Added code for palindrome partitioning problem under dynamic programming (#7222)

* Added code for palindrome partitioning problem under dynamic programming

* Updated return type for function

* Updated Line 24 according to suggestions

* Apply suggestions from code review

Co-authored-by: Caeden Perelli-Harris <[email protected]>

* Update palindrome_partitioning.py

* Update palindrome_partitioning.py

* is_palindromic

Co-authored-by: Christian Clauss <[email protected]>
Co-authored-by: Caeden Perelli-Harris <[email protected]>

Author: 3 people

dynamic_programming/palindrome_partitioning.py

@@ -0,0 +1,39 @@
+"""
+Given a string s, partition s such that every substring of the
+partition is a palindrome.
+Find the minimum cuts needed for a palindrome partitioning of s.
+
+Time Complexity: O(n^2)
+Space Complexity: O(n^2)
+For other explanations refer to: https://www.youtube.com/watch?v=_H8V5hJUGd0
+"""
+
+
+def find_minimum_partitions(string: str) -> int:
+    """
+    Returns the minimum cuts needed for a palindrome partitioning of string
+
+    >>> find_minimum_partitions("aab")
+    1
+    >>> find_minimum_partitions("aaa")
+    0
+    >>> find_minimum_partitions("ababbbabbababa")
+    3
+    """
+    length = len(string)
+    cut = [0] * length
+    is_palindromic = [[False for i in range(length)] for j in range(length)]
+    for i, c in enumerate(string):
+        mincut = i
+        for j in range(i + 1):
+            if c == string[j] and (i - j < 2 or is_palindromic[j + 1][i - 1]):
+                is_palindromic[j][i] = True
+                mincut = min(mincut, 0 if j == 0 else (cut[j - 1] + 1))
+        cut[i] = mincut
+    return cut[length - 1]
+
+
+if __name__ == "__main__":
+    s = input("Enter the string: ").strip()
+    ans = find_minimum_partitions(s)
+    print(f"Minimum number of partitions required for the '{s}' is {ans}")