diff --git a/algorithms/dynamic_programming/print_all_longest_common_subsequence.py b/algorithms/dynamic_programming/print_all_longest_common_subsequence.py
new file mode 100644
index 000000000..025a7873a
--- /dev/null
+++ b/algorithms/dynamic_programming/print_all_longest_common_subsequence.py
@@ -0,0 +1,62 @@
+"""
+Find All Distinct Longest Common Subsequences
+
+Complexity:
+    Time:  Exponential in the number of LCSs
+    Space: O(m * n) + O(number of LCSs)
+"""
+
+
+def print_all_longest_common_subsequence(s: str, t: str) -> list[str]:
+    """Compute all distinct longest common subsequences of two strings.
+
+    Args:
+        s: First string.
+        t: Second string.
+
+    Returns:
+        A list containing all distinct longest common subsequences.
+
+    Examples:
+        >>> print_all_longest_common_subsequence('rajroy', 'rxoyxraxj')
+            ['raj', 'roy']
+    """
+
+    n, m = len(s), len(t)
+    dp = [[0] * (m + 1) for _ in range(n + 1)]
+
+    for i in range(1, n + 1):
+        for j in range(1, m + 1):
+            if s[i - 1] == t[j - 1]:
+                dp[i][j] = 1 + dp[i - 1][j - 1]
+            else:
+                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
+
+    res = set()
+    memo = {}
+
+    def solve(i: int, j: int, current: str) -> None:
+        if i < 1 or j < 1:
+            res.add(current[::-1])
+            return
+
+        key = (i, j, current)
+        if key in memo:
+            res.update(memo[key])
+            return
+
+        if s[i - 1] == t[j - 1]:
+            solve(i - 1, j - 1, current + s[i - 1])
+        else:
+            if dp[i - 1][j] > dp[i][j - 1]:
+                solve(i - 1, j, current)
+            elif dp[i][j - 1] > dp[i - 1][j]:
+                solve(i, j - 1, current)
+            else:
+                solve(i - 1, j, current)
+                solve(i, j - 1, current)
+
+        memo[key] = set(res)
+
+    solve(n, m, "")
+    return list(res)
diff --git a/algorithms/dynamic_programming/print_any_longest_common_subsequence.py b/algorithms/dynamic_programming/print_any_longest_common_subsequence.py
new file mode 100644
index 000000000..455e1561c
--- /dev/null
+++ b/algorithms/dynamic_programming/print_any_longest_common_subsequence.py
@@ -0,0 +1,52 @@
+"""
+Print any longest common subsequence
+
+Complexity:
+    Time:  O(n * m)
+    Space: O(n * m)
+"""
+
+
+def print_any_longest_common_subsequence(s1: str, s2: str) -> str:
+    """Return any one longest common subsequence of two strings.
+
+    Args:
+        s1: First string.
+        s2: Second string.
+
+    Returns:
+        A string representing any one longest common subsequence.
+
+    Example:
+        >>> print_any_longest_common_subsequence('rajroy', 'rxoyxraxj')
+            "raj" or "roy"
+    """
+
+    n = len(s1)
+    m = len(s2)
+    dp = [[0] * (m + 1) for _ in range(n + 1)]
+
+    for ind1 in range(1, n + 1):
+        for ind2 in range(1, m + 1):
+            if s1[ind1 - 1] == s2[ind2 - 1]:
+                dp[ind1][ind2] = 1 + dp[ind1 - 1][ind2 - 1]
+            else:
+                dp[ind1][ind2] = max(dp[ind1 - 1][ind2], dp[ind1][ind2 - 1])
+
+    length = dp[n][m]
+    i, j = n, m
+    lcs_str = [""] * length
+    index = length - 1
+
+    while i > 0 and j > 0:
+        if s1[i - 1] == s2[j - 1]:
+            lcs_str[index] = s1[i - 1]
+            index -= 1
+            i -= 1
+            j -= 1
+        elif dp[i - 1][j] > dp[i][j - 1]:
+            i -= 1
+        else:
+            j -= 1
+
+    return "".join(lcs_str)