Symmetric Substring Division — Problem Statement & Solution Guide

StringsHardDP on Strings

TimeO(n^2)

SpaceO(n)

Problem Description

Given a string s, split it into one or more contiguous substrings such that every resulting substring is a palindrome (reads identically forward and backward). Determine the smallest possible number of cuts required to achieve such a partition. A cut is placed between two adjacent characters; if the whole string is already a palindrome, zero cuts are needed. Return this minimum cut count.

DSA Pattern Breakdown

"Symmetric Substring Division"

hard

WHY DOES IT MATTER?

Palindrome partitioning with minimum cuts is a fundamental DP pattern that teaches how to decompose a global optimum into optimal sub‑structures, a skill transferable to many string and array partitioning challenges encountered in high‑frequency trading platforms and large‑scale data pipelines.

OPTIMIZATION CHALLENGE

The breakthrough is recognizing that you can expand palindromes around every center in O(n) total time per center and update the DP cut array on the fly, turning an O(n^3) naive DP into an O(n^2) solution with only O(n) auxiliary space.

REAL-WORLD CONNECTION

Think of a distributed log replication system where each log segment must be self‑consistent (palindromic) before being shipped to a replica; minimizing the number of segment boundaries (cuts) reduces network overhead and latency, mirroring the algorithm's goal of fewer cuts for efficiency.

During an interview, pre‑compute a boolean palindrome table only if the language's memory budget permits; otherwise, use center expansion with a single DP array to stay within linear space and still achieve the optimal time bound.

COMPLEXITY AT A GLANCE

⏱ Time:O(n^2)

💾 Space:O(n)

Core Theory — Why This Approach?

The problem of finding the minimum number of cuts to partition a string into palindromic substrings is a classic example of dynamic programming on strings. A naive solution would examine every possible cut configuration, leading to exponential time because each character can either be a cut or not, which quickly becomes infeasible for strings longer than 20 characters. The optimal paradigm leverages two observations: (1) the sub‑problem "minimum cuts for the prefix ending at index i" can be expressed in terms of earlier prefixes, and (2) palindrome checking can be pre‑computed or expanded on‑the‑fly in O(1) amortized time per pair of indices. By constructing a DP array cut[i] that stores the minimum cuts needed for the substring s[0..i] and simultaneously expanding around each center to discover all palindromes that end at i, we reduce the overall complexity to quadratic time while using only linear extra space.

Why this approach outperforms the brute force is twofold. First, dynamic programming eliminates redundant recomputation of overlapping sub‑problems; each prefix’s answer is computed once and reused. Second, palindrome expansion (or a pre‑computed palindrome table) transforms the costly O(n) palindrome verification per cut into O(1) checks, collapsing the exponential search space into a deterministic O(n^2) scan. This makes the solution scalable to strings of length up to 10^5 in practice, which is the typical constraint in coding interviews and competitive programming platforms.

Interview Questions on This Problem

Q1How would you modify the minimum cut algorithm to also return the actual palindrome partitioning, not just the cut count?

Maintain a predecessor array prev[i] that stores the index j where the last palindrome ending at i starts when cut[i] is updated. After filling the DP tables, backtrack from the end of the string using prev to reconstruct the substrings in reverse order, then reverse the list to obtain the partition.

Q2Can the minimum cut problem be solved in O(n) time using Manacher's algorithm? Explain the trade‑offs.

Manacher's algorithm can compute all palindrome radii in O(n) time, providing immediate knowledge of every palindrome centered at each position. However, integrating this information into the DP for minimum cuts still requires O(n) updates per center in the worst case, leading to O(n^2) overall. The benefit is reduced constant factors and linear space, but true O(n) total time is not achievable because the DP recurrence inherently depends on the number of palindrome endpoints.

Q3Why is the minimum cut problem considered a "partitioning" problem rather than a "subsequence" problem, and how does that affect the choice of algorithm?

Partitioning requires the substrings to be contiguous, preserving the original order without gaps, whereas subsequence problems allow skipping characters. This contiguity enables palindrome expansion from each center and DP on prefixes, while subsequence variants would need different techniques such as longest palindromic subsequence DP, which operates on two‑dimensional state spaces.

Examples

Example 1

Input

ababa

Output

Explanation: The entire string "ababa" reads the same from left to right and right to left, so no division is necessary. Minimum cuts = 0.

Example 2

Input

aab

Output

Explanation: Possible palindrome partitions: "a|a|b" (2 cuts) and "aa|b" (1 cut). The latter uses the palindrome "aa" followed by "b", requiring only one cut, which is optimal.

Example 3

Input

abcde

Output

Explanation: No two‑character substring forms a palindrome, so each character must stand alone. The partition is "a|b|c|d|e" which uses 4 cuts (length‑1). This is the minimum.

Example 4

Input

cddpd

Output

Explanation: Palindromic substrings are: "c", "d", "dd", "p", "d". The best partition is "c|dd|p|d", requiring cuts after the first, third, and fourth characters – a total of 3 cuts. No arrangement can use fewer cuts because there is no palindrome that spans more than the "dd" segment.

Constraints

1 <= s.length <= 5000
s consists only of lowercase English letters ('a'–'z')
The algorithm should run in O(n^2) time or better and use O(n^2) or O(n) auxiliary space.

Optimal Approach & Strategy

Use dynamic programming with palindrome expansion: for each center expand outward, update a 1‑D cut array in O(1) per palindrome, achieving O(n^2) time and O(n) space.

Brute Force Approach

Enumerate every possible subset of cut positions, check each resulting substring for palindrome property, and keep the minimum cut count; this leads to exponential time complexity.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n^2)

function symmetricSubstringDivision(s) {
   let n = s.length;
   let dp = Array(n).fill(0).map(() => Array(n).fill(0));
   for (let i = 0; i < n; i++) {
      dp[i][i] = 1;
      if (i + 1 < n && s[i] === s[i + 1]) {
         dp[i][i + 1] = 1;
      }
   }
   for (let length = 3; length <= n; length++) {
      for (let i = 0; i <= n - length; i++) {
         let j = i + length - 1;
         if (s[i] === s[j] && (length === 3 || dp[i + 1][j - 1] === length - 2)) {
            dp[i][j] = length;
         }
      }
   }
   let max = 0;
   for (let i = 0; i < n; i++) {
      for (let j = i; j < n; j++) {
         max = Math.max(max, dp[i][j]);
      }
   }
   return max;
}

class Solution {
   public:
      int symmetricSubstringDivision(string s) {
         int n = s.length();
         vector<vector<int>> dp(n, vector<int>(n, 0));
         for (int i = 0; i < n; i++) {
            dp[i][i] = 1;
            if (i + 1 < n && s[i] == s[i + 1]) {
               dp[i][i + 1] = 1;
            }
         }
         for (int length = 3; length <= n; length++) {
            for (int i = 0; i <= n - length; i++) {
               int j = i + length - 1;
               if (s[i] == s[j] && (length == 3 || dp[i + 1][j - 1] == length - 2)) {
                  dp[i][j] = length;
               }
            }
         }
         int max = 0;
         for (int i = 0; i < n; i++) {
            for (int j = i; j < n; j++) {
               max = max(dp[i][j], max);
            }
         }
         return max;
      }
   };

class Solution {
   public int symmetricSubstringDivision(String s) {
      int n = s.length();
      int[][] dp = new int[n][n];
      for (int i = 0; i < n; i++) {
         dp[i][i] = 1;
         if (i + 1 < n && s.charAt(i) == s.charAt(i + 1)) {
            dp[i][i + 1] = 1;
         }
      }
      for (int length = 3; length <= n; length++) {
         for (int i = 0; i <= n - length; i++) {
            int j = i + length - 1;
            if (s.charAt(i) == s.charAt(j) && (length == 3 || dp[i + 1][j - 1] == length - 2)) {
               dp[i][j] = length;
            }
         }
      }
      int max = 0;
      for (int i = 0; i < n; i++) {
         for (int j = i; j < n; j++) {
            max = Math.max(max, dp[i][j]);
         }
      }
      return max;
   }
}

def symmetric_substring_division(s):
   n = len(s)
   dp = [[0] * n for _ in range(n)]
   for i in range(n):
      dp[i][i] = 1
      if i + 1 < n and s[i] == s[i + 1]:
         dp[i][i + 1] = 1
   for length in range(3, n + 1):
      for i in range(n - length + 1):
         j = i + length - 1
         if s[i] == s[j] and (length == 3 or dp[i + 1][j - 1] == length - 2):
            dp[i][j] = length
   max_divisions = 0
   for i in range(n):
      for j in range(i, n):
         max_divisions = max(max_divisions, dp[i][j])
   return max_divisions

function symmetricSubstringDivision(s) {
   let n = s.length;
   let dp = Array(n).fill(0).map(() => Array(n).fill(0));
   for (let i = 0; i < n; i++) {
      dp[i][i] = 1;
      if (i + 1 < n && s[i] === s[i + 1]) {
         dp[i][i + 1] = 1;
      }
   }
   for (let length = 3; length <= n; length++) {
      for (let i = 0; i <= n - length; i++) {
         let j = i + length - 1;
         if (s[i] === s[j] && (length === 3 || dp[i + 1][j - 1] === length - 2)) {
            dp[i][j] = length;
         }
      }
   }
   let max = 0;
   for (let i = 0; i < n; i++) {
      for (let j = i; j < n; j++) {
         max = Math.max(max, dp[i][j]);
      }
   }
   return max;
}

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Symmetric Substring Division Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Input

Output

Constraints

More Strings

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Symmetric Substring Division — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Solve in Interative Editor