diff --git a/dynamic_programming/longest_palindromic_subsequence.py b/dynamic_programming/longest_palindromic_subsequence.py
new file mode 100644
index 000000000000..362300f23992
--- /dev/null
+++ b/dynamic_programming/longest_palindromic_subsequence.py
@@ -0,0 +1,73 @@
+"""
+The longest palindromic subsequence (LPS) is the longest
+subsequence in a string that is the same when reversed.
+A subsequence is not the same as a substring.
+Unlike substrings, the characters of subsequences are not required
+to occupy consecutive positions.
+
+Explanation:
+https://www.techiedelight.com/longest-palindromic-subsequence-using-dynamic-programming/
+"""
+
+
+def longest_palindromic_subsequence(sequence: str, i: int, j: int, memo: dict) -> int:
+    """Find the longest palindromic subsequence in a given string
+
+    :param sequence: The sequence to search in
+    :type sequence: str
+    :param i: The initial lower index bounding the search area
+    :type i: int
+    :param j: The initial upper index bounding the search area
+    :type j: int
+    :param memo: The memoization table used to optimize this algorithm
+    :type memo: dict
+    :return: The length of the longest palindromic subsequence
+    :rtype: int
+
+    >>> longest_palindromic_subsequence("ABBDCACB", 0, 7)
+    5
+    >>> longest_palindromic_subsequence("nurses run", 0, 9)
+    9
+    >>> longest_palindromic_subsequence("Z", 0, 0)
+    1
+    """
+    # If the sequence only has one character, it is a palindrome
+    # This prevents a one-letter palindrome from being counted as 2
+    if i == j:
+        return 1
+
+    # Base case: If i is greater than j you have gone too far
+    if i > j:
+        return 0
+
+    # Each subproblem can be identified by the substring it works on.
+    # So, we can identify whether a subproblem has been solved by
+    # checking if it is in the memo table.
+    key = (i, j)
+
+    if key not in memo:
+
+        # If the first character of the current string equals the last
+        if sequence[i] == sequence[j]:
+            # Recur with the remaining substring and put the result in the memo table
+            memo[key] = (
+                longest_palindromic_subsequence(sequence, i + 1, j - 1, memo) + 2
+            )
+
+        # If the first and last characters are NOT equal, find the larger LPS between:
+        #   a.) the substring formed by removing the first character
+        #   b.) the substring formed by removing the last character
+        # Then, store in the memo table
+        else:
+            memo[key] = max(
+                longest_palindromic_subsequence(sequence, i + 1, j, memo),
+                longest_palindromic_subsequence(sequence, i, j - 1, memo),
+            )
+
+    return memo[key]
+
+
+if __name__ == "__main__":
+    sequence = input("Enter sequence string: ")
+    len_lps = longest_palindromic_subsequence(sequence, 0, len(sequence) - 1, {})
+    print(f'The length of the LPS in "{sequence}" is {len_lps}')