Minimum Segment Cost — Problem Statement & Solution Guide

StringsEasyDP on Strings

TimeO(n^2)

SpaceO(n^2)

Problem Description

Given a non‑empty string s consisting of lowercase English letters, partition s into one or more contiguous substrings such that every substring is a palindrome. 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.

Input: a single line containing the string s.

Output: a single integer representing the minimum number of cuts needed to split s into palindromic pieces.

DSA Pattern Breakdown

"Minimum Segment Cost"

easy

WHY DOES IT MATTER?

Palindrome partitioning exemplifies the "optimal substructure + overlapping subproblems" pattern, a cornerstone of dynamic programming. Mastering this pattern equips engineers to tackle a wide range of segmentation, cutting, and grouping problems that appear in text processing, bioinformatics, and compiler design.

OPTIMIZATION CHALLENGE

The breakthrough is recognizing that palindrome checks can be pre‑computed in quadratic time, turning an otherwise exponential recursion into a linear scan over previously solved sub‑problems. This reduces the search from 2^n possibilities to O(n^2) deterministic work.

REAL-WORLD CONNECTION

Think of a distributed log replication system where you must split a continuous stream into chunks that can be verified independently (e.g., via checksums). Minimizing the number of splits while ensuring each chunk is self‑consistent mirrors the palindrome‑cut problem.

During an interview, compute the palindrome table first; it isolates the combinatorial explosion. Then focus on the simple one‑dimensional DP for cuts—this separation keeps your code clean and avoids tangled nested loops.

COMPLEXITY AT A GLANCE

⏱ Time:O(n^2)

💾 Space:O(n^2)

Core Theory — Why This Approach?

The problem of minimizing cuts for palindrome partitioning is a classic dynamic‑programming challenge. A naive recursive solution explores every possible cut position, leading to an exponential O(2^n) search space because each character can either be a cut or not, which quickly becomes infeasible for strings longer than 20‑30 characters. The optimal paradigm leverages two observations: (1) the sub‑problem "minimum cuts for prefix s[0..i]" can be expressed in terms of earlier prefixes, and (2) whether a substring s[j..i] is a palindrome can be pre‑computed in O(1) after an O(n^2) preprocessing step. By filling a palindrome table P where P[j][i] is true if s[j..i] reads the same forward and backward, we can compute a cuts array C where C[i] = 0 if P[0][i] is true, otherwise C[i] = min_{j<i, P[j+1][i]} (C[j] + 1). This DP runs in O(n^2) time and O(n^2) space (or O(n) extra space if we compute palindromes on the fly). The resulting C[n‑1] gives the minimal number of cuts required.

Interview Questions on This Problem

Q1How would you modify the solution to also return one possible optimal palindrome partitioning, not just the cut count?

Maintain a predecessor array prev[i] that stores the index j where the optimal cut before i occurs (i.e., s[j+1..i] is a palindrome and C[i] = C[j] + 1). After DP finishes, backtrack from n‑1 using prev to reconstruct the substrings in reverse order.

Q2What is the time and space complexity if you compute palindrome information on the fly using expanding centers instead of a full O(n^2) table?

Expanding around each center still yields O(n^2) total time because each expansion touches each character at most O(n) times, but the space drops to O(n) for the cuts array and O(1) auxiliary space for the expansion loops.

Q3Can the algorithm be adapted to handle strings with uppercase letters and digits without changing its asymptotic complexity?

Yes. The palindrome check is character‑agnostic; you only need to treat the input as a generic array of symbols. The DP formulation and preprocessing remain O(n^2) time and O(n^2) (or O(n)) space regardless of the alphabet size.

Examples

Example 1

Input

ababa

Output

Explanation: The entire string "ababa" reads the same from left to right and right to left, so it forms a single palindrome. No cuts are necessary, thus the answer is 0.

Example 2

Input

aab

Output

Explanation: The optimal partition is "aa" | "b". "aa" is a palindrome and "b" is trivially a palindrome. Only one cut between the second and third characters is required, giving a minimum of 1 cut.

Example 3

Input

abcde

Output

Explanation: No two‑character substring of "abcde" is a palindrome. Consequently each character must stand alone. The partition is "a" | "b" | "c" | "d" | "e", which uses 4 cuts (between each consecutive pair). This is the smallest possible number of cuts.

Constraints

1 <= |s| <= 2000
s contains only characters 'a' through 'z'
The answer fits in a 32‑bit signed integer

Optimal Approach & Strategy

Pre‑compute palindrome substrings in O(n^2) time, then fill a DP array where each entry uses previously computed results to find the minimal cuts in O(n^2) total. This eliminates exponential recursion.

Brute Force Approach

Recursively try every possible cut position, checking each resulting substring for palindrome property, and keep the minimum cut count. This explores 2^n partitions and quickly becomes infeasible.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n^2)

function minSegmentCost(s) {
   let n = s.length;
   let dp = new Array(n).fill(0);
   dp[0] = 0;
   for (let i = 1; i < n; i++) {
       let min = Infinity;
       for (let j = 0; j <= i; j++) {
           if (isPalindrome(s.substring(j, i + 1))) {
               min = Math.min(min, dp[j] + 1);
           }
       }
       dp[i] = min;
   }
   return dp[n - 1];
}

function isPalindrome(s) {
   let n = s.length;
   for (let i = 0; i < n / 2; i++) {
       if (s[i] !== s[n - i - 1]) {
           return false;
       }
   }
   return true;
}

class Solution {
public:
   int minSegmentCost(string s) {
       int n = s.length();
       vector<int> dp(n, 0);
       dp[0] = 0;
       for (int i = 1; i < n; i++) {
           int min = INT_MAX;
           for (int j = 0; j <= i; j++) {
               if (isPalindrome(s.substr(j, i - j + 1))) {
                   min = min < dp[j] + 1 ? min : dp[j] + 1;
               }
           }
           dp[i] = min;
       }
       return dp[n - 1];
   }

   bool isPalindrome(string s) {
       int n = s.length();
       for (int i = 0; i < n / 2; i++) {
           if (s[i] != s[n - i - 1]) {
               return false;
           }
       }
       return true;
   }
};

public class Solution {
   public int minSegmentCost(String s) {
       int n = s.length();
       int[] dp = new int[n];
       dp[0] = 0;
       for (int i = 1; i < n; i++) {
           int min = Integer.MAX_VALUE;
           for (int j = 0; j <= i; j++) {
               if (isPalindrome(s.substring(j, i + 1))) {
                   min = Math.min(min, dp[j] + 1);
               }
           }
           dp[i] = min;
       }
       return dp[n - 1];
   }

   public boolean isPalindrome(String s) {
       int n = s.length();
       for (int i = 0; i < n / 2; i++) {
           if (s.charAt(i) != s.charAt(n - i - 1)) {
               return false;
           }
       }
       return true;
   }
}

def min_segment_cost(s):
   n = len(s)
   dp = [0] * n
   dp[0] = 0
   for i in range(1, n):
       min_val = float('inf')
       for j in range(i + 1):
           if is_palindrome(s[j:i + 1]):
               min_val = min(min_val, dp[j] + 1)
       dp[i] = min_val
   return dp[-1]

def is_palindrome(s):
   n = len(s)
   for i in range(n // 2):
       if s[i] != s[n - i - 1]:
           return False
   return True

function minSegmentCost(s) {
   let n = s.length;
   let dp = new Array(n).fill(0);
   dp[0] = 0;
   for (let i = 1; i < n; i++) {
       let min = Infinity;
       for (let j = 0; j <= i; j++) {
           if (isPalindrome(s.substring(j, i + 1))) {
               min = Math.min(min, dp[j] + 1);
           }
       }
       dp[i] = min;
   }
   return dp[n - 1];
}

function isPalindrome(s) {
   let n = s.length;
   for (let i = 0; i < n / 2; i++) {
       if (s[i] !== s[n - i - 1]) {
           return false;
       }
   }
   return true;
}

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Minimum Segment Cost Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Constraints

More Strings

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Minimum Segment Cost — 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