Min and max stacks

A stack supports push, pop, and top in O(1). Now an interviewer adds a fourth requirement: getMin returns the smallest value currently on the stack, also in O(1). Storing a single min field and updating it on push gets you most of the way there, until the moment the current minimum is popped. Then the field has to recompute the new minimum from the surviving elements, which is a linear scan. The constant-time bound breaks on every pop of the running min.

This is the chapter’s whole problem in one paragraph. The fix is older than the problem statement. In Algorithms, Sedgewick & Wayne give it as Exercise 1.3.58 with a one-line hint: “maintain two stacks, one which contains all of the items and another which contains the minima.”¹ The hint is the design technique. Pin a parallel auxiliary stack to the values stack and update both in lockstep, so the aggregate at any moment is the auxiliary’s top. Once you see it on the running minimum, the same technique scales to the running maximum, the running sum, the deferred bulk-update on the bottom-k elements, and the most-frequent-element-so-far. Each application changes the shape of the auxiliary; the pinning idea stays put.

The signature problem is LC 155 Min Stack, the canonical demonstration. Two more applications (LC 1381 with a lazy-increment auxiliary, LC 895 with a frequency-bucketed auxiliary) extend the technique into territory where the aggregate stops being a single integer. One STRETCH problem (LC 716 Max Stack, Premium) is included to mark the boundary where the technique stops applying. The monotonic-stack family lives next door in Monotonic stack and deque, and the boundary between this chapter and that one is the difference between augmenting a stack and evicting from one.

The API the interviewer is asking for

LC 155 wants four operations on a stack of integers, all in O(1):

top() and getMin() answer different questions. top() is recency. getMin() is an aggregate. A normal stack already does top() in O(1) because the top is at one end of an array; a normal stack does not do getMin() in O(1) because the minimum could live anywhere in the middle. The interesting design choice is what to store on push so that pop never has to scan.

The only thing the auxiliary records is the running minimum at each historical depth. When the values stack shrinks, the auxiliary shrinks with it; the minimum that was true two pushes ago, which the field-and-rescan design would have to recompute, sits exposed at the top of the auxiliary the moment the popped frame is gone.

Why a single min field cannot work

Consider four pushes in order: -2, 0, -3, 5. After all four, the running min is -3. Now pop. The top of the values stack was 5; the surviving values are [-2, 0, -3]; the running min is still -3. So far the field design is fine.

Pop again. The top was -3; the surviving values are [-2, 0]; the running min is now -2. The field, which was -3 a moment ago, has no way to know -2 without re-scanning the surviving values. There is no algebraic recurrence that turns a popped element back into the previous running min, because the previous min could be any element below it.²

The mirrored design records -2 explicitly at the second push, so when the third frame (the one that brought the min down to -3) is popped, the auxiliary’s new top is exactly the answer. No recomputation. The price is one extra integer of storage per element, paid in exchange for an unconditional pop on both stacks.

The design

Two stacks. Same length at all times. Pushes update both; pops shrink both; queries peek the relevant one.

class MinStack:
    def __init__(self):
        self.values = []
        self.mins = []

    def push(self, val):
        self.values.append(val)
        current = val if not self.mins else min(self.mins[-1], val)
        self.mins.append(current)

    def pop(self):
        self.values.pop()
        self.mins.pop()

    def top(self):
        return self.values[-1]

    def getMin(self):
        return self.mins[-1]

# LC 155. Min Stack
#           duplicate-min probe, and all-same probe all pass.
from typing import List


class MinStack:
    """LC 155 Min Stack with O(1) push, pop, top, getMin via mirrored aux stack."""

    def __init__(self) -> None:
        self.values: List[int] = []
        self.mins: List[int] = []

    def push(self, val: int) -> None:
        self.values.append(val)
        # min(prev, val) preserves duplicate minima on the aux stack.
        # Strict `val < self.mins[-1]` is the canonical bug; see chapter pitfalls.
        current = val if not self.mins else min(self.mins[-1], val)
        self.mins.append(current)

    def pop(self) -> None:
        self.values.pop()
        self.mins.pop()

    def top(self) -> int:
        return self.values[-1]

    def getMin(self) -> int:
        return self.mins[-1]

// LC 155. Min Stack
import java.util.ArrayDeque;
import java.util.Deque;

public final class Sol {

    /** LC 155 Min Stack: ArrayDeque, never java.util.Stack (synchronized Vector subclass). */
    public static final class MinStack {
        private final Deque<Integer> values = new ArrayDeque<>();
        private final Deque<Integer> mins = new ArrayDeque<>();

        public MinStack() {}

        public void push(int val) {
            values.push(val);
            int current = mins.isEmpty() ? val : Math.min(mins.peek(), val);
            mins.push(current);
        }

        public void pop() {
            values.pop();
            mins.pop();
        }

        public int top() {
            return values.peek();
        }

        public int getMin() {
            return mins.peek();
        }
    }

    private Sol() {}
}

// LC 155. Min Stack
#include <algorithm>
#include <stack>

namespace dsa {

class MinStack {
public:
    MinStack() = default;

    void push(int val) {
        values_.push(val);
        int current = mins_.empty() ? val : std::min(mins_.top(), val);
        mins_.push(current);
    }

    void pop() {
        values_.pop();
        mins_.pop();
    }

    int top() const { return values_.top(); }
    int getMin() const { return mins_.top(); }

private:
    std::stack<int> values_;
    std::stack<int> mins_;
};

}  // namespace dsa

// LC 155. Min Stack
package main

// MinStack supports O(1) Push, Pop, Top, GetMin via a parallel mins slice
// pinned to the values slice in lockstep.
type MinStack struct {
	values []int
	mins   []int
}

func NewMinStack() *MinStack { return &MinStack{} }

func (s *MinStack) Push(val int) {
	s.values = append(s.values, val)
	current := val
	if n := len(s.mins); n > 0 && s.mins[n-1] < current {
		current = s.mins[n-1]
	}
	s.mins = append(s.mins, current)
}

func (s *MinStack) Pop() {
	s.values = s.values[:len(s.values)-1]
	s.mins = s.mins[:len(s.mins)-1]
}

func (s *MinStack) Top() int    { return s.values[len(s.values)-1] }
func (s *MinStack) GetMin() int { return s.mins[len(s.mins)-1] }

The line worth staring at is current = val if not self.mins else min(self.mins[-1], val). The min(prev, val) form preserves duplicate minima on the auxiliary, which matters more than it looks. Strict if val < self.mins[-1]: self.mins.append(val) is the canonical bug. Duplicates of the current minimum never make it onto the auxiliary, the lengths drift apart, and after one pop the auxiliary is reporting a stale value.

Pop is unconditional on both stacks. There is no branch for “did this pop change the minimum”; the auxiliary’s predecessor is the new minimum because every push wrote it there. This is what makes the lockstep argument carry. CLRS §10.1 gives the array implementation of a stack directly: a single S.top integer indexes the most recently inserted slot, push writes at S[S.top + 1] and increments, pop reads S[S.top] and decrements.³ The mirrored design is two such triples, indexed in lockstep. All four operations remain O(1).

The Java implementation uses Deque<Integer> stack = new ArrayDeque<>() rather than java.util.Stack; the legacy class extends the synchronized Vector and pays the lock tax on every push and pop even in single-threaded code. Sedgewick & Wayne write that “Java has a built in library called Stack, but you should avoid using it”; Oracle’s JDK 21 Javadoc on ArrayDeque says verbatim “This class is likely to be faster than Stack when used as a stack.”⁴ The C++ implementation uses std::stack<int>, which is a container adapter on std::deque whose interface is intentionally narrower than the underlying container (no iteration, no random access), so the LIFO contract is enforced by the type rather than by convention. Go has no native stack; the slice idiom (append to push, s = s[:len(s)-1] to pop) is the canonical shape and is footgun-free for a stack because both operations work at the tail.

Walking the canonical sequence

Trace the LC 155 example: push -2, push 0, push -3, getMin, pop, top, getMin. Expected output: -3, then 0, then -2.

flowchart LR
    S0["start<br/>values: []<br/>mins: []"] -->|"push -2"| S1["values: [-2]<br/>mins: [-2]"]
    S1 -->|"push 0"| S2["values: [-2, 0]<br/>mins: [-2, -2]"]
    S2 -->|"push -3"| S3["values: [-2, 0, -3]<br/>mins: [-2, -2, -3]"]
    S3 -->|"getMin = -3"| S3
    S3 -->|"pop"| S4["values: [-2, 0]<br/>mins: [-2, -2]"]
    S4 -->|"top = 0"| S4
    S4 -->|"getMin = -2"| S4

The values stack and the auxiliary stack always have the same length. The auxiliary’s top is always the running minimum across the values stack as it stands right now. Together, those two invariants are the chapter.

What it actually costs

The space cost is one auxiliary integer per push, doubling storage from n to 2n words. In the comparison-based model this is provably optimal for the O(1) getMin requirement: without an auxiliary, getMin after a pop forces a re-scan of the surviving values, which is Omega(n) in the worst case.² Sedgewick algs4 Exercise 1.3.58 specifies exactly this design and asserts the constant-time bound directly.¹

A clever variant tries to halve the space by storing val - cur_min on a single stack and updating cur_min whenever the difference is negative. The design is real, it appears in top-voted LC Discuss threads, and it has a real bug. On signed 32-bit, when val = INT_MIN and cur_min = INT_MAX, the difference overflows. LC 155’s input range is exactly [-2^31, 2^31 - 1], so the overflow is reachable on legitimate inputs.⁵ The mirrored two-stack design pays an extra integer per push and is overflow-safe by construction. Pay the storage; keep the correctness.

Scaling the auxiliary: deferred updates and frequency buckets

The min-stack design depends on one property of the aggregate: there is a recurrence agg[i+1] = f(agg[i], val) that survives a pop because the predecessor is recoverable from the auxiliary’s top(). Min, max, sum, and count-of-key all satisfy it. Two LeetCode problems push the technique into territory where the auxiliary stops being a single parallel integer.

Lazy increments on the bottom-k (LC 1381)

LC 1381 asks the stack to support inc(k, val): add val to each of the bottom k elements. The naive approach is a for i in range(min(k, len(values))): values[i] += val loop, which is O(k) per call. LC 1381’s constraints (1 <= k, maxSize <= 1000, up to 1000 calls) leave the naive solution within the time limit, but the chapter’s lesson is the auxiliary-pinning technique, not the time limit.

The auxiliary changes role. Instead of recording an aggregate of past pushes, it records pending updates to apply to future pops. Maintain a parallel array inc_aux the same length as the values stack, zero-initialised. inc(k, val) writes val to inc_aux[min(k, len) - 1], one O(1) write, no matter how large k is. The increment sits at the boundary index, dormant until pop reaches it.

def push(self, x):
    if len(self.values) < self.max_size:
        self.values.append(x)
        self.inc_aux.append(0)

def pop(self):
    if not self.values:
        return -1
    i = len(self.values) - 1
    result = self.values[i] + self.inc_aux[i]
    if i > 0:
        self.inc_aux[i - 1] += self.inc_aux[i]   # propagate downward
    self.values.pop()
    self.inc_aux.pop()
    return result

def increment(self, k, val):
    if not self.values:
        return
    i = min(k, len(self.values)) - 1
    self.inc_aux[i] += val

On pop, the popped index’s pending increment is added to its predecessor before the pop returns, so the increment continues to apply to elements still on the stack. The invariant: inc_aux[i] is the cumulative pending increment that has not yet been applied to values[0..i] and will be applied lazily as elements are popped down to i.⁶ Each operation does a constant number of writes; all four are O(1).

The shape of the technique is identical to LC 155. A parallel auxiliary, pinned to the values stack in lockstep, carries the work that would otherwise force a linear scan. The only structural change is the direction of the work: LC 155’s auxiliary records past aggregates and is read on query; LC 1381’s auxiliary records pending updates and is read on pop. The lockstep is the constant.

Frequency buckets and a running max (LC 895)

LC 895 stretches the auxiliary further. pop() must return the most-frequent element pushed so far, breaking ties by recency (the most recent push of the max-frequency value wins). There is no recurrence on a single integer that captures “the most frequent”. A pop of a non-max-frequency value changes the frequency landscape only locally, but a pop of the max-frequency value can change which value is most frequent, and recomputing that from the surviving stack is Omega(n).

The auxiliary becomes two structures plus a running variable:

• freq[v]: the current count of value v. A hash map.
• group[k]: a stack of values whose count reached k at some push. A hash map of stacks. A value pushed three times appears in group[1], group[2], and group[3].
• maxFreq: the maximum count touched so far. A single integer.

def push(self, val):
    self.freq[val] = self.freq.get(val, 0) + 1
    f = self.freq[val]
    if f > self.maxFreq:
        self.maxFreq = f
    self.group.setdefault(f, []).append(val)

def pop(self):
    val = self.group[self.maxFreq].pop()
    self.freq[val] -= 1
    if not self.group[self.maxFreq]:
        self.maxFreq -= 1
    return val

Push increments the value’s frequency, appends it to the count-bucket at its new frequency, and bumps maxFreq if needed. Pop reads from group[maxFreq] (the most recent value at the maximum count, which gives the recency tiebreak for free), decrements the frequency, and decrements maxFreq if its bucket emptied.⁷ Each operation does a constant number of hash-map and stack ops.

The running maxFreq variable is the trick. Without it, pop would have to find the largest non-empty key in group, which is not an O(1) operation on a hash map. Tracking the max as a single integer that decrements only when its bucket empties is the auxiliary-pinning technique applied one level up: the running aggregate is again a single integer, but the structures it indexes into are richer.

The pattern across the three problems:

Three problems, one technique. The auxiliary’s shape tracks the aggregate’s complexity; the pinning (push and pop and query in lockstep with the primary) is invariant.

Where the technique stops applying

LC 716 Max Stack looks like a sibling of LC 155: same four operations, swap min for max. It would be, except for one extra requirement. LC 716 also wants popMax(), which removes the maximum from the stack regardless of where it sits, returning the most-recent copy on ties.

The lockstep auxiliary cannot do this. Removing a non-top element from the values stack is a O(n) splice; whatever the auxiliary records, it cannot avoid that splice, because the values stack must preserve every other element’s position. The canonical LC 716 solution drops the parallel-stack design entirely and uses a doubly linked list (for O(1) push and pop at the tail) plus an ordered multiset that maps each value to the list nodes carrying it. popMax reads the last key of the multiset in O(log n), removes the most recent matching list node in O(1), and erases the multiset entry in O(log n).⁸

flowchart TD
    A["stack with O(1) push, pop, top"] --> B{"fourth op operates on top only?"}
    B -- "yes: getMin, getMax, mostFrequent, lazy inc" --> C["lockstep auxiliary; O(1) per op<br/>LC 155, LC 1381, LC 895"]
    B -- "no: popMax removes a non-top" --> D["lockstep breaks<br/>doubly linked list + ordered multimap<br/>O(log n) per op<br/>LC 716"]

This boundary matters more than LC 716 itself. Whenever an interview question asks for an O(1) augmented operation on a stack, check what the operation touches. If it reads or removes only the top, the auxiliary-stack technique is on the table. If it reaches into the middle, the answer is a different data structure, and the question has just left the chapter.

Five places where this goes wrong

Confusing this with the monotonic-stack pattern

A reader who has met monotonic stacks first sometimes tries to apply that discipline to LC 155: push only if the new value is less than the current top, otherwise replace the top. This breaks LC 155’s top() contract. top() must return the most recently pushed value, not the running minimum, and replacing the top corrupts the values stack.

The two patterns differ on a single axis. The augmented-stack pattern (this chapter) keeps two stacks in lockstep with no eviction; every push is recorded, every pop pops both. The monotonic-stack pattern (Monotonic stack and deque) keeps one stack with eviction; pushes that violate the monotonic invariant pop offending elements before the new push lands. Augmented stacks preserve every push; monotonic stacks discard violators. If the values stack must answer top() correctly across an arbitrary push history, you are in this chapter. If the stack’s purpose is to expose “the next smaller element to the right” for each input, you are in 4.1.

Problem ladder

CORE (solve all three to claim chapter mastery)

• LC 155: Min Stack [Medium] • stack-with-aux-min-stack ★
• LC 1381: Design a Stack With Increment Operation [Medium] • stack-with-lazy-aux-increment (reference: Python only)
• LC 895: Maximum Frequency Stack [Hard] • stack-with-frequency-bucket-aux (reference: Python only)

STRETCH (optional)

• LC 716: Max Stack [Hard] • stack-with-doubly-linked-list-and-multimap (Premium; the boundary case where the chapter’s technique stops applying)

The pattern admits no canonical Easy LeetCode problem; an O(1) auxiliary aggregate query on a stack is structurally Medium-and-up. LC 155 is the spine and the only ladder problem with verified four-language reference code; LC 1381 and LC 895 carry inline Python sketches in the prose above. LC 716 is Premium-locked, listed as STRETCH only so the chapter can name the boundary at which the lockstep auxiliary stops working.

Where this leads

The augmented-stack pattern’s natural successors live two chapters away. Expression evaluation pairs a value stack with an operator stack: two stacks in lockstep again, but with different update rules and a richer state machine for operator precedence. Queue from stacks keeps the lockstep idea and inverts it: two stacks reproduce a FIFO queue’s contract by pushing on one and popping from the other.

Further afield, the LRU and LFU caches in Part 13 are the same auxiliary-pinning idea at one level higher complexity: a doubly linked list pinned to a hash map, with an eviction policy layered on top. And the lazy-increment auxiliary from LC 1381 is the simplest possible setting for the lazy-propagation pattern that powers segment trees with range updates in Part 11. Same technique, richer auxiliary, harder invariants.

References