kartikkukreja

OPT[0] = 0
for j = 1 to n:
  for i = 1 to j:
    compute e(i, j) for the segment pi, pi+1, ..., pj		
for j = 1 to n:
  OPT(j) = min((e(i,j) + C + OPT[i-1]) forall 1 ≤ i ≤ j)
return OPT[n]

kartikkukreja

OPT(j):
  if j == 0:
    return 0
  return max(wj + OPT(p(j)), OPT(j-1))

kartikkukreja

Let R be the set of all requests
Let d = 0 be the number of resources
while !R.empty():
  Choose a request i ∈ R that has the earliest start time
  if i can be assigned to some resource k <= d:
    assign request i to resource k
  else:
    allocate a new resource d+1
    assign request i to resource d+1
    d = d + 1

kartikkukreja

sort the n requests in order of start time
d = 1
assign request 1 to resource 1
min_pq.insert(1, f(i))

for i = 2 to n:
  j = min_pq.minIndex()
  if s(i) >= min_pq.keyOf(j):
    assign request i to resource j
    min_pq.increaseKey(j, f(i))

kartikkukreja

Let R be the set of all requests
schedule = {}
while !R.empty():
  Choose a request i ∈ R that has the smallest finish time
  Add i to schedule
  Delete all requests from R that are incompatible with i
return schedule

kartikkukreja

sort the n requests in order of finish time
schedule = {}
last_finish = 0
for i = 1 to n:
  if s(i) >= last_finish:
    Add i to schedule
    last_finish = f(i)
return schedule

kartikkukreja

Initially all m ∈ M and w ∈ W are free
while ∃ m who is free and hasn't proposed to every woman:
  choose m
  Let w be the highest-ranked woman in m's preference list to whom m has not yet proposed
  if w is free:
    (m, w) become engaged
  elif w is currently engaged to m':
    if w prefers m' to m:
      m remains free
    else w prefers m to m':

kartikkukreja

Assume graph G is connected. Otherwise, we can run the algorithm for each connected component of G.
Let q be an empty queue

Pick any vertex s ∈ V and color it Red
q.enqueue(s)

while !q.empty()
  u = q.dequeue()
  foreach v in u.adjList:
    if v.color is nil:

kartikkukreja

interpolation_search(A[0...n-1], x):
  i = 0, j = n-1, lo = A[i], hi = A[j]

  if x < lo: 		
    return -1 	
  if x >= hi:
    i = j

  while i < j: 		
    k = i + ((j - i) * (x - lo)) / (hi - lo) 		

kartikkukreja

foreach p of the remaining n-3 points in sorted order
  while (point next to top in S, S.top(), p) do not form a counter-clockwise turn
    S.pop()
  S.push(p)
return S