Stack
1/0
1.0x
Stack
O(1)
Space: O(n)
Pseudocode
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procedure push(S: stack, value)
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S.top ← S.top + 1
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S[S.top] ← value
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procedure pop(S: stack)
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if S.top < 0 then error "underflow"
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value ← S[S.top]
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S.top ← S.top - 1
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return value
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procedure peek(S: stack)
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return S[S.top]
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