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

Time Complexity: Palindrome Partitioning DP Minimum Cuts

O(n²)

Understanding Time Complexity

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

Specifically, how the number of steps grows when checking all possible partitions and palindromes.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function minCut(s: string): number {
  const n = s.length;
  const dp = new Array(n).fill(0);
  const palindrome = Array.from({ length: n }, () => new Array(n).fill(false));

  for (let end = 0; end < n; end++) {
    let minCuts = end;
    for (let start = 0; start <= end; start++) {
      if (s[start] === s[end] && (end - start < 2 || palindrome[start + 1][end - 1])) {
        palindrome[start][end] = true;
        minCuts = start === 0 ? 0 : Math.min(minCuts, dp[start - 1] + 1);
      }
    }
    dp[end] = minCuts;
  }
  return dp[n - 1];
}
    

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops checking all substring pairs (start, end) and palindrome checks.
• How many times: Outer loop runs n times; inner loop runs up to n times for each outer loop, totaling about n×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: The work grows roughly by the square of the input size, so doubling the input makes the work about four times bigger.

Final Time Complexity

Time Complexity: O(n²)

This means the time needed grows roughly with the square of the string length, because we check all substring pairs.

Common Mistake

[X] Wrong: "Since we only loop once over the string, the time is O(n)."

[OK] Correct: We actually have a nested loop checking all substrings, so the time grows much faster than just one pass.

Interview Connect

Understanding this time complexity helps you explain how dynamic programming reduces repeated work when checking palindromes, a common pattern in coding challenges.

Self-Check

"What if we used a simple recursive approach without memoization? 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 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: 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: Focus on splitting string into palindrome parts with fewest cuts [OK]

Common Mistakes:

• Confusing minimum cuts with longest palindrome substring
• Counting total palindrome substrings instead of cuts
• Thinking about maximum palindrome length

2. Which of the following TypeScript function signatures correctly represents a function to compute minimum palindrome partition cuts for a string s?

easy

A. function minCut(s: string): string[]

B. function minCut(s: number): string

C. function minCut(s: string): number

D. function minCut(s: boolean): number

Solution

1. Step 1: Identify input and output types

   Input is a string s, output is a number representing minimum cuts.

2. Step 2: Match correct signature

   Only function minCut(s: string): number matches input string and numeric output.

3. Final Answer:

   function minCut(s: string): number -> Option C

4. Quick Check:

   Input string, output number for cuts [OK]

Hint: Input string, output number for minimum cuts [OK]

Common Mistakes:

• Using wrong input type like number or boolean
• Returning array or string instead of number
• Mixing input and output types

3. What is the output of the following TypeScript code for input "aab"?

function minCut(s: string): number {
  const n = s.length;
  const dp = new Array(n).fill(0);
  const palindrome = Array.from({ length: n }, () => new Array(n).fill(false));

  for (let i = 0; i < n; i++) {
    let minCuts = i;
    for (let j = 0; j <= i; j++) {
      if (s[j] === s[i] && (i - j < 2 || palindrome[j + 1][i - 1])) {
        palindrome[j][i] = true;
        minCuts = j === 0 ? 0 : Math.min(minCuts, dp[j - 1] + 1);
      }
    }
    dp[i] = minCuts;
  }
  return dp[n - 1];
}
console.log(minCut("aab"));

medium

A. 0

B. 1

C. 2

D. 3

Solution

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

   Substrings: "a" (palindrome), "aa" (palindrome), "b" (palindrome). Minimum cuts for "aab" is 1 (cut between "aa" and "b").

2. Step 2: Check dp array updates

   dp[0]=0 ("a"), dp[1]=0 ("aa"), dp[2]=1 (cut before "b").

3. Final Answer:

   1 -> Option B

4. Quick Check:

   Minimum cuts for "aab" = 1 [OK]

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

Common Mistakes:

• Assuming no cuts needed for "aab"
• Counting total palindromes instead of cuts
• Incorrect dp index updates

4. Identify the bug in this TypeScript code snippet for minimum palindrome partition cuts:

function minCut(s: string): number {
  const n = s.length;
  const dp = new Array(n).fill(0);
  const palindrome = Array.from({ length: n }, () => new Array(n).fill(false));

  for (let i = 0; i < n; i++) {
    dp[i] = i;
    for (let j = 0; j <= i; j++) {
      if (s[j] === s[i] && (i - j < 2 || palindrome[j + 1][i - 1])) {
        palindrome[j][i] = true;
        dp[i] = Math.min(dp[i], dp[j - 1] + 1);
      }
    }
  }
  return dp[n - 1];
}

medium

A. dp[i] should be initialized to 0 instead of i

B. Inner loop should run from j = 0 to j <= i

C. palindrome array initialization is incorrect

D. dp[j] should be dp[j - 1] when updating dp[i]

Solution

1. Step 1: Analyze dp update logic

   When substring s[j..i] is palindrome, dp[i] should consider dp[j-1] + 1 cuts, not dp[j] + 1.

2. Step 2: Identify correct index for dp

   dp[j-1] represents cuts before substring start; dp[j] is incorrect and causes off-by-one error.

3. Final Answer:

   dp[j] should be dp[j - 1] when updating dp[i] -> Option D

4. Quick Check:

   Use dp[j-1] for correct minimum cuts [OK]

Hint: Use dp[j-1] when substring starts at j to avoid off-by-one error [OK]

Common Mistakes:

• Using dp[j] instead of dp[j-1]
• Incorrect loop bounds causing missed palindromes
• Wrong initialization of dp array

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

hard

A. 2

B. 1

C. 3

D. 0

Solution

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

   Palindromes include "a", "b", "cc", "bcb", "c". Partitioning as "a" | "bccb" | "c" is possible with 2 cuts.

2. Step 2: Calculate minimum cuts

   Minimum cuts = 2 because splitting into "a" | "bccb" | "c" yields all palindromes with 2 cuts.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Minimum cuts for "abccbc" = 2 [OK]

Hint: Look for largest palindrome substrings to minimize cuts [OK]

Common Mistakes:

• Assuming 0 cuts because string is not fully palindrome
• Counting each character as separate palindrome causing more cuts
• Missing palindrome "bccb" in middle