Palindrome Partitioning DP Minimum Cuts in DSA C - Execution Trace

Concept Flow - Palindrome Partitioning DP Minimum Cuts

Start with full string

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Check all substrings palindrome?

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Build palindrome table

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Initialize cuts array

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For each index i

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For each j <= i

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If substring j to i is palindrome

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Update cuts[i

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Repeat until i reaches end

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Return cuts for full string - 1

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Done

We first find all palindrome substrings, then use that info to find minimum cuts needed by checking all partitions.

Execution Sample

DSA C

int minCut(char* s) {
  int n = strlen(s);
  bool pal[n][n];
  int cuts[n];
  // Build palindrome table
  // Compute cuts
  return cuts[n-1];
}

This code finds minimum cuts to partition string s into palindromes using DP.

Execution Table

Step	Operation	Substring Checked	Palindrome Table Update	Cuts Array Update	Visual State
1	Initialize palindrome table	-	pal[i][i] = true for all i	cuts[i] = i for all i	pal: diagonal true, cuts: [0,1,2,3,4]
2	Check substring length 2	s[0..1] = 'ab'	pal[0][1] = false	cuts unchanged	pal: pal[0][1]=false
3	Check substring length 2	s[1..2] = 'bb'	pal[1][2] = true	cuts[2] = min(cuts[2], cuts[0]+1) = min(2,0+1)=1	cuts: [0,1,1,3,4]
4	Check substring length 3	s[0..2] = 'abb'	pal[0][2] = false	cuts unchanged	pal: pal[0][2]=false
5	Check substring length 3	s[1..3] = 'bba'	pal[1][3] = false	cuts unchanged	pal: pal[1][3]=false
6	Check substring length 4	s[0..3] = 'abba'	pal[0][3] = true	cuts[3] = min(cuts[3], (j == 0 ? 0 : cuts[j-1]) + 1) = 0 (since cuts[-1] treated as -1)	cuts: [0,1,1,0,4]
7	Check substring length 5	s[0..4] = 'abbac'	pal[0][4] = false	cuts unchanged	pal: pal[0][4]=false
8	Check substring length 2	s[3..4] = 'ac'	pal[3][4] = false	cuts unchanged	pal: pal[3][4]=false
9	Check substring length 1	s[4..4] = 'c'	pal[4][4] = true	cuts[4] = min(cuts[4], cuts[3]+1) = min(4,0+1)=1	cuts: [0,1,1,0,1]
10	End	-	-	-	Minimum cuts = cuts[n-1] = 1

💡 All substrings checked, cuts array finalized with minimum cuts for full string.

Variable Tracker

Variable	Start	After Step 1	After Step 3	After Step 6	After Step 9	Final
pal[0][0]	false	true	true	true	true	true
pal[1][2]	false	false	true	true	true	true
pal[0][3]	false	false	false	true	true	true
cuts[0]	0	0	0	0	0	0
cuts[2]	2	2	1	1	1	1
cuts[3]	3	3	3	0	0	0
cuts[4]	4	4	4	4	1	1

Key Moments - 3 Insights

Why do we initialize cuts[i] = i at the start?

Because in the worst case, each character is a palindrome itself, so maximum cuts needed is i (cut before each character). See step 1 in execution_table.

How do we handle the case when j=0 in cuts update?

When j=0 and substring s[0..i] is palindrome, cuts[i] can be 0 because no cut is needed before the first character. This is shown in step 6 where cuts[3] becomes 0.

Why do we check palindrome substrings before updating cuts?

Because cuts depend on palindrome partitions. We only update cuts if substring s[j..i] is palindrome. This dependency is clear in steps 3 and 6.

Visual Quiz - 3 Questions

Look at the execution_table at step 3, what is the value of cuts[2] after update?

A 2

B 0

C 1

D 3

💡 Hint

Refer to cuts array update column in step 3 of execution_table.

At which step does cuts[3] become 0?

A Step 6

B Step 4

C Step 5

D Step 7

💡 Hint

Check the cuts array update and palindrome table update columns in execution_table.

If substring s[1..2] was not palindrome, how would cuts[2] change at step 3?

A cuts[2] would become 0

B cuts[2] would remain 2

C cuts[2] would become 1

D cuts[2] would become 3

💡 Hint

Look at how cuts[2] updates only if palindrome is true in step 3.

Concept Snapshot

Palindrome Partitioning Minimum Cuts:
- Use DP to find all palindrome substrings.
- Initialize cuts[i] = i (max cuts).
- For each i, check all j <= i:
  if s[j..i] palindrome, cuts[i] = min(cuts[i], cuts[j-1]+1).
- Result is cuts[n-1].
- Handles no-cut case when substring from start is palindrome.

Full Transcript

This visualization shows how to find the minimum cuts needed to partition a string into palindromes using dynamic programming. First, we build a table to mark all palindrome substrings. Then, we initialize an array cuts where cuts[i] represents the minimum cuts needed for substring s[0..i]. We start with the worst case where cuts[i] = i. For each index i, we check all substrings ending at i. If substring s[j..i] is palindrome, we update cuts[i] to the minimum of its current value and cuts[j-1] + 1. Special case is when j=0, meaning substring s[0..i] is palindrome, so cuts[i] can be zero. The final answer is cuts[n-1]. The execution table traces these steps with substring checks, palindrome table updates, and cuts array updates. Key moments clarify why initialization is done and how updates depend on palindrome checks. The quiz tests understanding of cuts updates and palindrome checks.