Palindrome Partitioning DP Minimum Cuts in DSA C

Mental Model

We want to split a string into the fewest pieces so that each piece reads the same forwards and backwards.

Analogy: Imagine cutting a ribbon into the smallest number of palindromic segments, like cutting a patterned ribbon where each piece shows a symmetrical design.

s: a b a c d c b a
cuts: 0 1 2 3 4 5 6 7 8
palindrome checks: true/false for each substring

Dry Run Walkthrough

Input: string: "aab"

Goal: Find the minimum cuts needed to split "aab" into palindromic substrings

Step 1: Check all substrings for palindrome

palindrome table:
[0][0]=true (a)
[0][1]=true (aa)
[0][2]=false (aab)
[1][1]=true (a)
[1][2]=false (ab)
[2][2]=true (b)

Why: Knowing which substrings are palindromes helps decide where to cut

Step 2: Initialize cuts array with max cuts (index)

cuts: [0, 1, 2]

Why: Start assuming worst case: cut before every character

Step 3: At index 1, substring "aa" is palindrome, update cuts[1] to 0

cuts: [0, 0, 2]

Why: No cut needed if substring from start is palindrome

Step 4: At index 2, check substrings ending at 2: "b" palindrome, cuts[2] = cuts[1] + 1 = 1 "ab" not palindrome "aab" not palindrome

cuts: [0, 0, 1]

Why: Minimum cuts updated based on palindrome substrings ending at current index

Result:

cuts array: [0, 0, 1]
Minimum cuts needed: 1

Annotated Code

DSA C

#include <stdio.h>
#include <string.h>
#include <stdbool.h>

// Function to check palindrome substrings using DP
void computePalindromeTable(char *s, bool palindrome[][100], int n) {
    for (int i = 0; i < n; i++) {
        palindrome[i][i] = true; // single char palindrome
    }
    for (int length = 2; length <= n; length++) {
        for (int start = 0; start <= n - length; start++) {
            int end = start + length - 1;
            if (s[start] == s[end]) {
                if (length == 2) {
                    palindrome[start][end] = true; // two chars same
                } else {
                    palindrome[start][end] = palindrome[start + 1][end - 1]; // depends on inner substring
                }
            } else {
                palindrome[start][end] = false;
            }
        }
    }
}

// Function to find minimum cuts for palindrome partitioning
int minCut(char *s) {
    int n = strlen(s);
    if (n == 0) return 0;
    bool palindrome[100][100] = {{false}};
    int cuts[100];

    computePalindromeTable(s, palindrome, n);

    for (int i = 0; i < n; i++) {
        cuts[i] = i; // max cuts
        for (int j = 0; j <= i; j++) {
            if (palindrome[j][i]) {
                if (j == 0) {
                    cuts[i] = 0; // whole substring is palindrome
                } else if (cuts[j - 1] + 1 < cuts[i]) {
                    cuts[i] = cuts[j - 1] + 1; // update min cuts
                }
            }
        }
    }
    return cuts[n - 1];
}

int main() {
    char s[] = "aab";
    int result = minCut(s);
    printf("Minimum cuts needed: %d\n", result);
    return 0;
}

palindrome[i][i] = true; // single char palindrome

mark single characters as palindrome

if (s[start] == s[end]) { ... palindrome[start][end] = palindrome[start + 1][end - 1]; }

check palindrome status of substring based on inner substring

cuts[i] = i; // max cuts

initialize cuts with worst case (cut before every character)

if (palindrome[j][i]) { if (j == 0) cuts[i] = 0; else cuts[i] = min(cuts[i], cuts[j - 1] + 1); }

update cuts if substring s[j..i] is palindrome

Output

Minimum cuts needed: 1

Complexity Analysis

Time: O(n^2) because we check all substrings for palindrome and compute cuts for each index

Space: O(n^2) for palindrome table and O(n) for cuts array

vs Alternative: Naive approach checks palindrome on the fly causing O(n^3) time; DP reduces it to O(n^2)

Edge Cases

empty string

returns 0 cuts as no partition needed

DSA C

int n = strlen(s); if n == 0 return 0;

string with all same characters, e.g. "aaaa"

returns 0 cuts since whole string is palindrome

DSA C

if (j == 0) cuts[i] = 0;

string with no palindrome longer than 1, e.g. "abc"

returns n-1 cuts, cutting before every character

DSA C

cuts[i] = i; // max cuts initialization

When to Use This Pattern

When asked to split a string into palindromic parts with minimum cuts, use DP with palindrome substring precomputation to optimize.

Common Mistakes

Mistake: Checking palindrome substrings repeatedly without DP causing slow O(n^3) time
Fix: Precompute palindrome table with DP to check in O(1) time

Mistake: Not updating cuts correctly when substring from start is palindrome
Fix: Set cuts[i] = 0 if substring s[0..i] is palindrome

Summary

Find the fewest cuts needed to split a string into palindromic substrings.

Use when you want to partition strings minimizing cuts with palindrome constraints.

Precompute palindrome substrings to quickly decide where to cut and update minimum cuts.