Palindrome Partitioning DP Minimum Cuts in DSA Go - 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 = min(cuts_left + cuts_right + 1)

↓

Store result in DP table

↓

Return min cuts for full string

We check substrings for palindrome, then use DP to find minimum cuts by trying all partitions and storing results.

Execution Sample

DSA Go

func minCut(s string) int {
 n := len(s)
 dp := make([]int, n)
 palindrome := make([][]bool, n)
 // fill palindrome and dp
 return dp[n-1]
}

This code finds the minimum cuts needed to partition a string into palindromes using DP.

Execution Table

Step	Operation	Substring Checked	Is Palindrome?	DP Array State	Min Cuts for substring
1	Initialize dp and palindrome arrays	-	-	[0,0,0,0,0]	-
2	Check substring s[0:0] = 'a'	a	Yes	[0,0,0,0,0]	0
3	Check substring s[0:1] = 'ab'	ab	No	[0,1,0,0,0]	1
4	Check substring s[1:1] = 'b'	b	Yes	[0,1,0,0,0]	0
5	Check substring s[0:2] = 'aba'	aba	Yes	[0,1,0,0,0]	0
6	Update dp[2] considering partitions	-	-	[0,1,0,0,0]	0
7	Check substring s[3:3] = 'c'	c	Yes	[0,1,0,0,0]	0
8	Check substring s[2:3] = 'ac'	ac	No	[0,1,0,1,0]	1
9	Check substring s[0:3] = 'abac'	abac	No	[0,1,0,1,0]	1
10	Update dp[3] considering partitions	-	-	[0,1,0,1,0]	1
11	Check substring s[4:4] = 'a'	a	Yes	[0,1,0,1,0]	0
12	Check substring s[3:4] = 'ca'	ca	No	[0,1,0,1,2]	1
13	Check substring s[2:4] = 'aca'	aca	Yes	[0,1,0,1,2]	0
14	Update dp[4] considering partitions	-	-	[0,1,0,1,1]	1
15	Final min cuts for full string s[0:4]	abaca	-	[0,1,0,1,1]	1

💡 All substrings checked and dp array filled; minimum cuts found for full string.

Variable Tracker

Variable	Start	After Step 2	After Step 5	After Step 10	After Step 15
dp	[0,0,0,0,0]	[0,0,0,0,0]	[0,1,0,0,0]	[0,1,0,1,0]	[0,1,0,1,1]
palindrome[0][0]	false	true	true	true	true
palindrome[0][2]	false	false	true	true	true
palindrome[2][4]	false	false	false	false	true

Key Moments - 3 Insights

Why does dp[2] become 0 even though substring s[0:2] has partitions?

Because substring 'aba' is a palindrome (step 5), so no cuts needed, dp[2] is set to 0 (step 6).

Why do we check smaller substrings first before bigger ones?

Because dp depends on results of smaller substrings; we must know if smaller parts are palindrome to compute bigger ones (see steps 2-5).

How does the algorithm decide the minimum cuts at each position?

It tries all partitions and picks the minimum cuts from dp values plus one cut between partitions (steps 6, 10, 14).

Visual Quiz - 3 Questions

Look at the execution table at step 5, is substring 'aba' a palindrome?

A Not checked yet

B No

C Yes

D Only partially palindrome

💡 Hint

Check 'Is Palindrome?' column at step 5 in execution_table.

At which step does dp[3] get updated to 1?

A Step 10

B Step 7

C Step 8

D Step 12

💡 Hint

Look at 'DP Array State' and 'Min Cuts' columns around step 10.

If substring s[2:4] was not palindrome, how would dp[4] change?

A dp[4] would be 1

B dp[4] would be greater than 1

C dp[4] would be 0

D dp[4] would be unchanged

💡 Hint

Check step 13 and 14 where palindrome status affects dp[4].

Concept Snapshot

Palindrome Partitioning Minimum Cuts:
- Use DP array dp[i] for min cuts up to i
- palindrome[i][j] marks if s[i..j] is palindrome
- dp[i] = min(dp[j] + 1) for all j < i if s[j+1..i] palindrome
- If s[0..i] palindrome, dp[i] = 0
- Result is dp[n-1]

Full Transcript

This visualization shows how to find the minimum cuts needed to partition a string into palindromes using dynamic programming. We start by checking all substrings for palindrome property and store results in a 2D array. Then, we use a 1D dp array where dp[i] represents the minimum cuts needed for substring s[0..i]. For each position, we try all partitions and update dp[i] with the minimum cuts found. The execution table traces each substring check and dp update step by step, showing how the dp array evolves. Key moments clarify why palindrome checks are done first and how dp values are computed. The visual quiz tests understanding of palindrome checks and dp updates. The concept snapshot summarizes the approach in concise 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 maximum length of a palindrome substring in s

C. The total number of palindrome substrings in s

D. The longest palindrome substring in s

Solution

1. Step 1: Understand the problem goal

   The problem asks for the fewest cuts needed to split the string into parts where each part is a palindrome.

2. Step 2: Differentiate from other palindrome problems

   It is not about longest palindrome or counting palindromes, but minimizing cuts for palindrome partitions.

3. Final Answer:

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

4. Quick Check:

   Minimum cuts = minimum partitions [OK]

Hint: Focus on cuts that split string into palindrome parts [OK]

Common Mistakes:

• Confusing minimum cuts with longest palindrome length
• Thinking problem counts palindrome substrings
• Assuming problem finds longest palindrome substring

2. Which of the following Go code snippets correctly initializes a 2D boolean slice isPalindrome for a string s of length n?

easy

A. isPalindrome := [][]bool{{false}}
for i := 0; i < n; i++ {
isPalindrome[i][i] = true
}

B. var isPalindrome [][]bool
isPalindrome = make([]bool, n*n)

C. isPalindrome := make([][]bool, n)
for i := range isPalindrome {
isPalindrome[i] = make([]bool, n)
}

D. isPalindrome := make([]bool, n)
for i := 0; i < n; i++ {
isPalindrome[i] = true
}

Solution

1. Step 1: Understand 2D slice initialization in Go

   To create a 2D slice, first allocate the outer slice, then allocate each inner slice separately.

2. Step 2: Check each option for correct 2D boolean slice creation

   isPalindrome := make([][]bool, n) for i := range isPalindrome { isPalindrome[i] = make([]bool, n) } correctly creates a slice of slices with length n and initializes each inner slice with length n.

3. Final Answer:

   isPalindrome := make([][]bool, n) for i := range isPalindrome { isPalindrome[i] = make([]bool, n) -> Option C

4. Quick Check:

   2D slice needs nested make calls [OK]

Hint: Use nested make calls for 2D slices in Go [OK]

Common Mistakes:

• Trying to create 2D slice with single make call
• Confusing slice of bool with slice of slices
• Not initializing inner slices before use

3. What is the output of the following Go code snippet that calculates minimum cuts for palindrome partitioning of s = "aab"?

func minCut(s string) int {
    n := len(s)
    isPalindrome := make([][]bool, n)
    for i := range isPalindrome {
        isPalindrome[i] = make([]bool, n)
    }
    for i := n - 1; i >= 0; i-- {
        for j := i; j < n; j++ {
            if s[i] == s[j] && (j - i < 2 || isPalindrome[i+1][j-1]) {
                isPalindrome[i][j] = true
            }
        }
    }
    cuts := make([]int, n)
    for i := 0; i < n; i++ {
        minCut := i
        for j := 0; j <= i; j++ {
            if isPalindrome[j][i] {
                if j == 0 {
                    minCut = 0
                } else if cuts[j-1]+1 < minCut {
                    minCut = cuts[j-1] + 1
                }
            }
        }
        cuts[i] = minCut
    }
    return cuts[n-1]
}

fmt.Println(minCut("aab"))

medium

A. 0

B. 1

C. 2

D. 3

Solution

1. Step 1: Identify palindromic substrings in "aab"

   "a", "a", "b" are palindromes; "aa" is palindrome; "aab" is not.

2. Step 2: Calculate minimum cuts using DP logic

   Minimum cuts needed: cut between "aa" and "b" -> 1 cut.

3. Final Answer:

   1 -> Option B

4. Quick Check:

   Minimum cuts for "aab" = 1 [OK]

Hint: Check palindrome substrings, then count cuts between them [OK]

Common Mistakes:

• Assuming no cuts needed for "aab"
• Counting cuts as number of palindrome substrings
• Misunderstanding DP array meaning

4. The following Go code attempts to compute minimum cuts for palindrome partitioning but has a bug. What is the error?

func minCut(s string) int {
    n := len(s)
    isPalindrome := make([][]bool, n)
    for i := range isPalindrome {
        isPalindrome[i] = make([]bool, n)
    }
    for i := 0; i < n; i++ {
        for j := i; j < n; j++ {
            if s[i] == s[j] && (j - i < 2 || isPalindrome[i+1][j-1]) {
                isPalindrome[i][j] = true
            }
        }
    }
    cuts := make([]int, n)
    for i := 0; i < n; i++ {
        cuts[i] = i
        for j := 0; j <= i; j++ {
            if isPalindrome[j][i] {
                if j == 0 {
                    cuts[i] = 0
                } else if cuts[j-1]+1 < cuts[i] {
                    cuts[i] = cuts[j-1] + 1
                }
            }
        }
    }
    return cuts[n-1]
}

medium

A. The function returns cuts[0] instead of cuts[n-1]

B. The cuts array is not initialized with correct default values

C. The condition s[i] == s[j] is incorrect for palindrome check

D. The palindrome check loop should run from i = n-1 down to 0, not 0 to n-1

Solution

1. Step 1: Analyze palindrome DP filling order

   Palindrome DP requires filling from end to start (i from n-1 down to 0) to use isPalindrome[i+1][j-1] correctly.

2. Step 2: Identify impact of incorrect loop order

   Looping i from 0 to n-1 causes isPalindrome[i+1][j-1] to be uninitialized or incorrect, leading to wrong results.

3. Final Answer:

   The palindrome check loop should run from i = n-1 down to 0, not 0 to n-1 -> Option D

4. Quick Check:

   Fill palindrome DP bottom-up from right [OK]

Hint: Fill palindrome DP from right to left for correct subproblems [OK]

Common Mistakes:

• Filling palindrome DP from left to right
• Ignoring dependency on isPalindrome[i+1][j-1]
• Incorrect initialization of cuts array

5. Given the string s = "ababbbabbababa", what is the minimum number of cuts needed to partition it into palindromic substrings using the DP minimum cuts approach?

hard

A. 4

B. 6

C. 5

D. 3

Solution

1. Step 1: Understand the string and palindrome partitions

   The string "ababbbabbababa" can be partitioned into palindromes with minimum cuts by checking all substrings using DP.

2. Step 2: Apply DP minimum cuts algorithm

   Using the palindrome DP table and cuts array, the minimum cuts calculated for this string is 4.

3. Final Answer:

   4 -> Option A

4. Quick Check:

   Minimum cuts for "ababbbabbababa" = 4 [OK]

Hint: Use DP palindrome table and cuts array to find minimum cuts [OK]

Common Mistakes:

• Underestimating cuts by ignoring some palindrome partitions
• Overcounting cuts by splitting unnecessarily
• Not using DP to optimize palindrome checks