Dynamic Programming | Set 17 (Palindrome Partitioning) | GeeksforGeeks
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.
// i is the starting index and j is the ending index. i must be passed as 0 and j as n-1
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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.
// 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
O(n^2)
In above approach, we can calculating minimum cut while finding all palindromic substring. If we finding all palindromic substring 1st and then we calculate minimum cut, time complexity will reduce to O(n2).
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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];
}
O(n^3)
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 ( 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_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];}