Create beautiful_matrix.py

week-1/presented code/beautiful_matrix.py

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+def make_it_beautiful(arr:list[int]):
+    """
+    Returns TRUE if array can be made beautiful and prints its beautiful version.
+    ELSE returns FALSE if array can not be made beautiful.
+    """
+
+    # MAIN IDEA: AN ARRAY CAN BE BEAUTIFUL AS LONG AS ALL THE ELEMENTS OF  
+    #THE ARRAY ARE NOT THE SAME.
+
+    # For that, we can compare any two arbitrary elements
+
+    if arr[0] == arr[1]: # Comparing first and second element
+        # Are we missing a case here... I think so
+        # CAN YOU TELL WHICH INPUT CAN HALT OUR CODE....?
+        return False
+
+
+    # HINT:
+    # ^^^^
+
+    # See which of the test cases can cause an issue:
+
+    # [3, 2, 5, 10]
+    # [2, 2, 4, 8]
+    # [3, 5, 6, 8]
+    # [47]
+    # [8, 8, 6, 9]
+    # [7, 6, 5]
+    # [39, 1000]
+
+
+
+
+    # YA, [47] will cause us some trouble, Can you tell why?
+
+
+
+
+
+
+
+
+    # Think what will happen if single elment array is passed?
+    # How should we tackle it
+
+    # One way is to write an explicit case for a single like:
+
+    if len(arr) <= 1:
+        print(arr)
+        return True
+
+    # We can do better than that by comparing the first and last element,
+    #this way we can avoid giving explicit condition for single element array
+
+    if arr[0] == arr[len(arr)-1]: # for cpp, we will pass array length as an argument
+        return False
+
+    # Modulo Operation(%), a%b will return the remainder when a was divided by b,
+    #returned value will always be >= 0 but <b.
+    #looks like we can use it to access elements of the array without causing any errors.
+
+    # But this again creates one problem, it will return false for [a]
+    #which is already beautiful, giving us wrong answer.
+
+    # Another alternative could be:
+    m = 108
+    n = 22
+    if arr[m%len(arr)] == arr[n%len(arr)]: #comparing two random elements, no condition for m and n
+        return False
+
+    #  It has its own set of issues, can you guess what?
+
+    # Moral of the story: we can not avoid giving explicit condition in this case.
+
+    # NOW WE ARE LEFT WITH THE TASK TO PRINT THE BEAUTIFUL VERSION AND RETURN TRUE
+
+    # The first approach that hits the mind is printing the array in
+        # DECREASING Order.
+    arr.sort()
+    print(arr[::-1])
+    # Guess what, it will not always give always correct answer
+    # like for arr = [4, 7, 7, 3]
+
+
+
+
+
+
+
+
+
+
+
+    # The Solution I propose is to print the biggest element first
+    #followed by the smallest element and then print the remaining
+    #elements in which ever way you want, sorting the array beforehand will make things easy
+    arr.sort()
+    size = len(arr)
+
+    print(arr[size-1], end=" ")
+    print(arr[0], end=" ")
+    print(*arr[1:size-1])