Added new algorithm to solve the famous celebrity problem using stack

Problem Description : Find the celebrity in a party.

    A celebrity is a person who:
    - Is known by everyone else
    - Knows no one else

    Args:
        mat: n x n adjacency matrix where mat[i][j] == 1 if person i knows person j.

    Returns:
        Index of the celebrity if one exists, else -1.

Author: ShivamKhunger

data_structures/stacks/celebrity_problem.py

@@ -0,0 +1,61 @@
+from __future__ import annotations
+
+# Example input (matrix where mat[i][j] == 1 if person i knows person j)
+mat = [
+    [1, 1, 0],
+    [0, 1, 0],
+    [0, 1, 1],
+]
+
+# Expected output for this input
+expect = 1  # person at index 1 is the celebrity
+
+
+def celebrity(mat: list[list[int]]) -> int:
+    """
+    Function to find the celebrity in a party.
+
+    A celebrity is a person who is:
+    - Known by everyone else
+    - Knows no one else
+    - At max there can be only one celebrity
+    Args:
+        mat: n x n matrix, where mat[i][j] == 1 if person i knows person j.
+
+    Returns:
+        Index of the celebrity if one exists, else -1.
+
+    Example:
+    >>> celebrity(mat) == expect
+    True
+    """
+    n = len(mat)
+    st: list[int] = list(range(n))  # push everybody in the stack
+
+    # Find a potential celebrity
+    while len(st) > 1:
+        a = st.pop()
+        b = st.pop()
+
+        if mat[a][b]:
+            st.append(b)  # a knows b so a can't be celebrity
+        else:
+            st.append(a)  # a doesn't know b so b can't be celebrity
+
+    # Potential candidate
+    c = st.pop()
+
+    # Verify candidate
+    for i in range(n):
+        if i == c:
+            continue
+        if mat[c][i] or not mat[i][c]:
+            return -1
+    return c
+
+
+if __name__ == "__main__":
+    from doctest import testmod
+
+    testmod()
+    print("Celebrity is:", celebrity(mat))