# Dynamic Programming | Set 17 (Palindrome Partitioning)

Given a string, a partitioning of the string is a palindrome partitioning if every substring of the partition is a palindrome. For example, “aba|b|bbabb|a|b|aba” is a palindrome partitioning of “ababbbabbababa”. Determine the fewest cuts needed for palindrome partitioning of a given string. For example, minimum 3 cuts are needed for “ababbbabbababa”. The three cuts are “a|babbbab|b|ababa”. If a string is palindrome, then minimum 0 cuts are needed. If a string of length n containing all different characters, then minimum n-1 cuts are needed.

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

This problem is a variation of Matrix Chain Multiplication problem. If the string is palindrome, then we simply return 0. Else, like the Matrix Chain Multiplication problem, we try making cuts at all possible places, recursively calculate the cost for each cut and return the minimum value.

Let the given string be str and minPalPartion() be the function that returns the fewest cuts needed for palindrome partitioning. following is the optimal substructure property.

// i is the starting index and j is the ending index. i must be passed as 0 and j as n-1
minPalPartion(str, i, j) = 0 if i == j. // When string is of length 1.
minPalPartion(str, i, j) = 0 if str[i..j] is palindrome.

// If none of the above conditions is true, then minPalPartion(str, i, j) can be
// calculated recursively using the following formula.
minPalPartion(str, i, j) = Min { minPalPartion(str, i, k) + 1 +
minPalPartion(str, k+1, j) }
where k varies from i to j-1

Following is Dynamic Programming solution. It stores the solutions to subproblems in two arrays P[][] and C[][], and reuses the calculated values.

## C/C++

// Dynamic Programming Solution for Palindrome Partitioning Problem
#include <stdio.h>
#include <string.h>
#include <limits.h>

// A utility function to get minimum of two integers
int min (int a, int b) { return (a < b)? a : b; }

// Returns the minimum number of cuts needed to partition a string
// such that every part is a palindrome
int minPalPartion(char *str)
{
// Get the length of the string
int n = strlen(str);

/* Create two arrays to build the solution in bottom up manner
C[i][j] = Minimum number of cuts needed for palindrome partitioning
of substring str[i..j]
P[i][j] = true if substring str[i..j] is palindrome, else false
Note that C[i][j] is 0 if P[i][j] is true */
int C[n][n];
bool P[n][n];

int i, j, k, L; // different looping variables

// Every substring of length 1 is a palindrome
for (i=0; i<n; i++)
{
P[i][i] = true;
C[i][i] = 0;
}

/* L is substring length. Build the solution in bottom up manner by
considering all substrings of length starting from 2 to n.
The loop structure is same as Matrx Chain Multiplication problem (
See https://www.geeksforgeeks.org/archives/15553 )*/
for (L=2; L<=n; L++)
{
// For substring of length L, set different possible starting indexes
for (i=0; i<n-L+1; i++)
{
j = i+L-1; // Set ending index

// If L is 2, then we just need to compare two characters. Else
// need to check two corner characters and value of P[i+1][j-1]
if (L == 2)
P[i][j] = (str[i] == str[j]);
else
P[i][j] = (str[i] == str[j]) && P[i+1][j-1];

// IF str[i..j] is palindrome, then C[i][j] is 0
if (P[i][j] == true)
C[i][j] = 0;
else
{
// Make a cut at every possible location starting from i to j,
// and get the minimum cost cut.
C[i][j] = INT_MAX;
for (k=i; k<=j-1; k++)
C[i][j] = min (C[i][j], C[i][k] + C[k+1][j]+1);
}
}
}

// Return the min cut value for complete string. i.e., str[0..n-1]
return C[0][n-1];
}

// Driver program to test above function
int main()
{
char str[] = "ababbbabbababa";
printf("Min cuts needed for Palindrome Partitioning is %d",
minPalPartion(str));
return 0;
}

## Java

// Java Code for Palindrome Partitioning
// Problem
public class GFG
{
// Returns the minimum number of cuts needed
// to partition a string such that every
// part is a palindrome
static int minPalPartion(String str)
{
// Get the length of the string
int n = str.length();

/* Create two arrays to build the solution
in bottom up manner
C[i][j] = Minimum number of cuts needed
for palindrome partitioning
of substring str[i..j]
P[i][j] = true if substring str[i..j] is
palindrome, else false
Note that C[i][j] is 0 if P[i][j] is
true */
int[][] C = new int[n][n];
boolean[][] P = new boolean[n][n];

int i, j, k, L; // different looping variables

// Every substring of length 1 is a palindrome
for (i = 0; i < n; i++)
{
P[i][i] = true;
C[i][i] = 0;
}

/* L is substring length. Build the solution in
bottom up manner by considering all substrings
of length starting from 2 to n. The loop
structure is same as Matrx Chain Multiplication
problem (
See https://www.geeksforgeeks.org/archives/15553 )*/
for (L = 2; L <= n; L++)
{
// For substring of length L, set different
// possible starting indexes
for (i = 0; i < n - L + 1; i++)
{
j = i + L - 1; // Set ending index

// If L is 2, then we just need to
// compare two characters. Else need to
// check two corner characters and value
// of P[i+1][j-1]
if (L == 2)
P[i][j] = (str.charAt(i) ==
str.charAt(j));
else
P[i][j] = (str.charAt(i) ==
str.charAt(j)) && P[i+1][j-1];

// IF str[i..j] is palindrome, then
// C[i][j] is 0
if (P[i][j] == true)
C[i][j] = 0;
else
{
// Make a cut at every possible
// localtion starting from i to j,
// and get the minimum cost cut.
C[i][j] = Integer.MAX_VALUE;
for (k = i; k <= j - 1; k++)
C[i][j] = Integer.min(C[i][j],
C[i][k] + C[k+1][j] + 1);
}
}
}

// Return the min cut value for complete
// string. i.e., str[0..n-1]
return C[0][n-1];
}

// Driver program to test above function
public static void main(String args[])
{
String str = "ababbbabbababa";
System.out.println("Min cuts needed for "+
"Palindrome Partitioning is "+
minPalPartion(str));
}
}
// This code is contributed by Sumit Ghosh

## C#

// C# Code for Palindrome Partitioning
// Problem
using System;

class GFG
{
// Returns the minimum number of cuts needed
// to partition a string such that every
// part is a palindrome
static int minPalPartion(String str)
{
// Get the length of the string
int n = str.Length;

/* Create two arrays to build the solution
in bottom up manner
C[i][j] = Minimum number of cuts needed
for palindrome partitioning
of substring str[i..j]
P[i][j] = true if substring str[i..j] is
palindrome, else false
Note that C[i][j] is 0 if P[i][j] is
true */
int[,] C = new int[n, n];
bool[,] P = new bool[n, n];

int i, j, k, L; // different looping variables

// Every substring of length 1 is a palindrome
for (i = 0; i < n; i++)
{
P[i, i] = true;
C[i, i] = 0;
}

/* L is substring length. Build the solution in
bottom up manner by considering all substrings
of length starting from 2 to n. The loop
structure is same as Matrx Chain Multiplication
problem (
See https://www.geeksforgeeks.org/archives/15553 )*/
for (L = 2; L <= n; L++)
{
// For substring of length L, set different
// possible starting indexes
for (i = 0; i < n - L + 1; i++)
{
j = i + L - 1; // Set ending index

// If L is 2, then we just need to
// compare two characters. Else need to
// check two corner characters and value
// of P[i+1][j-1]
if (L == 2)
P[i,j] = (str[i] == str[j]);
else
P[i,j] = (str[i] == str[j]) &&
P[i + 1, j - 1];

// IF str[i..j] is palindrome, then
// C[i][j] is 0
if (P[i, j] == true)
C[i, j] = 0;
else
{
// Make a cut at every possible
// localtion starting from i to j,
// and get the minimum cost cut.
C[i, j] = int.MaxValue;
for (k = i; k <= j - 1; k++)
C[i, j] = Math.Min(C[i, j], C[i, k]
+ C[k + 1, j] + 1);
}
}
}

// Return the min cut value for complete
// string. i.e., str[0..n-1]
return C[0, n - 1];
}

// Driver program
public static void Main()
{
String str = "ababbbabbababa";
Console.Write("Min cuts needed for "+
"Palindrome Partitioning is "+
minPalPartion(str));
}
}

// This code is contributed by Sam007

Output:

Min cuts needed for Palindrome Partitioning is 3

Time Complexity: O(n3)

An optimization to above approach
In above approach, we can calculate minimum cut while finding all palindromic substring. If we find all palindromic substring 1st and then we calculate minimum cut, time complexity will reduce to O(n2).
Thanks for Vivek for suggesting this optimization.

## C++

// Dynamic Programming Solution for Palindrome Partitioning Problem
#include <stdio.h>
#include <string.h>
#include <limits.h>

// A utility function to get minimum of two integers
int min (int a, int b) { return (a < b)? a : b; }

// Returns the minimum number of cuts needed to partition a string
// such that every part is a palindrome
int minPalPartion(char *str)
{
// Get the length of the string
int n = strlen(str);

/* Create two arrays to build the solution in bottom up manner
C[i] = Minimum number of cuts needed for palindrome partitioning
of substring str[0..i]
P[i][j] = true if substring str[i..j] is palindrome, else false
Note that C[i] is 0 if P[0][i] is true */
int C[n];
bool P[n][n];

int i, j, k, L; // different looping variables

// Every substring of length 1 is a palindrome
for (i=0; i<n; i++)
{
P[i][i] = true;
}

/* L is substring length. Build the solution in bottom up manner by
considering all substrings of length starting from 2 to n. */
for (L=2; L<=n; L++)
{
// For substring of length L, set different possible starting indexes
for (i=0; i<n-L+1; i++)
{
j = i+L-1; // Set ending index

// If L is 2, then we just need to compare two characters. Else
// need to check two corner characters and value of P[i+1][j-1]
if (L == 2)
P[i][j] = (str[i] == str[j]);
else
P[i][j] = (str[i] == str[j]) && P[i+1][j-1];
}
}

for (i=0; i<n; i++)
{
if (P[0][i] == true)
C[i] = 0;
else
{
C[i] = INT_MAX;
for(j=0;j<i;j++)
{
if(P[j+1][i] == true && 1+C[j]<C[i])
C[i]=1+C[j];
}
}
}

// Return the min cut value for complete string. i.e., str[0..n-1]
return C[n-1];
}

// Driver program to test above function
int main()
{
char str[] = "ababbbabbababa";
printf("Min cuts needed for Palindrome Partitioning is %d",
minPalPartion(str));
return 0;
}

## Java

// Java Code for Palindrome Partitioning
// Problem
public class GFG
{
// Returns the minimum number of cuts needed
// to partition a string such that every part
// is a palindrome
static int minPalPartion(String str)
{
// Get the length of the string
int n = str.length();

/* Create two arrays to build the solution
in bottom up manner
C[i] = Minimum number of cuts needed for
palindrome partitioning of substring
str[0..i]
P[i][j] = true if substring str[i..j] is
palindrome, else false
Note that C[i] is 0 if P[0][i] is true */
int[] C = new int[n];
boolean[][] P = new boolean[n][n];

int i, j, k, L; // different looping variables

// Every substring of length 1 is a palindrome
for (i = 0; i < n; i++)
{
P[i][i] = true;
}

/* L is substring length. Build the solution
in bottom up manner by considering all substrings
of length starting from 2 to n. */
for (L = 2; L <= n; L++)
{
// For substring of length L, set different
// possible starting indexes
for (i = 0; i < n - L + 1; i++)
{
j = i + L - 1; // Set ending index

// If L is 2, then we just need to
// compare two characters. Else need to
// check two corner characters and value
// of P[i+1][j-1]
if (L == 2)
P[i][j] = (str.charAt(i) ==
str.charAt(j));
else
P[i][j] = (str.charAt(i) ==
str.charAt(j)) && P[i+1][j-1];
}
}

for (i = 0; i < n; i++)
{
if (P[0][i] == true)
C[i] = 0;
else
{
C[i] = Integer.MAX_VALUE;
for(j = 0; j < i; j++)
{
if(P[j+1][i] == true && 1 +
C[j] < C[i])
C[i] = 1 + C[j];
}
}
}

// Return the min cut value for complete
// string. i.e., str[0..n-1]
return C[n-1];
}

// Driver program to test above function
public static void main(String args[])
{
String str = "ababbbabbababa";
System.out.println("Min cuts needed for "+
"Palindrome Partitioning"+
" is "+ minPalPartion(str));
}
}
// This code is contributed by Sumit Ghosh

## C#

// C# Code for Palindrome Partitioning
// Problem
using System;

class GFG
{

// Returns the minimum number of cuts needed
// to partition a string such that every part
// is a palindrome
static int minPalPartion(String str)
{
// Get the length of the string
int n = str.Length;

/* Create two arrays to build the solution
in bottom up manner
C[i] = Minimum number of cuts needed for
palindrome partitioning of substring
str[0..i]
P[i][j] = true if substring str[i..j] is
palindrome, else false
Note that C[i] is 0 if P[0][i] is true */
int[] C = new int[n];
bool[,] P = new bool[n,n];

int i, j, L; // different looping variables

// Every substring of length 1 is a palindrome
for (i = 0; i < n; i++)
{
P[i,i] = true;
}

/* L is substring length. Build the solution
in bottom up manner by considering all substrings
of length starting from 2 to n. */
for (L = 2; L <= n; L++)
{
// For substring of length L, set different
// possible starting indexes
for (i = 0; i < n - L + 1; i++)
{
j = i + L - 1; // Set ending index

// If L is 2, then we just need to
// compare two characters. Else need to
// check two corner characters and value
// of P[i+1][j-1]
if (L == 2)
P[i,j] = (str[i] == str[j]);
else
P[i,j] = (str[i] == str[j]) && P[i+1,j-1];
}
}

for (i = 0; i < n; i++)
{
if (P[0,i] == true)
C[i] = 0;
else
{
C[i] = int.MaxValue;
for(j = 0; j < i; j++)
{
if(P[j+1,i] == true && 1 + C[j] < C[i])
C[i] = 1 + C[j];
}
}
}

// Return the min cut value for complete
// string. i.e., str[0..n-1]
return C[n-1];
}

// Driver program
public static void Main()
{
String str = "ababbbabbababa";
Console.Write("Min cuts needed for "+
"Palindrome Partitioning"+
" is "+ minPalPartion(str));
}
}

// This code is contributed by Sam007

Output:

Min cuts needed for Palindrome Partitioning is 3

Time Complexity: O(n2)