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#刷题交流# 今天来打卡啦，给你一个整数数组 nums 和一个整数 k，请你用一个字符串返回其中出现频率前 k 高的元素。请按升序排列。
def solution(nums: list, k: int) -> str:
# from collections import Counter
# c = Counter(nums)
# return ','.join([str(v[0]) for v in c.most_common(k)])
# 统计每个元素的频率
import heapq
from collections import Counter
frequency = Counter(nums)
# 使用小根堆来存储频率最高的 k 个元素
heap = []
for num, freq in frequency.items():
heapq.heappush(heap, (freq, num))
if len(heap) > k:
heapq.heappop(heap)
# 提取堆中的元素并按升序排序
result = [num for freq, num in heap]
result.sort()
return ','.join([str(num) for num in result])
if __name__ == '__main__':
print(solution(nums=[1, 1, 1, 2, 2, 3], k=2) == "1,2")
print(solution(nums=[1], k=1) == "1")
print(solution(nums=[4, 4, 4, 2, 2, 2, 3, 3, 1], k=2) == "2,4")
def solution(nums: list, k: int) -> str:
# from collections import Counter
# c = Counter(nums)
# return ','.join([str(v[0]) for v in c.most_common(k)])
# 统计每个元素的频率
import heapq
from collections import Counter
frequency = Counter(nums)
# 使用小根堆来存储频率最高的 k 个元素
heap = []
for num, freq in frequency.items():
heapq.heappush(heap, (freq, num))
if len(heap) > k:
heapq.heappop(heap)
# 提取堆中的元素并按升序排序
result = [num for freq, num in heap]
result.sort()
return ','.join([str(num) for num in result])
if __name__ == '__main__':
print(solution(nums=[1, 1, 1, 2, 2, 3], k=2) == "1,2")
print(solution(nums=[1], k=1) == "1")
print(solution(nums=[4, 4, 4, 2, 2, 2, 3, 3, 1], k=2) == "2,4")
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#刷题交流# day17打卡：小R拿到了一个长度为n的数组，其中每个元素都是一个正整数。小R发现每次可以删除某个数组中某个数的一位数字，这样可以逐步将所有数字变为0。他想知道，要将数组中所有数字都变为0，最少需要多少步？
def solution(n: int, a: list) -> int:
assert n == len(a)
s = "".join(str(x) for x in a)
return len(s) - s.count('0')
if __name__ == '__main__':
print(solution(n = 5,a = [10, 13, 22, 100, 30]) == 7)
print(solution(n = 3,a = [5, 50, 505]) == 4)
print(solution(n = 4,a = [1000, 1, 10, 100]) == 4)
def solution(n: int, a: list) -> int:
assert n == len(a)
s = "".join(str(x) for x in a)
return len(s) - s.count('0')
if __name__ == '__main__':
print(solution(n = 5,a = [10, 13, 22, 100, 30]) == 7)
print(solution(n = 3,a = [5, 50, 505]) == 4)
print(solution(n = 4,a = [1000, 1, 10, 100]) == 4)
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#刷题交流# 刷题打卡：小F得到了一个特殊的字符串，这个字符串只包含字符A、S、D、F，其长度总是4的倍数。他的任务是通过尽可能少的替换，使得A、S、D、F这四个字符在字符串中出现的频次相等。求出实现这一条件的最小子串长度。
from collections import Counter
def solution(input:str)->int:
s = input
n = len(s)
assert n % 4 == 0
cnt = Counter(s)
if all(cnt[c] <= n // 4 for c in cnt):
return 0
l, r = 0, 0
res = n
while r < n:
cnt[s[r]] -= 1
r += 1
while all(cnt[c] <= n // 4 for c in cnt) and l < r:
res = min(res, r - l)
cnt[s[l]] += 1
l += 1
return res
if __name__ == '__main__':
print(solution(input = "ADDF") == 1)
print(solution(input = "ASAFASAFADDD") == 3)
print(solution(input = "SSDDFFFFAAAS") == 1)
print(solution(input = "AAAASSSSDDDDFFFF") == 0)
print(solution(input = "AAAADDDDAAAASSSS") == 4)
from collections import Counter
def solution(input:str)->int:
s = input
n = len(s)
assert n % 4 == 0
cnt = Counter(s)
if all(cnt[c] <= n // 4 for c in cnt):
return 0
l, r = 0, 0
res = n
while r < n:
cnt[s[r]] -= 1
r += 1
while all(cnt[c] <= n // 4 for c in cnt) and l < r:
res = min(res, r - l)
cnt[s[l]] += 1
l += 1
return res
if __name__ == '__main__':
print(solution(input = "ADDF") == 1)
print(solution(input = "ASAFASAFADDD") == 3)
print(solution(input = "SSDDFFFFAAAS") == 1)
print(solution(input = "AAAASSSSDDDDFFFF") == 0)
print(solution(input = "AAAADDDDAAAASSSS") == 4)
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#刷题交流# Day14刷题打卡，小U计划进行一场从地点A到地点B的徒步旅行，旅行总共需要 M 天。为了在旅途中确保安全，小U每天都需要消耗一份食物。在路程中，小U会经过一些补给站，这些补给站分布在不同的天数上，且每个补给站的食物价格各不相同。
小U需要在这些补给站中购买食物，以确保每天都有足够的食物。现在她想知道，如何规划在不同补给站的购买策略，以使她能够花费最少的钱顺利完成这次旅行。
M：总路程所需的天数。
N：路上补给站的数量。
p：每个补给站的描述，包含两个数字 A 和 B，表示第 A 天有一个补给站，并且该站每份食物的价格为 B 元。
保证第0天一定有一个补给站，并且补给站是按顺序出现的。
def solution(m:int, n:int, p:list)->int:
assert n == len(p) and m >= p[-1][0] and p[0][0] == 0
cost = 0
min_price, min_price_index = p[0][1], 0
for j in range(1, n):
a, b = p[j]
if b < min_price:
cost += min_price * (a - min_price_index)
min_price = b
min_price_index = a
cost += min_price * (m - min_price_index)
return cost
if __name__ == '__main__':
print(solution(m = 5 ,n = 4 ,p = [[0, 2], [1, 3], [2, 1], [3, 2]]) == 7)
print(solution(m = 6 ,n = 5 ,p = [[0, 1], [1, 5], [2, 2], [3, 4], [5, 1]]) == 6)
print(solution(m = 4 ,n = 3 ,p = [[0, 3], [2, 2], [3, 1]]) == 9)
小U需要在这些补给站中购买食物，以确保每天都有足够的食物。现在她想知道，如何规划在不同补给站的购买策略，以使她能够花费最少的钱顺利完成这次旅行。
M：总路程所需的天数。
N：路上补给站的数量。
p：每个补给站的描述，包含两个数字 A 和 B，表示第 A 天有一个补给站，并且该站每份食物的价格为 B 元。
保证第0天一定有一个补给站，并且补给站是按顺序出现的。
def solution(m:int, n:int, p:list)->int:
assert n == len(p) and m >= p[-1][0] and p[0][0] == 0
cost = 0
min_price, min_price_index = p[0][1], 0
for j in range(1, n):
a, b = p[j]
if b < min_price:
cost += min_price * (a - min_price_index)
min_price = b
min_price_index = a
cost += min_price * (m - min_price_index)
return cost
if __name__ == '__main__':
print(solution(m = 5 ,n = 4 ,p = [[0, 2], [1, 3], [2, 1], [3, 2]]) == 7)
print(solution(m = 6 ,n = 5 ,p = [[0, 1], [1, 5], [2, 2], [3, 4], [5, 1]]) == 6)
print(solution(m = 4 ,n = 3 ,p = [[0, 3], [2, 2], [3, 1]]) == 9)
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#刷题交流# 分享一道今天刷的題，猜猜是哪一道啊。
def solution(n: int, a: list) -> int:
max_contribution = 0
# 遍历所有可能的下标对 (i, j)
for i in range(n):
for j in range(n):
if i != j:
# 计算直接距离
direct_dist = abs(i - j)
# 计算环形距离
circular_dist = n - direct_dist
# 选择最短距离
min_dist = min(direct_dist, circular_dist)
# 计算贡献值
contribution = (a[i] + a[j]) * min_dist
# 更新最大贡献值
if contribution > max_contribution:
max_contribution = contribution
return max_contribution
if __name__ == '__main__':
print(solution(n = 3, a = [1, 2, 3]) == 5)
print(solution(n = 4, a = [4, 1, 2, 3]) == 12)
print(solution(n = 5, a = [1, 5, 3, 7, 2]) == 24)
def solution(n: int, a: list) -> int:
max_contribution = 0
# 遍历所有可能的下标对 (i, j)
for i in range(n):
for j in range(n):
if i != j:
# 计算直接距离
direct_dist = abs(i - j)
# 计算环形距离
circular_dist = n - direct_dist
# 选择最短距离
min_dist = min(direct_dist, circular_dist)
# 计算贡献值
contribution = (a[i] + a[j]) * min_dist
# 更新最大贡献值
if contribution > max_contribution:
max_contribution = contribution
return max_contribution
if __name__ == '__main__':
print(solution(n = 3, a = [1, 2, 3]) == 5)
print(solution(n = 4, a = [4, 1, 2, 3]) == 12)
print(solution(n = 5, a = [1, 5, 3, 7, 2]) == 24)
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#刷题交流# 小U决定在一个m×n 的地图上行走。地图中的每个位置都有一个高度，表示地形的高低。小U只能在满足以下条件的情况下移动：只能上坡或者下坡，不能走到高度相同的点。。def solution(m: int, n: int, a: list) -> int:
directions = [(-1, 0), (1, 0), (0, -1), (0, 1)]
visited = [[False] * n for _ in range(m)]
def dfs(x: int, y: int, is_up: bool) -> int:
visited[x][y] = True
max_steps = 0
for dx, dy in directions:
nx, ny = x + dx, y + dy
if 0 <= nx < m and 0 <= ny < n and not visited[nx][ny]:
if is_up and a[nx][ny] > a[x][y]:
max_steps = max(max_steps, 1 + dfs(nx, ny, False))
elif not is_up and a[nx][ny] < a[x][y]:
max_steps = max(max_steps, 1 + dfs(nx, ny, True))
visited[x][y] = False
return max_steps
max_moves = 0
for i in range(m):
for j in range(n):
max_moves = max(max_moves, dfs(i, j, True))
max_moves = max(max_moves, dfs(i, j, False)) return max_moves
directions = [(-1, 0), (1, 0), (0, -1), (0, 1)]
visited = [[False] * n for _ in range(m)]
def dfs(x: int, y: int, is_up: bool) -> int:
visited[x][y] = True
max_steps = 0
for dx, dy in directions:
nx, ny = x + dx, y + dy
if 0 <= nx < m and 0 <= ny < n and not visited[nx][ny]:
if is_up and a[nx][ny] > a[x][y]:
max_steps = max(max_steps, 1 + dfs(nx, ny, False))
elif not is_up and a[nx][ny] < a[x][y]:
max_steps = max(max_steps, 1 + dfs(nx, ny, True))
visited[x][y] = False
return max_steps
max_moves = 0
for i in range(m):
for j in range(n):
max_moves = max(max_moves, dfs(i, j, True))
max_moves = max(max_moves, dfs(i, j, False)) return max_moves
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![[可怜]](http://lf-web-assets.juejin.cn/obj/juejin-web/xitu_juejin_web/img/jj_emoji_5.ece2a96.png)
#刷题交流# def solution(array):
# 初始化候选者和计数器
candidate = None
count = 0
# 第一阶段：找到可能的候选者
for num in array:
if count == 0:
candidate = num
count = 1
elif num == candidate:
count += 1
else:
count -= 1
# 第二阶段：验证候选者是否是多数元素
if array.count(candidate) > len(array) // 2:
return candidate
return 0 # 如果没有满足条件的元素，返回0
if __name__ == "__main__":
# 测试用例
print(solution([1, 3, 8, 2, 3, 1, 3, 3, 3]) == 3)
print(solution([1, 1, 1, 2, 3, 4, 1]) == 1)
print(solution([5, 5, 5, 5, 5]) == 5)
print(solution([6, 6, 6, 7, 7, 7, 7]) == 7)
# 初始化候选者和计数器
candidate = None
count = 0
# 第一阶段：找到可能的候选者
for num in array:
if count == 0:
candidate = num
count = 1
elif num == candidate:
count += 1
else:
count -= 1
# 第二阶段：验证候选者是否是多数元素
if array.count(candidate) > len(array) // 2:
return candidate
return 0 # 如果没有满足条件的元素，返回0
if __name__ == "__main__":
# 测试用例
print(solution([1, 3, 8, 2, 3, 1, 3, 3, 3]) == 3)
print(solution([1, 1, 1, 2, 3, 4, 1]) == 1)
print(solution([5, 5, 5, 5, 5]) == 5)
print(solution([6, 6, 6, 7, 7, 7, 7]) == 7)
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#刷题交流# 多米诺骨牌def solution(num, data):
forces = [0] * num
force = 0
for i in range(num):
if data[i] == 'R':
force = num
elif data[i] == 'L':
force = 0
else:
force = max(force - 1, 0)
forces[i] += force
force = 0
for i in range(num - 1, -1, -1):
if data[i] == 'L':
force = num
elif data[i] == 'R':
force = 0
else:
force = max(force - 1, 0)
forces[i] -= force
upright_positions = []
for i in range(num):
if forces[i] == 0: # 平衡点
upright_positions.append(i + 1) # 位置从 1 开始计数
# 返回结果
if upright_positions:
return f"{len(upright_positions)}:{','.join(map(str, upright_positions))}"
else:
return "0"
if __name__ == "__main__":
print(solution(14, ".L.R...LR..L..") == "4:3,6,13,14")
print(solution(5, "R....") == "0")
print(solution(1, ".") == "1:1")
forces = [0] * num
force = 0
for i in range(num):
if data[i] == 'R':
force = num
elif data[i] == 'L':
force = 0
else:
force = max(force - 1, 0)
forces[i] += force
force = 0
for i in range(num - 1, -1, -1):
if data[i] == 'L':
force = num
elif data[i] == 'R':
force = 0
else:
force = max(force - 1, 0)
forces[i] -= force
upright_positions = []
for i in range(num):
if forces[i] == 0: # 平衡点
upright_positions.append(i + 1) # 位置从 1 开始计数
# 返回结果
if upright_positions:
return f"{len(upright_positions)}:{','.join(map(str, upright_positions))}"
else:
return "0"
if __name__ == "__main__":
print(solution(14, ".L.R...LR..L..") == "4:3,6,13,14")
print(solution(5, "R....") == "0")
print(solution(1, ".") == "1:1")
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#刷题交流# 有点抽象的刷一下
def solution(a, b):
# 初始化两个指针，分别指向两个数组的末尾
i, j = len(a) - 1, len(b) - 1
result = []
# 从数组尾部开始寻找交集
while i >= 0 and j >= 0:
if a[i] == b[j]:
# 如果找到相同的元素，加入结果数组
result.append(a[i])
i -= 1
j -= 1
elif a[i] > b[j]:
# 如果 a[i] 较大，向前移动 i
i -= 1
else:
# 如果 b[j] 较大，向前移动 j
j -= 1
return result
def solution(a, b):
# 初始化两个指针，分别指向两个数组的末尾
i, j = len(a) - 1, len(b) - 1
result = []
# 从数组尾部开始寻找交集
while i >= 0 and j >= 0:
if a[i] == b[j]:
# 如果找到相同的元素，加入结果数组
result.append(a[i])
i -= 1
j -= 1
elif a[i] > b[j]:
# 如果 a[i] 较大，向前移动 i
i -= 1
else:
# 如果 b[j] 较大，向前移动 j
j -= 1
return result
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