Palindrome Partitioning DP Minimum Cuts in DSA Go

Mental Model

We want to split a string into the fewest parts so that each part reads the same forwards and backwards.

Analogy: Imagine cutting a ribbon into pieces where each piece is a perfect mirror image of itself. We want to make as few cuts as possible.

s: a b a c d c b a
cuts: [0] [1] [2] [3] [4] [5] [6] [7] [8]
palindrome check matrix:
  T F T F F F F T
  F T F F F F T F
  T F T F F T F T
  F F F T F F F F
  F F F F T F F F
  F F T F F T F F
  F T F F F F T F
  T F T F F F F T

Dry Run Walkthrough

Input: string: "abacdcba"

Goal: Find the minimum number of cuts needed to split the string so every part is a palindrome

Step 1: Check all single characters as palindromes

palindrome matrix diagonal set to true: a(0), b(1), a(2), c(3), d(4), c(5), b(6), a(7)

Why: Every single character is a palindrome by itself

Step 2: Check substrings of length 2 and mark palindromes

palindrome matrix updated: 'ab'(0-1) false, 'ba'(1-2) false, 'ac'(2-3) false, 'cd'(3-4) false, 'dc'(4-5) false, 'cb'(5-6) false, 'ba'(6-7) false

Why: Two characters are palindrome only if they are equal

Step 3: Check substrings length 3 and more using DP relation

palindrome matrix updated: 'aba'(0-2) true, 'cdc'(3-5) true, 'bcd'(4-6) false, 'dcb'(5-7) false

Why: A substring is palindrome if first and last chars match and inner substring is palindrome

Step 4: Calculate minimum cuts for each prefix using palindrome info

cuts array: cuts[0]=0, cuts[1]=1, cuts[2]=0, cuts[3]=1, cuts[4]=2, cuts[5]=2, cuts[6]=1, cuts[7]=0

Why: If substring from start to i is palindrome, cuts=0 else min cuts from previous partitions +1

Result:

Minimum cuts needed: 0
Final cuts array: 0 1 0 1 2 2 1 0

Annotated Code

DSA Go

package main

import (
	"fmt"
)

func minCut(s string) int {
	n := len(s)
	if n == 0 {
		return 0
	}

	// palindrome[i][j] will be true if s[i..j] is palindrome
	palindrome := make([][]bool, n)
	for i := range palindrome {
		palindrome[i] = make([]bool, n)
	}

	// cuts[i] = minimum cuts needed for s[0..i]
	cuts := make([]int, n)

	for i := 0; i < n; i++ {
		minCuts := i // max cuts possible
		for j := 0; j <= i; j++ {
			if s[j] == s[i] && (i-j < 2 || palindrome[j+1][i-1]) {
				palindrome[j][i] = true
				if j == 0 {
					minCuts = 0
				} else if cuts[j-1]+1 < minCuts {
					minCuts = cuts[j-1] + 1
				}
			}
		}
		cuts[i] = minCuts
	}

	return cuts[n-1]
}

func main() {
	s := "abacdcba"
	fmt.Println(minCut(s))
}

for i := 0; i < n; i++ {

iterate over end index of substring

for j := 0; j <= i; j++ {

iterate over start index of substring

if s[j] == s[i] && (i-j < 2 || palindrome[j+1][i-1]) {

check if substring s[j..i] is palindrome

palindrome[j][i] = true

mark substring as palindrome

if j == 0 { minCuts = 0 } else if cuts[j-1]+1 < minCuts { minCuts = cuts[j-1] + 1 }

update minimum cuts needed for s[0..i]

cuts[i] = minCuts

store minimum cuts for prefix ending at i

Output

0

Complexity Analysis

Time: O(n^2) because we check all substrings and update cuts for each prefix

Space: O(n^2) for palindrome matrix plus O(n) for cuts array

vs Alternative: Naive approach tries all partitions recursively with exponential time; DP reduces it to polynomial time

Edge Cases

empty string

returns 0 cuts since no partitions needed

DSA Go

if n == 0 { return 0 }

string with all same characters like "aaaa"

returns 0 cuts because whole string is palindrome

DSA Go

if j == 0 { minCuts = 0 }

string with no palindrome longer than 1 like "abc"

returns n-1 cuts because each character is separate palindrome

DSA Go

minCuts := i

When to Use This Pattern

When asked to partition a string into palindromes with minimum cuts, use DP with palindrome substring checks to optimize.

Common Mistakes

Mistake: Not checking the inner substring palindrome condition causing wrong palindrome detection
Fix: Add condition (i-j < 2 || palindrome[j+1][i-1]) to ensure inner substring is palindrome

Mistake: Initializing cuts array incorrectly leading to wrong minimum cuts
Fix: Initialize minCuts to i (max cuts) and update only when better partition found

Summary

Finds minimum cuts to split string into palindrome substrings.

Use when you need to partition strings into palindromes with fewest cuts.

Key insight: use DP to check palindrome substrings and track minimum cuts for prefixes.

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