Stacks: keep unresolved workLESSON 1.05 · 5 OF 23 IN CHAPTER
PART A / Data structures and algorithms
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LESSON 1.05 · 5 OF 23 IN CHAPTERWorked lesson

Stacks: keep unresolved work

“For [73,74,71,75], return how long each day waits for a strictly warmer one.”

The output is [1,2,1,0]. Equal temperatures do not qualify. Identify the unresolved days before trying to optimize repeated forward scans.

The coding-practice chapter will apply this tool to complete problems. Here, focus on the mechanism and trace how its state changes.

Working example: For each temperature, return days until a strictly warmer day. [73,74,71,75] → [1,2,1,0].

Stacks: keep unresolved work

The idea: Store unresolved indices in non-increasing temperature order. A new warmer value resolves colder indices at the top.

First, what is a stack?

A stack keeps the most recently added item on top. Python lists can push with append and pop with pop. Here the stack holds day positions, not temperatures: we must eventually return how many days each earlier position waited. The temperatures at those positions are non-increasing from the bottom to the top of the stack.

temperatures = [73, 74, 71, 75]
waiting = []                  # indices still waiting for a warmer day
answer = [0] * len(temperatures)
for day, temperature in enumerate(temperatures):
    while waiting and temperatures[waiting[-1]] < temperature:
        earlier = waiting.pop()
        answer[earlier] = day - earlier
    waiting.append(day)
print(answer)  # [1, 2, 1, 0]

At 74, day 0 is resolved. At 75, both 71 (day 2) and 74 (day 1) are popped. Equal temperatures stay on the stack because the requirement is strictly warmer, not at least as warm. The animation's moving indices represent unresolved days waiting for an answer.

Day / temperature Stack after day (indices) Answer so far
0 / 73 [0] [0, 0, 0, 0]
1 / 74 [1] [1, 0, 0, 0]
2 / 71 [1, 2] [1, 0, 0, 0]
3 / 75 [3] [1, 2, 1, 0]

Check the mechanism

Predict each expected result, then trace the state that produces it. Explain the boundary case before opening the reference.

Cost: Scan forward for each day: O(n²). Monotonic stack: O(n) time and O(n) space.

01 · Try this input

Input / starting state
[73, 74, 71, 75]
Expected result
[1, 2, 1, 0]

Why: Day 3 resolves days 1 and 2.

02 · Try this input

Input / starting state
[70, 70]
Expected result
[0, 0]

Why: Equal is not warmer.

03 · Try this input

Input / starting state
[75, 74, 73]
Expected result
[0, 0, 0]

Why: No later warmer day.

04 · Try this input

Input / starting state
[70, 60, 80]
Expected result
[2, 1, 0]

Why: One arrival resolves two waiting days.

05 · Try this input

Input / starting state
[]
Expected result
[]

Why: No days.

The while loop looks nested, but each of n indices enters the stack once and leaves it at most once, so total push/pop work is O(n). At worst a decreasing list leaves all n indices waiting: O(n) extra space. Trying every later day for every start is O(n²).

Pass before moving on: Prove linear total work by counting pushes and pops, even with the nested loop.

Change strictly warmer to warmer-or-equal. Which comparison changes?

After attempting: reference and explanation

Compare daily_temperatures in algorithms.py (download file, source below). Use pattern notes for the invariant and contracts for complexity edge cases. Reimplement tomorrow without copying.

algorithms.py · algorithms.py
"""Reference solutions. Try the exercises in README.md before opening this file."""
from collections import Counter, OrderedDict, deque
from heapq import heappop, heappush, nlargest


def two_sum(nums, target):
    visited = {}
    for j, value in enumerate(nums):
        complement = target - value
        if complement in visited:
            return visited[complement], j
        visited.setdefault(value, j)
    return None


def longest_unique(text):
    left = best = 0
    last_seen = {}
    for right, char in enumerate(text):
        left = max(left, last_seen.get(char, -1) + 1)
        best = max(best, right - left + 1)
        last_seen[char] = right
    return best


def subarray_sum(nums, target):
    prefix_counts = {0: 1}
    prefix = result = 0
    for value in nums:
        prefix += value
        result += prefix_counts.get(prefix - target, 0)
        prefix_counts[prefix] = prefix_counts.get(prefix, 0) + 1
    return result


def lower_bound(nums, target):
    lo, hi = 0, len(nums)
    while lo < hi:
        mid = lo + (hi - lo) // 2
        if nums[mid] < target:
            lo = mid + 1
        else:
            hi = mid
    return lo


def merge_intervals(intervals):
    """Closed intervals; touching endpoints merge. Does not mutate input."""
    out = []
    for start, end in sorted(intervals):
        if start > end:
            raise ValueError('reversed interval')
        if out and start <= out[-1][1]:
            out[-1][1] = max(out[-1][1], end)
        else:
            out.append([start, end])
    return out


def top_k_frequent(nums, k):
    if k < 0:
        raise ValueError('k must be nonnegative')
    counts = Counter(nums)
    # Higher count wins; smaller value breaks ties deterministically.
    return nlargest(k, counts, key=lambda value: (counts[value], -value))


def islands(grid):
    """Rectangular list of lists of '0'/'1'; input remains unchanged."""
    if not grid:
        return 0
    cols = len(grid[0])
    if any(len(row) != cols or any(x not in ('0', '1') for x in row) for row in grid):
        raise ValueError('expected rectangular binary grid')
    seen = set()
    count = 0
    for r, row in enumerate(grid):
        for c, value in enumerate(row):
            if value != '1' or (r, c) in seen:
                continue
            count += 1
            seen.add((r, c))
            queue = deque([(r, c)])
            while queue:
                x, y = queue.popleft()
                for a, b in ((x-1, y), (x+1, y), (x, y-1), (x, y+1)):
                    if 0 <= a < len(grid) and 0 <= b < cols and grid[a][b] == '1' and (a, b) not in seen:
                        seen.add((a, b))  # Mark on enqueue, not dequeue.
                        queue.append((a, b))
    return count


def course_order(n, prerequisites):
    """Each pair is (course, prerequisite); [] means a cycle or no courses."""
    if n < 0:
        raise ValueError('negative course count')
    adjacency = [set() for _ in range(n)]
    degree = [0] * n
    for course, prerequisite in prerequisites:
        if not 0 <= course < n or not 0 <= prerequisite < n:
            raise ValueError('course outside graph')
        if course not in adjacency[prerequisite]:
            adjacency[prerequisite].add(course)
            degree[course] += 1
    queue = deque(i for i, d in enumerate(degree) if d == 0)
    order = []
    while queue:
        node = queue.popleft()
        order.append(node)
        for neighbor in adjacency[node]:
            degree[neighbor] -= 1
            if degree[neighbor] == 0:
                queue.append(neighbor)
    return order if len(order) == n else []


def shortest_paths(n, edges, source):
    """Directed nonnegative weighted graph; unreachable distances are infinity."""
    if not 0 <= source < n:
        raise ValueError('invalid source')
    graph = [[] for _ in range(n)]
    for u, v, weight in edges:
        if not 0 <= u < n or not 0 <= v < n or weight < 0:
            raise ValueError('invalid edge')
        graph[u].append((v, weight))
    distance = [float('inf')] * n
    distance[source] = 0
    heap = [(0, source)]
    while heap:
        cost, node = heappop(heap)
        if cost != distance[node]:
            continue
        for neighbor, weight in graph[node]:
            candidate = cost + weight
            if candidate < distance[neighbor]:
                distance[neighbor] = candidate
                heappush(heap, (candidate, neighbor))
    return distance


class LRU:
    """Capacity in entries; None denotes a miss. Not thread-safe."""
    def __init__(self, capacity):
        if capacity < 0:
            raise ValueError('negative capacity')
        self.capacity = capacity
        self.data = OrderedDict()

    def get(self, key):
        if key not in self.data:
            return None
        self.data.move_to_end(key)
        return self.data[key]

    def put(self, key, value):
        if self.capacity == 0:
            return
        self.data[key] = value
        self.data.move_to_end(key)
        if len(self.data) > self.capacity:
            self.data.popitem(last=False)


def min_coins(coins, amount):
    if amount < 0 or any(c <= 0 for c in coins):
        raise ValueError('nonnegative amount and positive coins required')
    dp = [0] + [amount + 1] * amount
    for total in range(1, amount + 1):
        for coin in coins:
            if coin <= total:
                dp[total] = min(dp[total], dp[total-coin] + 1)
    return -1 if dp[amount] > amount else dp[amount]


def daily_temperatures(temperatures):
    answer = [0] * len(temperatures)
    stack = []
    for i, temp in enumerate(temperatures):
        while stack and temperatures[stack[-1]] < temp:
            previous = stack.pop()
            answer[previous] = i - previous
        stack.append(i)
    return answer


def word_exists(board, word):
    """4-neighbor word search; do not reuse a cell or mutate the board."""
    if not word:
        return True
    if not board:
        return False
    cols = len(board[0])
    if any(len(row) != cols for row in board):
        raise ValueError('ragged board')
    def visit(r, c, index, used):
        if not (0 <= r < len(board) and 0 <= c < cols) or (r, c) in used or board[r][c] != word[index]:
            return False
        if index == len(word) - 1:
            return True
        used.add((r, c))
        found = any(visit(a, b, index+1, used) for a, b in ((r+1,c),(r-1,c),(r,c+1),(r,c-1)))
        used.remove((r, c))
        return found
    return any(visit(r, c, 0, set()) for r in range(len(board)) for c in range(cols))


class Trie:
    END = object()

    def __init__(self):
        self.root = {}

    def insert(self, word):
        node = self.root
        for char in word:
            node = node.setdefault(char, {})
        node[self.END] = True

    def contains(self, word):
        node = self.root
        for char in word:
            if char not in node:
                return False
            node = node[char]
        return self.END in node