These algorithms are coded in a python-like pseudocode, but in some cases the pseudocode is actually valid python!


\[f(n) \in O(g(n)) \Longleftrightarrow \exists N, c_1, c_2 >0 \text{ s.t. } \forall n > N: 0 \leq g(n) \leq k g(n)\]

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.

Sorting Algorithms

Insertion Sort

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  
\[T(n) = 2 (1 + 2 + \cdots + n - 1) = n(n-1)\]

Selection sort

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.

Binary insertion sort

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
                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:


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
            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)

Order statistics

A quicksort-like algorithm can be used to compute the median and, more generally, order statistics.

  1. Select a pivot and partition the array into subarrays of size $p$ and $n-p$
  2. If $k < p$, recursively look for the kth item in the lower partition.
  3. Otherwise recurse into the upper partition to find rank $k-p$

This has recurrence:

\[T(n) = f(n/2) + kn = O(n)\]

However, the worst case is $O(n^2)$ as with quicksort. There exists a guaranteed linear time algorithm but it is much more complicated.


  1. Turn the array into a max-heap in $O(n)$
  2. Swap last item with max, reduce heapsize, then heapify down.
  3. Repeat until heapsize is 0.
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:
    j = largest child of iRoot
    swap(iRoot, j)
    heapify(a, j, iEnd)

Runtime $O(n \lg n)$

Counting sort

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 = [0] * max(a)
    for x in a:
        count[x] += 1
    # cumulate
    for i in range(1, len(count)):
        count[i] += count[i-1]
    sorted = [0] * 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:
    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.

Radix sort

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

Dynamic programming tends to be useful when problems have the following features:

  1. There exist many choices each with some ‘score’
  2. The optimal solution is composed of optimal solutions to subproblems
  3. The subproblems overlap.

Memoization is a technique often used in top-down dynamic programming:

Greedy algorithms

Other solution strategies

Graph algorithms


Used to traverse or search a graph.

def dfs(g, s):
    for v in g.vertices:
        v.visited = False
    s.visited = True

    stack = Stack()
    while not stack.empty():
        v = stack.pop()
        for w in v.neighbours:
            if not w.visited:
            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()
    while not q.empty():
        v = q.pop()
        for w in v.neighbours():
            if not w.visited:
            w.visited = True


def dijkstra(g, s):
    for v in g.vertices:
        v.distance = infinity
    s.distance = 0
    pq = PriorityQueue(sortkey = lambda v: v.distance)
    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:

Bellman equation

\[W_{ij} = \begin{cases} 0, & i = j\\ \text{weight}(i \to j), & \text{if there is an edge} \\ \infty, & \text{otherwise} \\ \end{cases}\] \[\begin{aligned} M_{ij}^{(1)} &= W_{ij} \\ M_{ij}^{(l)} &= \min_k \{ W_{ik} + M_{kj}^{(l-1)}\} \end{aligned}\] \[M_{ij}^{(l)} = (W_{i1} + M_{1j}^{(l-1)}) \wedge (W_{i2} + M_{2j}^{(l-1)}) \wedge \cdots \wedge (W_{in} + M_{nj}^{(l-1)})\]


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()

$O(VE)$ runtime.

Johnson’s algorithm

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:
    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:
    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: $O(E \log E) = O(E \log V)$.

Topological sort

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)
    # 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
            # 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:

        delta = infinity # bottleneck
        for each edge (v1, v2) in p:
            if edge.forwards:
                delta = min(delta, c(v1 -> v2) - f(v1 -> v2))
                delta = min(delta, f(v2 -> v1))
        # Augment flow
        for each edge (v1, v2) in p:
            if edge.forwards: 
                f(v1 -> v2) += delta
                f(v2 -> v1) -= delta

Runtime is $O(Ef^*)$

Geometrical algorithms

Geometrical algorithms notes

Graham’s scan

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()):
    return S

Runtime is $O(n \log n)$ from sorting the points.