These algorithms are coded in a python-like pseudocode, but in some cases the pseudocode is actually valid python!
More informally, we can define other asymptotic notation as follows:
In the analysis of algorithms, many assumptions about hardware and basic operations must be made, e.g array access is constant time.
Maintain a sorted section of the array, then insert new items into the correct position.
def insertion_sort(a): for i in range(1, len(a)): # assert first i positions sorted j = i - 1 while j >= 0 and a[j] > a[j+1]: swap(j, j+1) j -= 1
At each iteration, find the minimum of the remaining array and swap it to the current index.
def selection_sort(a): for i in range(len(a)): swap(i, argmin(a[i:end]))
$O(n^2)$ and unstable. Its only advantage is that it is easy to analyse.
Same as insertion sort, except we find the correct position using binary partitioning.
def binary_insertion_sort(a): for i in range(1, len(a)): hi = i lo = 0 while lo < hi: j = (hi + lo) // 2 if a[i] > a[j]: lo = j + 1 else: hi = j # Swap a[i] into the right place tmp = a[i] for j from i - 1 down to (hi - 1): a[j+1] = a[j] a[hi] = tmp
Binary insertion sort will be preferred to insertion sort when comparisons are expensive, but the swapping costs mean that it is still $O(n^2)$.
In each pass, go through the list swapping adjacent elements as needed. If no swaps are done in a pass, the array is sorted.
def bubblesort(a): while True: didSwap = False for i in range(len(a) - 1): if a[i] > a[i+1]: swap(i, i+1) didSwap = True if not didSwap: break
Divide and conquer algorithm that splits the list in two then recursively sorts each half, before merging sorted lists.
def mergesort(a, lo, hi): if lo < hi: mid = (lo + hi) // 2 mergesort(a, lo, mid) mergesort(a, mid+1, hi) merge(a, lo, mid, hi) def merge(a, lo, mid, hi): # both these subarrays are sorted l = a[lo: mid] r = a[mid+1 : hi] aux =  * (len(l) + len(r)) i = lo j = mid + 1 for k in range(len(aux)): if i > mid: # fill using right only aux[k] = aux[j] j += 1 else if j > hi: # fill using left only aux[k] = a[i] i += 1 else if a[i] <= a[j]: # otherwise compare aux[k] = a[i] i += 1 else: aux[k] = a[j] j += 1
def mergesort(a): step = 1 while (step < len(a)): for lo in range(0, len(a), 2*step): mid = lo + step - 1; hi = min(lo + 2*step - 1, len(a) - 1); merge(aux, lo, mid, hi);
Choose the last item as the pivot, then partition the array into items ≤ the pivot and items > pivot. Put the pivot in the middle then recursively sort left and right.
def quicksort(a): pivot = a[len(a) - 1] i = 0 j = len(a) - 2 while i <= j: if a[i] > pivot and a[j] <= pivot: swap(i, j) i += 1 j -= 1 else if a[i] <= pivot: i += 1 else: j -= 1 # ASSERT i == j + 1 # ASSERT all items to the left of i <= pivot swap(j, len(a) - 1) quicksort(a[0:j]) quicksort(a[j+1:end])
A quicksort-like algorithm can be used to compute the median and, more generally, order statistics.
This has recurrence:
However, the worst case is $O(n^2)$ as with quicksort. There exists a guaranteed linear time algorithm but it is much more complicated.
def heapsort(a): for i in range(len(a) // 2, 0 included): heapify(a[i], i, len(a)) for k in range(len(a), 1): # a[0:k] is a max-heap # a[k:end] is sorted swap(0, k - 1) heapify(a, 0, k-1) def heapify(a, iRoot, iEnd): if a[iRoot] satisfies max-heap: return j = largest child of iRoot swap(iRoot, j) heapify(a, j, iEnd)
Runtime $O(n \lg n)$
Counting sort does not require comparisons. Assuming that the inputs are positive integers within some range, it counts the number of each element, then finds the cumulative sum, from which we can identify exactly where a given element should go.
def counting_sort(a): count =  * max(a) for x in a: count[x] += 1 # cumulate for i in range(1, len(count)): count[i] += count[i-1] sorted =  * len(a) for i in range(len(a) - 1, 0 included): idx = count[a[i]] - 1 sorted[idx] = a[i] count[a[i]] -= 1 return sorted
Bucketsort creates an array of buckets (linked lists), with the assumption that elements will fall into these buckets uniformly. We can then run insertion sort within each bucket before concatenating the buckets into a sorted array.
def bucket_sort(a): # assuming that elements are drawn uniformly from [0,1] n = len(a) bucketWidth = 1/n buckets = new  of length n for x in a: idx = int(x / bucketWidth) bucket[idx].next = x sorted =  for b in buckets: insertion_sort(b) sorted.append(b) return b
We use insertion sort because each bucket should contain only one element on average. But the worst case is still $O(n^2)$ as a result.
Assuming all elements have the same number of digits, we use a stable sort each column starting from the least significant digit.
def radix_sort(a, d): for i in range(1, d): stable_sort(a on digit i)
$O(n)$ if counting sort is used for digits.
Dynamic programming tends to be useful when problems have the following features:
Memoization is a technique often used in top-down dynamic programming:
Used to traverse or search a graph.
def dfs(g, s): for v in g.vertices: v.visited = False s.visited = True stack = Stack() stack.push(s) while not stack.empty(): v = stack.pop() for w in v.neighbours: if not w.visited: stack.push(w) w.visited = True
Used to traverse or search a graph.
def bfs(g, s): for v in g.vertices: v.visited = False s.visited = True q = new Queue() q.push(s) while not q.empty(): v = q.pop() for w in v.neighbours(): if not w.visited: q.push(w) w.visited = True
previousfield for each node, then walk back from the target to the start.
distancefield contains the minimum distance from
sto that vertex.
def dijkstra(g, s): for v in g.vertices: v.distance = infinity s.distance = 0 pq = PriorityQueue(sortkey = lambda v: v.distance) pq.push(s) while not pq.empty(): v = pq.popmin() for (w, edgecost) in v.neighbours: dist = v.distance + edgecost if dist < w.distance: w.distance = dist if w in pq: pq.decreasekey(w) else: pq.push(w)
def bellman_ford(g, s): for v in g.vertices: v.minweight = infinity s.minweight = 0 # relax all edges repeat len(g.vertices) - 1 times: for (u, v, c) in g.edges: if v.minweight > (u.minweight + c): v.minweight = u.minweight + c # check for negative cycles in last pass for (u, v, c) in v.edges: if v.minweight > u.minweight + c: raise NegativeWeightCycle()
def johnson(g): h = new Graph() h.add_vertex(s, weights=[0, 0, 0,...]) bellman_ford(g, s) # Make edges positive for (u, v) in g.edges: w(u -> v) = h.u.distance + w(u -> v) - h.v.distance for v in g.vertices: dijkstra(g, v)
Runtime $O(VE + V^2 log V)$
def prim(g, s): for v in g.vertices: v.distance = infinity v.in_tree = False s.come_from = None s.distance = 0 s.in_tree = True pq = new PriorityQueue(sortkey = lambda v: v.distance) while not pq.empty(): v = pq.popmin() v.in_tree = True for (w, edgeweight) in v.neighbours: if edgeweight < w.distance and (not w.in_tree): w.distance = edgeweight w.come_from = v if w in pq: pq.decreasekey() else: pq.push(w) return [(w, w.come_from) for w in g.vertices excluding s]
Runtime the same as Dijkstra, i.e $O(E + V \log V)$.
def kruskal(g): tree_edges =  partition = DisjointSet() for v in g.vertices: partition.add_singleton(v) edges = sorted(g.edges, sortkey = edgeweight) for (u, v, edgeweight) in edges: p = partition.get_set_with(u) q = partition.get_set_with(v) if p != q: tree_edges.append((u, v)) parition.merge(p, q) return tree_edges
Runtime is dominated by the sort: $O(E \log E) = O(E \log V)$.
def topological_sort(g): for v in g.vertices: v.visited = False # v.colour = "white" totalorder =  for v in g.vertices: if not v.visited: visit(v, totalorder) return totalorder def visit(v, totalorder): v.visited = True # v.colour = "grey" for w in v.neighbours: if not w.visited: visit(w, totalorder) totalorder.prepend(v) # v.colour = "black"
Same runtime as DFS: $O(V+E)$
def find_augmenting_path(g): # helper graph h = new Graph(g.vertices) for each pair of vertices (v,w) in g: if f(v -> w) < c(v -> w): h.add_forward_edge(v -> w) if f(w -> v) > 0: h.add_backward_edge(v -> w) if h contains path(s to t): return path else: # no more paths return None def ford_fulkerson(g, s, t): # zero flow initially for (u, v) in g.edges: f(u -> v) = 0 while True: p = find_augmenting_path() if p is None: break delta = infinity # bottleneck for each edge (v1, v2) in p: if edge.forwards: delta = min(delta, c(v1 -> v2) - f(v1 -> v2)) else: delta = min(delta, f(v2 -> v1)) # Augment flow for each edge (v1, v2) in p: if edge.forwards: f(v1 -> v2) += delta else: f(v2 -> v1) -= delta
Runtime is $O(Ef^*)$
Used to find the corner points on a convex hull.
def graham_scan(points): let r0 be the lowest point r = [r1, r2, r3, ..., rn] = sort(points, sortkey=r.angle) S = new Stack([r1, r2, r3]) for i in range(3, n): while r[i] is not on the left of the segment(S.first(), S.second()): S.pop() S.push(r[i]) return S
Runtime is $O(n \log n)$ from sorting the points.