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DSA Javascriptprogramming~5 mins

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

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Time Complexity: Palindrome Partitioning DP Minimum Cuts
O(n³)
Understanding Time Complexity

We want to know how the time needed grows when we find the minimum cuts to split a string into palindromes.

How does the work increase as the string gets longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.


function minCut(s) {
  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 the string into palindrome substrings using dynamic programming.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

  • Primary operation: Nested loops checking all substring pairs (start to end) and palindrome checks.
  • How many times: Outer loop runs n times; inner loop runs up to n times for each outer loop, so roughly n x 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
10About 100 checks
100About 10,000 checks
1000About 1,000,000 checks

Pattern observation: Doubling the input size roughly quadruples the work because of the nested loops.

Final Time Complexity

Time Complexity: O(n³)

This means the time needed grows roughly with the cube of the string length.

Common Mistake

[X] Wrong: "The algorithm runs in linear time because it just scans the string twice."

[OK] Correct: The nested loops check all substring pairs, which means the work grows much faster than just scanning once or twice.

Interview Connect

Understanding this time complexity helps you explain how dynamic programming reduces repeated work and why nested loops matter in string problems.

Self-Check

"What if we precompute palindrome substrings separately before the main loop? How would that affect the time complexity?"

Practice

(1/5)
1. What does the term minimum cuts mean in the context of palindrome partitioning?
easy
A. The number of characters to remove to make the string a palindrome
B. The longest palindrome substring in the string
C. The total number of palindromic substrings in the string
D. The smallest number of cuts needed to split the string into palindromic substrings

Solution

  1. Step 1: Understand palindrome partitioning

    Palindrome partitioning splits a string into substrings where each substring is a palindrome.
  2. Step 2: Define minimum cuts

    Minimum cuts means the least number of cuts needed to divide the string so that every part is a palindrome.
  3. Final Answer:

    The smallest number of cuts needed to split the string into palindromic substrings -> Option D
  4. Quick Check:

    Minimum cuts = smallest cuts for palindrome parts [OK]
Hint: Minimum cuts split string into palindrome parts with fewest cuts [OK]
Common Mistakes:
  • Confusing minimum cuts with longest palindrome substring
  • Thinking minimum cuts count characters removed
  • Mixing total palindrome substrings with minimum cuts
2. Which of the following is the correct way to initialize a 2D boolean array isPalindrome for a string s of length n in JavaScript?
easy
A. const isPalindrome = Array.from({length: n}, () => Array(n).fill(false));
B. const isPalindrome = Array(n).fill(Array(n).fill(false));
C. const isPalindrome = new Array(n).fill(new Array(n).fill(false));
D. const isPalindrome = Array(n).map(Array(n).fill(false));

Solution

  1. Step 1: Understand 2D array initialization in JavaScript

    Using Array.from with a mapping function creates independent inner arrays.
  2. Step 2: Compare options

    const isPalindrome = Array.from({length: n}, () => Array(n).fill(false)); correctly creates separate inner arrays filled with false. Options A and C create references to the same inner array, causing bugs. const isPalindrome = Array(n).map(() => Array(n).fill(false)); fails because map on uninitialized array does not work as expected.
  3. Final Answer:

    const isPalindrome = Array.from({length: n}, () => Array(n).fill(false)); -> Option A
  4. Quick Check:

    Use Array.from with function for independent inner arrays [OK]
Hint: Use Array.from with function to avoid shared references [OK]
Common Mistakes:
  • Using fill with array directly causing shared references
  • Using map on uninitialized arrays
  • Not initializing inner arrays separately
3. What is the output of the following code snippet for s = "aab"?
function minCut(s) {
  const n = s.length;
  const isPalindrome = Array.from({length: n}, () => Array(n).fill(false));
  for (let i = 0; i < n; i++) {
    isPalindrome[i][i] = true;
  }
  for (let length = 2; length <= n; length++) {
    for (let start = 0; start <= n - length; start++) {
      let end = start + length - 1;
      if (s[start] === s[end]) {
        if (length === 2) {
          isPalindrome[start][end] = true;
        } else {
          isPalindrome[start][end] = isPalindrome[start + 1][end - 1];
        }
      }
    }
  }
  const cuts = Array(n).fill(0);
  for (let i = 0; i < n; i++) {
    let minCut = i;
    for (let j = 0; j <= i; j++) {
      if (isPalindrome[j][i]) {
        minCut = j === 0 ? 0 : Math.min(minCut, cuts[j - 1] + 1);
      }
    }
    cuts[i] = minCut;
  }
  return cuts[n - 1];
}
console.log(minCut(s));
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

    Minimum cuts needed: split "aab" as "aa" | "b" with 1 cut.
  3. Final Answer:

    1 -> Option B
  4. Quick Check:

    Minimum cuts for "aab" = 1 [OK]
Hint: Check palindrome substrings, then find fewest cuts [OK]
Common Mistakes:
  • Assuming no cuts needed for non-palindrome string
  • Counting total palindromic substrings instead of cuts
  • Misindexing palindrome checks
4. The following code attempts to compute minimum cuts for palindrome partitioning but has a bug. What is the error?
function minCut(s) {
  const n = s.length;
  const isPalindrome = Array(n).fill(Array(n).fill(false));
  for (let i = 0; i < n; i++) {
    isPalindrome[i][i] = true;
  }
  // ... rest of palindrome logic ...
  return 0;
}
medium
A. The function does not return the minimum cuts value
B. Loop for setting isPalindrome[i][i] should start from 1, not 0
C. isPalindrome initialization shares inner arrays causing incorrect updates
D. The string length n is not calculated correctly

Solution

  1. Step 1: Analyze isPalindrome array initialization

    Using fill with an array creates references to the same inner array, so updating one row updates all rows.
  2. Step 2: Understand impact on palindrome checks

    This causes incorrect palindrome marking because all rows share the same inner array, corrupting data.
  3. Final Answer:

    isPalindrome initialization shares inner arrays causing incorrect updates -> Option C
  4. Quick Check:

    Shared inner arrays cause bugs in 2D boolean array [OK]
Hint: Avoid fill with arrays; use Array.from with function [OK]
Common Mistakes:
  • Ignoring shared reference problem in 2D arrays
  • Thinking loop start index causes bug
  • Overlooking missing return value as main error
5. Given the string s = "ababbbabbababa", what is the minimum number of cuts needed to partition it into palindromic substrings using dynamic programming?
hard
A. 3
B. 5
C. 4
D. 6

Solution

  1. Step 1: Understand palindrome partitions for "ababbbabbababa"

    This string can be partitioned into palindromes with minimum cuts by checking all substrings and using DP to find minimal cuts.
  2. Step 2: Use known result or DP approach

    The minimum cuts for this string is 3, as it can be split into palindromic parts like "a|babbbab|b|ababa" or similar partitions.
  3. Final Answer:

    3 -> Option A
  4. Quick Check:

    Minimum cuts for complex palindrome string = 3 [OK]
Hint: Use DP palindrome checks and cuts array to find minimum cuts [OK]
Common Mistakes:
  • Assuming minimum cuts equals number of palindromic substrings
  • Miscounting cuts by off-by-one errors
  • Not considering overlapping palindrome substrings
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