Added Palindrome Partitioning Solution in Dynamic Programming

Author: Alok Negi

20. Dynamic Programming/Palindrome Partitioning.cpp

@@ -0,0 +1,54 @@
+// Dynamic Programming Solution for Palindrome Partitioning Problem
+#include <iostream>
+#include <bits/stdc++.h>
+#include <string.h>
+using namespace std;
+
+int min(int a, int b) { return (a < b) ? a : b; }
+
+int minPalPartion(char* str)
+{
+
+	int n = strlen(str);
+	int C[n];
+	bool P[n][n];
+
+	int i, j, k, L; // different looping variables
+
+	for (i = 0; i < n; i++) {
+		P[i][i] = true;
+	}
+
+	for (L = 2; L <= n; L++) {
+        for (i = 0; i < n - L + 1; i++) {
+			j = i + L - 1; // Set ending index
+
+			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 C[n - 1];
+}
+
+// Driver program to test above function
+int main()
+{
+	char str[] = "ababbbabbababa";
+	cout <<"Min cuts needed for Palindrome Partitioning is " << minPalPartion(str);
+	return 0;
+}