_bytes[j] && (j - i < 2 || palindrome[i + 1][j - 1]) {
                palindrome[i][j] = true;
                dp[j + 1] = dp[j + 1].min(dp[i] + 1);
            }
        }
    }
    dp[n]
}

println!("{}", min_cut("aab"));

medium

A. 3

B. 2

C. 1

D. 0

Solution

1. Step 1: Trace palindrome table for "aab"

   Substrings: "a" (0,0), "aa" (0,1), "aab" (0,2), "a" (1,1), "ab" (1,2), "b" (2,2). "aa" and single letters are palindrome.

2. Step 2: Calculate dp array values

   dp[0] = -1, dp[1] = 0, dp[2] = 1, dp[3] = 2 initially. Updates: For "aa", dp[2] = min(dp[2], dp[0] + 1) = 0. For "b", dp[3] = min(dp[3], dp[2] + 1) = 1.

3. Final Answer:

   1 -> Option C

4. Quick Check:

   Minimum cuts for "aab" = 1 [OK]

Hint: Check palindrome substrings and update dp minimum cuts [OK]

Common Mistakes:

• Not initializing dp correctly
• Ignoring palindrome length 1 or 2 cases
• Off-by-one errors in dp indexing

4. Identify the bug in the following Rust code snippet for palindrome partitioning minimum cuts:

fn min_cut(s: &str) -> i32 {
    let n = s.len();
    let s_bytes = s.as_bytes();
    let mut dp = vec![0; n + 1];
    let mut palindrome = vec![vec![false; n]; n];
    for i in 0..n {
        dp[i] = i as i32;
    }
    for i in 0..n {
        for j in i..n {
            if s_bytes[i] == s_bytes[j] && (j - i < 2 || palindrome[i + 1][j - 1]) {
                palindrome[i][j] = true;
                dp[j + 1] = dp[j + 1].min(dp[i] + 1);
            }
        }
    }
    dp[n]
}

medium

A. Palindrome check condition is incorrect

B. dp array initialization should start from -1 for dp[0]

C. Loop ranges for i and j are reversed

D. s_bytes should be converted to chars instead of bytes

Solution

1. Step 1: Check dp initialization

   dp[i] should be initialized as i - 1 to handle zero cuts correctly, but here dp[i] = i which is off by one.

2. Step 2: Understand impact of dp initialization

   Incorrect initialization causes wrong minimum cuts calculation because dp[0] should be -1 to represent no cuts before start.

3. Final Answer:

   dp array initialization should start from -1 for dp[0] -> Option B

4. Quick Check:

   dp[0] = -1 needed for correct minimum cuts [OK]

Hint: Initialize dp[0] = -1 for correct minimum cuts [OK]

Common Mistakes:

• Initializing dp with zeros or i instead of i-1
• Misunderstanding palindrome base cases
• Incorrect loop boundaries

5. You want to find the minimum cuts for the string "abccbaabccba" using the palindrome partitioning DP approach. Which of the following statements is TRUE about the expected minimum cuts and the DP process?

hard

A. The minimum cuts will be 0 because the entire string is a palindrome

B. The minimum cuts will be 1 because the string can be split into two palindromes

C. The minimum cuts will be 2 because the string contains two palindromic halves

D. The minimum cuts will be 3 because multiple small palindromes are needed

Solution

1. Step 1: Check if the entire string is palindrome

   "abccbaabccba" reads the same forwards and backwards, so it is a palindrome.

2. Step 2: Understand minimum cuts for a full palindrome

   If the whole string is palindrome, no cuts are needed, so minimum cuts = 0.

3. Final Answer:

   The minimum cuts will be 0 because the entire string is a palindrome -> Option A

4. Quick Check:

   Full palindrome means zero cuts [OK]

Hint: If whole string palindrome, minimum cuts = 0 [OK]

Common Mistakes:

• Assuming cuts needed even if full palindrome
• Splitting unnecessarily into halves
• Ignoring palindrome check for entire string