Palindrome Partitioning DP Minimum Cuts in DSA Rust - Execution Trace

Concept Flow - Palindrome Partitioning DP Minimum Cuts

Start with full string

↓

Check all substrings palindrome?

↓

If palindrome: cuts=0

↓

Else: Try all partitions

↓

Calculate min cuts using DP

↓

Return min cuts for full string

We check substrings for palindrome, then use DP to find minimum cuts needed to partition the string into palindromes.

Execution Sample

DSA Rust

fn is_palindrome(s: &str, mut l: usize, mut r: usize) -> bool {
    while l < r {
        if s.as_bytes()[l] != s.as_bytes()[r] {
            return false;
        }
        l += 1;
        r -= 1;
    }
    true
}

fn min_cut(s: &str) -> usize {
    let n = s.len();
    let mut cuts = vec![0; n + 1];
    for i in 1..=n {
        cuts[i] = cuts[i - 1] + 1;
        if is_palindrome(s, 0, i - 1) {
            cuts[i] = 0;
        } else {
            for j in 1..i {
                if is_palindrome(s, j, i - 1) {
                    cuts[i] = std::cmp::min(cuts[i], cuts[j] + 1);
                }
            }
        }
    }
    cuts[n] - 1
}

This code calculates the minimum cuts needed to partition the string into palindromes.

Execution Table

Step	Operation	Substring Checked	Palindrome?	Cuts Array State	Visual State
1	Initialize cuts array	-	-	[0, 0, 0, 0, 0, 0]	cuts: [0,0,0,0,0,0] (length+1)
2	Check substring 'a'	'a'	Yes	[0, 0, 0, 0, 0, 0]	No cut needed for 'a'
3	Check substring 'ab'	'ab'	No	[0, 0, 0, 0, 0, 0]	Not palindrome
4	Update cuts for 'ab'	'ab'	-	[0, 0, 1, 0, 0, 0]	cuts[2] = min cuts for 'ab' = 1
5	Check substring 'aba'	'aba'	Yes	[0, 0, 1, 0, 0, 0]	Palindrome 'aba' no cut
6	Update cuts for 'aba'	'aba'	-	[0, 0, 1, 0, 0, 0]	cuts[3] = 0 because 'aba' palindrome
7	Check substring 'abab'	'abab'	No	[0, 0, 1, 0, 0, 0]	Not palindrome
8	Update cuts for 'abab'	'abab'	-	[0, 0, 1, 0, 1, 0]	cuts[4] = 1
9	Check substring 'ababa'	'ababa'	Yes	[0, 0, 1, 0, 1, 0]	Palindrome 'ababa' no cut
10	Update cuts for 'ababa'	'ababa'	-	[0, 0, 1, 0, 1, 0]	cuts[5] = 0
11	Final min cuts	full string	-	[0, 0, 1, 0, 1, 0]	Minimum cuts = 0
12	Exit	-	-	-	All substrings checked, min cuts found

💡 All substrings checked and minimum cuts array finalized.

Variable Tracker

Variable	Start	After Step 4	After Step 6	After Step 8	After Step 10	Final
cuts	[0,0,0,0,0,0]	[0,0,1,0,0,0]	[0,0,1,0,0,0]	[0,0,1,0,1,0]	[0,0,1,0,1,0]	[0,0,1,0,1,0]
substring	-	'ab'	'aba'	'abab'	'ababa'	full string

Key Moments - 3 Insights

Why do we initialize cuts array with zeros and length n+1?

The cuts array stores minimum cuts needed for substrings up to each index. Length n+1 allows indexing for empty substring at start. Initially zero means no cuts assumed.

Why is cuts[3] zero when substring 'aba' is palindrome?

Because 'aba' is palindrome, no cuts are needed to partition it. So cuts[3] remains zero as shown in step 6 of execution_table.

Why do we update cuts only when substring is not palindrome?

If substring is palindrome, cuts needed is zero. If not, we try all partitions and update cuts with minimum cuts found, as in steps 4 and 8.

Visual Quiz - 3 Questions

Look at the execution_table at step 4, what is the cuts array state?

A [0, 0, 1, 0, 0, 0]

B [0, 1, 0, 0, 0, 0]

C [1, 0, 0, 0, 0, 0]

D [0, 0, 0, 1, 0, 0]

💡 Hint

Refer to the cuts array state column at step 4 in execution_table.

At which step does the substring 'ababa' get recognized as palindrome?

A Step 7

B Step 9

C Step 5

D Step 3

💡 Hint

Check the substring checked and palindrome? columns in execution_table.

If the substring 'abab' was palindrome, how would cuts[4] change at step 8?

A cuts[4] would be 2

B cuts[4] would remain 1

C cuts[4] would be 0

D cuts[4] would be unchanged from cuts[3]

💡 Hint

If substring is palindrome, cuts needed is zero as shown in steps 5 and 9.

Concept Snapshot

Palindrome Partitioning Minimum Cuts:
- Use DP to find min cuts to split string into palindromes.
- cuts[i] = min cuts for substring s[0..i-1].
- If s[j..i-1] palindrome, cuts[i] = min(cuts[i], cuts[j] + 1).
- Initialize cuts with max cuts (worst case).
- Result is cuts[n] - 1 or cuts[n] depending on indexing.

Full Transcript

This visualization shows how to find the minimum cuts needed to partition a string into palindromes using dynamic programming. We start by initializing an array to store minimum cuts for substrings. We check each substring if it is a palindrome. If yes, no cuts are needed. If not, we try all partitions and update the minimum cuts accordingly. The execution table tracks each step, substring checked, palindrome status, and the cuts array state. Key moments clarify why cuts array is initialized with zeros and why palindrome substrings need zero cuts. The visual quiz tests understanding of cuts array updates and palindrome recognition steps.

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