Stack

1/0
1.0x

Stack

O(1)Space: O(n)

Pseudocode

1procedure push(S: stack, value)
2 S.top ← S.top + 1
3 S[S.top] ← value
4
5procedure pop(S: stack)
6 if S.top < 0 then error "underflow"
7 value ← S[S.top]
8 S.top ← S.top - 1
9 return value
10
11procedure peek(S: stack)
12 return S[S.top]