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-rw-r--r--users/wpcarro/scratch/facebook/move-zeroes-to-end.py62
1 files changed, 62 insertions, 0 deletions
diff --git a/users/wpcarro/scratch/facebook/move-zeroes-to-end.py b/users/wpcarro/scratch/facebook/move-zeroes-to-end.py
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index 000000000000..1535b5a9faac
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+++ b/users/wpcarro/scratch/facebook/move-zeroes-to-end.py
@@ -0,0 +1,62 @@
+from collections import deque
+
+def move_zeroes_to_end_quadratic(xs):
+    """
+    This solution is suboptimal. It runs in quadratic time, and it uses constant
+    space.
+    """
+    i = 0
+    while i < len(xs) - 1:
+        if xs[i] == 0:
+            j = i + 1
+            while j < len(xs) and xs[j] == 0:
+                j += 1
+            if j >= len(xs):
+                break
+            xs[i], xs[j] = xs[j], xs[i]
+        i += 1
+
+def move_zeroes_to_end_linear(xs):
+    """
+    This solution is clever. It runs in linear time proportionate to the number
+    of elements in `xs`, and has linear space proportionate to the number of
+    consecutive zeroes in `xs`.
+    """
+    q = deque()
+    for i in range(len(xs)):
+        if xs[i] == 0:
+            q.append(i)
+        else:
+            if q:
+                j = q.popleft()
+                xs[i], xs[j] = xs[j], xs[i]
+                q.append(i)
+
+def move_zeroes_to_end_linear_constant_space(xs):
+    """
+    This is the optimal solution. It runs in linear time and uses constant
+    space.
+    """
+    i = 0
+    for j in range(len(xs)):
+        if xs[j] != 0:
+            xs[i], xs[j] = xs[j], xs[i]
+            i += 1
+
+
+################################################################################
+# Tests
+################################################################################
+
+xss = [
+    [1, 2, 0, 3, 4, 0, 0, 5, 0],
+    [0, 1, 2, 0, 3, 4],
+    [0, 0],
+]
+
+f = move_zeroes_to_end_linear_constant_space
+
+for xs in xss:
+    print(xs)
+    f(xs)
+    print(xs)