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+# Exercice 2
+
+---
+
+## `function_1`
+
+Deux boucles = O(n\*m)
+
+## `function_2`
+
+While qui parcours le tableau = O(n)
+
+## `function_3`
+
+Un seul process d'information = O(1)
+
+# Exercice 3
+
+---
+
+[[exercice_3.py]]
diff --git a/exercice_3.py b/exercice_3.py
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+def sort1d(arr):
+  """
+  Cet algo à une complexité de O(n log n)
+  """
+
+  if len(arr) < 2:
+    return arr
+
+  mid = len(arr) // 2
+  left = sort1d(arr[:mid])
+  right = sort1d(arr[mid:])
+
+  result = []
+  i = j = 0
+
+  while i < len(left) and j < len(right):
+    lv = left[i] if type(left[i]) is int else sum(left[i])
+    rv = right[j] if type(right[j]) is int else sum(right[j])
+
+    if lv < rv:
+      result.append(left[i])
+      i += 1
+    else:
+      result.append(right[j])
+      j += 1
+
+  result.extend(left[i:])
+  result.extend(right[j:])
+  return result
+
+def sort2d(arr):
+  """
+  Cet algo à une complexité de O(m n log n + m log m)
+  """
+
+  sorted_rows = [sort1d(row) for row in arr]
+  return sort1d(sorted_rows)
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