diff --git a/Python/algorithms/string_algorithms/are_string_occurences_equal.py b/Python/algorithms/string_algorithms/are_string_occurences_equal.py
new file mode 100644
index 00000000..21d9570c
--- /dev/null
+++ b/Python/algorithms/string_algorithms/are_string_occurences_equal.py
@@ -0,0 +1,47 @@
+from typing import Dict
+from collections import Counter
+
+def are_occurrences_equal(s: str) -> bool:
+    """
+    Return True if every character in the string `s` appears the same number
+    of times, otherwise False.
+
+    Approach:
+      - Use Counter to count occurrences of each character.
+      - Convert counts to a set and check if its length is 1.
+
+    Time Complexity: O(n)
+    Space Complexity: O(k) where k is number of distinct characters
+
+    Args:
+        s: input string
+
+    Returns:
+        bool: True if all character frequencies are equal, else False.
+
+    Examples:
+        >>> are_occurrences_equal("abacbc")
+        True  # a:2, b:2, c:2
+
+        >>> are_occurrences_equal("aaabb")
+        False # a:3, b:2
+    """
+    if not s:
+        return True  # empty string — trivially equal frequencies
+
+    counts = Counter(s)
+    return len(set(counts.values())) == 1
+
+
+if __name__ == "__main__":
+    examples = [
+        ("abacbc", True),
+        ("aaabb", False),
+        ("a", True),
+        ("", True),
+        ("abcabc", True),
+        ("aabbccddd", False),
+    ]
+    for inp, expected in examples:
+        out = are_occurrences_equal(inp)
+        print(f"{inp!r} -> {out} (expected {expected})")