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There are 3n piles of coins of varying size, you and your friends will take piles of coins as follows:
In each step, you will choose any 3 piles of coins (not necessarily consecutive).
Of your choice, Alice will pick the pile with the maximum number of coins.
You will pick the next pile with maximum number of coins.
Your friend Bob will pick the last pile.
Repeat until there are no more piles of coins.
Given an array of integers piles where piles[i] is the number of coins in the ith pile.
Return the maximum number of coins which you can have.
Example 1:
Input: piles = [2,4,1,2,7,8]
Output: 9
Explanation: Choose the triplet (2, 7, 8), Alice Pick the pile with 8 coins, you the pile with 7 coins and Bob the last one.
Choose the triplet (1, 2, 4), Alice Pick the pile with 4 coins, you the pile with 2 coins and Bob the last one.
The maximum number of coins which you can have are: 7 + 2 = 9.
On the other hand if we choose this arrangement (1, 2, 8), (2, 4, 7) you only get 2 + 4 = 6 coins which is not optimal.
Example 2:
Input: piles = [2,4,5]
Output: 4
Example 3:
Input: piles = [9,8,7,6,5,1,2,3,4]
Output: 18
Constraints:
3 <= piles.length <= 10^5
piles.length % 3 == 0
1 <= piles[i] <= 10^4
|
# brute force
class Solution:
def maxCoins(self, piles: List[int]) -> int:
res = 0
piles.sort(reverse = 1)
for i in range(0, (len(piles) // 3) * 2, 2):
res += piles[i + 1]
return res
|
There are 3n piles of coins of varying size, you and your friends will take piles of coins as follows:
In each step, you will choose any 3 piles of coins (not necessarily consecutive).
Of your choice, Alice will pick the pile with the maximum number of coins.
You will pick the next pile with maximum number of coins.
Your friend Bob will pick the last pile.
Repeat until there are no more piles of coins.
Given an array of integers piles where piles[i] is the number of coins in the ith pile.
Return the maximum number of coins which you can have.
Example 1:
Input: piles = [2,4,1,2,7,8]
Output: 9
Explanation: Choose the triplet (2, 7, 8), Alice Pick the pile with 8 coins, you the pile with 7 coins and Bob the last one.
Choose the triplet (1, 2, 4), Alice Pick the pile with 4 coins, you the pile with 2 coins and Bob the last one.
The maximum number of coins which you can have are: 7 + 2 = 9.
On the other hand if we choose this arrangement (1, 2, 8), (2, 4, 7) you only get 2 + 4 = 6 coins which is not optimal.
Example 2:
Input: piles = [2,4,5]
Output: 4
Example 3:
Input: piles = [9,8,7,6,5,1,2,3,4]
Output: 18
Constraints:
3 <= piles.length <= 10^5
piles.length % 3 == 0
1 <= piles[i] <= 10^4
|
# brute force
class Solution:
def maxCoins(self, piles: List[int]) -> int:
res = 0
piles.sort(reverse = 1)
for i in range(0, (len(piles) // 3) * 2, 2):
res += piles[i + 1]
print(res)
return res
|
There are 3n piles of coins of varying size, you and your friends will take piles of coins as follows:
In each step, you will choose any 3 piles of coins (not necessarily consecutive).
Of your choice, Alice will pick the pile with the maximum number of coins.
You will pick the next pile with maximum number of coins.
Your friend Bob will pick the last pile.
Repeat until there are no more piles of coins.
Given an array of integers piles where piles[i] is the number of coins in the ith pile.
Return the maximum number of coins which you can have.
Example 1:
Input: piles = [2,4,1,2,7,8]
Output: 9
Explanation: Choose the triplet (2, 7, 8), Alice Pick the pile with 8 coins, you the pile with 7 coins and Bob the last one.
Choose the triplet (1, 2, 4), Alice Pick the pile with 4 coins, you the pile with 2 coins and Bob the last one.
The maximum number of coins which you can have are: 7 + 2 = 9.
On the other hand if we choose this arrangement (1, 2, 8), (2, 4, 7) you only get 2 + 4 = 6 coins which is not optimal.
Example 2:
Input: piles = [2,4,5]
Output: 4
Example 3:
Input: piles = [9,8,7,6,5,1,2,3,4]
Output: 18
Constraints:
3 <= piles.length <= 10^5
piles.length % 3 == 0
1 <= piles[i] <= 10^4
|
class Solution:
def maxCoins(self, piles: List[int]) -> int:
piles.sort()
piles = piles[::-1]
return sum(piles[x] for x in range(1, 2 * len(piles) // 3, 2))
|
There are 3n piles of coins of varying size, you and your friends will take piles of coins as follows:
In each step, you will choose any 3 piles of coins (not necessarily consecutive).
Of your choice, Alice will pick the pile with the maximum number of coins.
You will pick the next pile with maximum number of coins.
Your friend Bob will pick the last pile.
Repeat until there are no more piles of coins.
Given an array of integers piles where piles[i] is the number of coins in the ith pile.
Return the maximum number of coins which you can have.
Example 1:
Input: piles = [2,4,1,2,7,8]
Output: 9
Explanation: Choose the triplet (2, 7, 8), Alice Pick the pile with 8 coins, you the pile with 7 coins and Bob the last one.
Choose the triplet (1, 2, 4), Alice Pick the pile with 4 coins, you the pile with 2 coins and Bob the last one.
The maximum number of coins which you can have are: 7 + 2 = 9.
On the other hand if we choose this arrangement (1, 2, 8), (2, 4, 7) you only get 2 + 4 = 6 coins which is not optimal.
Example 2:
Input: piles = [2,4,5]
Output: 4
Example 3:
Input: piles = [9,8,7,6,5,1,2,3,4]
Output: 18
Constraints:
3 <= piles.length <= 10^5
piles.length % 3 == 0
1 <= piles[i] <= 10^4
|
class Solution:
def maxCoins(self, piles: List[int]) -> int:
if len(piles) == 3:
return piles[1]
# print(len(piles)%2)
# print(list((i, a) for i, a in enumerate(sorted(piles)[len(piles)//3:]) if i % 2 == len(piles)%2))
return sum(a for a in sorted(piles)[len(piles)//3::2])
|
There are 3n piles of coins of varying size, you and your friends will take piles of coins as follows:
In each step, you will choose any 3 piles of coins (not necessarily consecutive).
Of your choice, Alice will pick the pile with the maximum number of coins.
You will pick the next pile with maximum number of coins.
Your friend Bob will pick the last pile.
Repeat until there are no more piles of coins.
Given an array of integers piles where piles[i] is the number of coins in the ith pile.
Return the maximum number of coins which you can have.
Example 1:
Input: piles = [2,4,1,2,7,8]
Output: 9
Explanation: Choose the triplet (2, 7, 8), Alice Pick the pile with 8 coins, you the pile with 7 coins and Bob the last one.
Choose the triplet (1, 2, 4), Alice Pick the pile with 4 coins, you the pile with 2 coins and Bob the last one.
The maximum number of coins which you can have are: 7 + 2 = 9.
On the other hand if we choose this arrangement (1, 2, 8), (2, 4, 7) you only get 2 + 4 = 6 coins which is not optimal.
Example 2:
Input: piles = [2,4,5]
Output: 4
Example 3:
Input: piles = [9,8,7,6,5,1,2,3,4]
Output: 18
Constraints:
3 <= piles.length <= 10^5
piles.length % 3 == 0
1 <= piles[i] <= 10^4
|
class Solution:
def maxCoins(self, piles: List[int]) -> int:
def dps():
def remove_from_first(T, s):
assert(len(s) == 3)
L = list(T)
for x in s:
if x in L:
L.remove(x)
return tuple(L)
from itertools import combinations
from functools import lru_cache
@lru_cache(None)
def dp(T):
if len(T) == 0: return 0
if len(T) == 3: return sum(T) - max(T) - min(T)
mxl = 0
for cm in combinations(T, 3):
S = remove_from_first(T, cm)
mxl = max(mxl, dp(S) + sum(cm) - max(cm) - min(cm))
return mxl
return dp(tuple(piles))
# return dps()
# def bu_dp():
def iter_appr():
assert(len(piles)%3==0)
maxheap, minheap = [], []
counter = 0
from heapq import heappush, heappop
for x in piles:
heappush(maxheap, -x)
heappush(minheap, x)
alice, me, bob = 0, 0, 0
while maxheap and minheap and counter < len(piles):
alice += -heappop(maxheap)
me += -heappop(maxheap)
bob += heappop(minheap)
counter += 3
return me
return iter_appr()
|
There are 3n piles of coins of varying size, you and your friends will take piles of coins as follows:
In each step, you will choose any 3 piles of coins (not necessarily consecutive).
Of your choice, Alice will pick the pile with the maximum number of coins.
You will pick the next pile with maximum number of coins.
Your friend Bob will pick the last pile.
Repeat until there are no more piles of coins.
Given an array of integers piles where piles[i] is the number of coins in the ith pile.
Return the maximum number of coins which you can have.
Example 1:
Input: piles = [2,4,1,2,7,8]
Output: 9
Explanation: Choose the triplet (2, 7, 8), Alice Pick the pile with 8 coins, you the pile with 7 coins and Bob the last one.
Choose the triplet (1, 2, 4), Alice Pick the pile with 4 coins, you the pile with 2 coins and Bob the last one.
The maximum number of coins which you can have are: 7 + 2 = 9.
On the other hand if we choose this arrangement (1, 2, 8), (2, 4, 7) you only get 2 + 4 = 6 coins which is not optimal.
Example 2:
Input: piles = [2,4,5]
Output: 4
Example 3:
Input: piles = [9,8,7,6,5,1,2,3,4]
Output: 18
Constraints:
3 <= piles.length <= 10^5
piles.length % 3 == 0
1 <= piles[i] <= 10^4
|
class Solution:
def maxCoins(self, piles: List[int]) -> int:
ans = 0
piles.sort(reverse = True)
for i in range(len(piles)//3):
ans += piles[i*2+1]
return ans
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
from heapq import *
from collections import Counter
class Solution:
def minSetSize(self, arr: List[int]) -> int:
counter = Counter(arr)
size = len(arr)
# unique elements (remove half of them)
if len(counter) == size:
return (size - 1) // 2 + 1
max_heap = [(-freq, value) for value, freq in list(counter.items())]
heapify(max_heap)
removed = 0 # number of elements removed
removed_size = 0 # size of the remvoved set
while removed < size//2:
count, value = heappop(max_heap)
count = -count # change the count back to +ve
removed += count
removed_size += 1
return removed_size
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
freq = Counter(arr)
final_length = 0
half = len(arr)/2
reduce = 0
for n, f in freq.most_common():
final_length+=f
reduce+=1
if final_length>=half:
return reduce
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
counts=collections.Counter(arr)
counts=[count for number,count in counts.most_common()]
set_size=0
tot=0
for x in counts:
tot+=x
set_size+=1
if tot>=(len(arr)//2):
break
return set_size
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
hash_map = collections.Counter(arr)
hash_map = {k:v for k,v in sorted(hash_map.items(),key=lambda item: item[1])}
total_size = len(arr)
values = list(hash_map.values())[::-1]
# print(values)
ans=0
i=0
while total_size>len(arr)//2:
total_size-=values[i]
i+=1
ans+=1
return ans
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
N = len(arr)
target = math.ceil(N/2)
dic = collections.Counter(arr)
lst = [(num,cnt) for num,cnt in dic.items()]
lst.sort(key = lambda x: [-x[1],x[0]])
ans = 0
count = 0
for num,cnt in lst:
count += cnt
ans += 1
if count >= target:
return ans
return ans
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
from collections import Counter
class Solution:
def minSetSize(self, arr: List[int]) -> int:
cnt = Counter(arr)
temp = sorted([(v, k) for k, v in list(cnt.items())])
curr = 0
for i in range(1, len(temp)+1):
curr += temp[-i][0]
if curr >= len(arr)/2:
return i
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
from collections import defaultdict
from collections import OrderedDict
class Solution:
def createCount(self,arr):
res = defaultdict(lambda:0)
for num in arr:
res[num]+=1
res = sorted(list(res.items()),key = lambda x:x[1],reverse = True)
res = collections.OrderedDict(res)
return res
def minSetSize(self, arr: List[int]) -> int:
total = 0
num = 0
res = self.createCount(arr)
for key in res:
if total>=len(arr)//2:
return num
total+=res[key]
num+=1
return num
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
dic = {}
n_arr = []
res = 0
length = len(arr)
for i in arr:
if i in list(dic.keys()):
dic[i] += 1
else:
dic[i] = 1
for i in dic:
n_arr.append([dic[i], i])
n_arr = sorted(n_arr)
for i in range (len(n_arr) -1, -1, -1):
length -= n_arr[i][0]
res+=1
if length <= len(arr)//2:
break
return res
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
L = len(arr)
count = collections.Counter(arr)
freq = sorted(count.values())
size = 0
res = 0
for i in range(len(freq)-1, -1, -1):
size += freq[i]
res += 1
if size >= L/2:
return res
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
import math
from collections import Counter
a=Counter(arr)
b=list(a.values())
b.sort(reverse = True)
n=len(arr)
summ=0
for i in range(len(b)):
summ+=b[i]
if summ>=int(math.floor(0.5*n)):
return i+1
return None
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
from collections import Counter
class Solution:
def minSetSize(self, arr: List[int]) -> int:
freq = Counter(arr)
h = []
for num, freq in freq.items():
print(num, freq)
heappush(h, (-freq, num))
count = 0
ret = 0
while count < (len(arr) // 2):
count += -heappop(h)[0]
ret += 1
return ret
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
from collections import OrderedDict
class Solution:
def minSetSize(self, arr: List[int]) -> int:
fq = {}
for i in arr:
fq[i] = fq.get(i, 0) + 1
sorted_m = OrderedDict(sorted(list(fq.items()), key=lambda k:(k[1], k[0]), reverse=True))
sm = 0
req = len(arr)//2 + len(arr)%2
ans = 0
for i in sorted_m:
sm += sorted_m[i]
ans += 1
if sm>=req:
break
return ans
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, array: List[int]) -> int:
hashMap = dict()
for item in array:
if item not in hashMap:
hashMap[item] = 0
hashMap[item] += 1
length = len(array)
itemToBeRemoved = 0
halfOflength = len(array) // 2
for value in sorted(list(hashMap.values()), reverse = True):
length -= value
itemToBeRemoved += 1
if length <= halfOflength:
return itemToBeRemoved
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
import collections
class Solution:
def minSetSize(self, arr: List[int]) -> int:
count = sorted(list(collections.Counter(arr).values()), reverse=True)
toRemove = 0
size = 0
for c in count:
toRemove += c
size += 1
if toRemove >= len(arr) / 2:
break
return size
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
counter = collections.Counter(arr)
sums = [0 for i in range(len(counter))]
curr = 0
half = len(arr) // 2
if len(arr) % 2 == 1:
half += 1
res = len(counter)
for i, key in enumerate(sorted(counter.keys(), key=lambda x:counter[x], reverse=True)):
curr += counter[key]
if curr >= half:
res = min(res, i + 1)
return res
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
if len(set(arr))==1:
return 1
if len(set(arr))==len(arr):
if len(arr)%2==0:
return int(len(arr)/2)
else:
return (len(arr)//2) +1
u=list(set(arr))
d_u={}
for ele in u:
d_u[ele]= arr.count(ele)
d_u=sorted(list(d_u.items()), key=lambda x: x[1], reverse=True)
#print(d_u)
s=0
size=0
for ele in d_u:
size+=1
s+= ele[1]
if (s>= len(arr)/2) & (len(arr)%2==0):
return size
if (len(arr)%2==1) & (s>= (len(arr)//2+1) ):
return size
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
half = len(arr) / 2
counts = {}
for val in arr:
if val in counts.keys():
counts[val] += 1
else:
counts[val] = 1
count = 0
included = 0
for key in sorted(counts, key=counts.get, reverse=True):
count += counts[key]
included += 1
if (count >= half):
return included
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
d={}
for i in arr:
if i not in d:
d[i]=0
d[i]+=1
d=list(sorted(d.items(), key = lambda kv:kv[1],reverse=True))
sum=0
cnt=0
for i in range(len(d)):
if sum>=(len(arr)/2):
break
else:
sum=sum+d[i][1]
cnt+=1
return cnt
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
'''
Keep a dict of counts and add all of them to a heap. Then we remove from heap until size <= half and we return size of set that we have to remove
'''
from heapq import heapify, heappush, heappop
class Solution:
def minSetSize(self, arr: List[int]) -> int:
counts = collections.Counter(arr)
maxheap = []
for i, freq in list(counts.items()):
maxheap.append((-freq,i))
heapify(maxheap)
curN = len(arr)
sol = 0
while curN > len(arr) / 2:
freq, cur = heappop(maxheap)
curN += freq
sol += 1
return sol
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
half = len(arr)//2
count = ans = 0
counts = Counter(arr)
for k, v in counts.most_common():
ans += v
count+=1
if ans >= half:
return count
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
n = len(arr)//2
freq = {}
for i in arr:
if i in freq:
freq[i] += 1
else:
freq[i] = 1
res = [i for i in freq.values()]
res.sort(reverse = True)
total = 0
count = 0
for i in res:
total += i
count += 1
if total >= n:
return count
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
from collections import defaultdict
class Solution:
def minSetSize(self, arr: List[int]) -> int:
counter = defaultdict(int)
for a in arr:
counter[a] += 1
nums = list(counter.keys())
nums.sort(key = lambda x: counter[x], reverse = True)
count = 0
total = 0
half = (len(arr)+1) // 2
for n in nums:
total += counter[n]
count += 1
if total >= half: return count
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
import collections
class Solution:
def minSetSize(self, arr: List[int]) -> int:
d = collections.Counter(arr)
vals = sorted(list(d.values()),reverse=True)
ans = 0
s = 0
for i in range(len(vals)):
ans += 1
s += vals[i]
if s >= len(arr)//2:
return ans
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
H ={}
for x in arr:
H[x]=H.get(x,0)-1
L = list(H.values())
heapq.heapify(L)
t = 0
a = 0
# print(L)
M = len(arr)//2
while t < M:
x = heapq.heappop(L)
t+=(-x)
a+=1
return a
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
from collections import Counter
class Solution:
def minSetSize(self, arr: List[int]) -> int:
cnt = Counter(arr)
temp = sorted(list(cnt.keys()), key = lambda x: cnt[x])
curr = 0
for i in range(1, len(temp)+1):
curr += cnt[temp[-i]]
if curr >= len(arr)/2:
return i
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
import collections
class Solution:
def minSetSize(self, arr: List[int]) -> int:
freq = collections.Counter(arr)
comp = [(freq[key],key) for key in freq]
comp.sort(reverse=True)
target = len(arr)/2
index,curr_sum = 0,0
while index < len(comp) and curr_sum < target:
curr_sum += comp[index][0]
index +=1
return index
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
import numpy as np
import collections
class Solution:
def minSetSize(self, arr: List[int]) -> int:
dict_gens = collections.Counter(arr)
generators = sorted(dict_gens.values())
min_deactivate = np.ceil(len(arr)/2)
turn_off = 0
i = 1
while turn_off < min_deactivate:
turn_off += generators[-i]
i += 1
return i-1
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
if len(arr) == 0:
return 0
if len(arr) < 3:
return 1
counter = {} # Could use defaultdict to save a few lines of code
for n in arr:
if n in counter:
counter[n] += 1
else:
counter[n] = 1
counts = sorted(counter.values())
total_ints = len(arr)
set_size = 0
while total_ints > len(arr)//2:
set_size += 1
total_ints -= counts.pop()
return set_size
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
n, target = len(arr), len(arr)//2
counts = {}
for i in arr:
if i in counts:
counts[i] += 1
else:
counts[i] = 1
counts = sorted(list(counts.values()), reverse=True)
numRemoved = 0
for i in counts:
n -= i
numRemoved += 1
if n <= target:
return numRemoved
return numRemoved
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr) -> int:
counter = {}
for val in arr:
if val not in counter:
counter[val] = 0
counter[val] += 1
values = []
for val in counter:
values.append(counter[val])
values.sort(reverse=True)
total = 0
nhalf = len(arr) // 2
new_n = len(arr)
i = 0
while new_n > nhalf:
total += 1
new_n -= values[i]
i += 1
return total
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
length = len(arr)
counter = {}
for num in arr:
try:
counter[num] += 1
except:
counter[num] = 1
array = sorted(list(counter.items()), key = lambda x:(-x[1], x[0]))
# print(array)
res = 0
i = 0
while res * 2 < length:
res += array[i][1]
i += 1
return i
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
order = collections.Counter(arr).most_common()
ans = 0
sum = 0
mid = len(arr)//2 if len(arr)%2==0 else (len(arr)//2 +1)
for i,j in order:
sum+=j
ans+=1
if sum >= mid:
break
return ans
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
counter = collections.Counter(arr)
heap = [(-freq, num) for num, freq in list(counter.items())]
heapq.heapify(heap)
target = len(arr) // 2
ans = len(heap)
while target > 0:
target += heapq.heappop(heap)[0]
return ans - len(heap)
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
import collections
class Solution:
def minSetSize(self, arr: List[int]) -> int:
lst = []
dic = collections.Counter(arr)
keys = sorted(dic,key = lambda x: dic[x],reverse = True)
for n in keys:
if len(lst) >= len(arr)//2:
return len(set(lst))
else:
for _ in range(dic[n]):
lst.append(n)
if len(lst) >= len(arr)//2:
return len(set(lst))
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
counts = dict()
# count all distinct values
for i in arr:
counts[i] = counts.get(i, 0) + 1
#greedy solution
#sort counts in desc order
total_count=0
for index, count in enumerate(sorted(list(counts.values()), reverse=True)):
total_count += count
if total_count >= len(arr) // 2:
return index + 1
return 0
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
d = {}
for num in arr:
if num in d:
d[num] += 1
else:
d[num] = 1
val = [d[i] for i in d]
val.sort(reverse = True)
count, total = 0,0
for each in val:
total += each
count += 1
if total >= len(arr)//2:
break
return count
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
arr.sort()
counts = []
current_run = 1
for i in range(1, len(arr)):
if arr[i] == arr[i - 1]:
current_run += 1
continue
counts.append(current_run)
current_run = 1
counts.append(current_run)
# Reverse sort the counts.
counts.sort(reverse=True)
# Remove numbers until at least half are removed.
numbers_removed_from_arr = 0
set_size = 0
for count in counts:
numbers_removed_from_arr += count
set_size += 1
if (numbers_removed_from_arr >= len(arr) // 2):
break
return set_size
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
arrCounter = Counter(arr)
sortedList = sorted([(key, arrCounter[key]) for key in arrCounter], key = lambda x:(-x[1], x))
arrLen = len(arr)
target = arrLen//2
ans = 0
for _, aNum in sortedList:
arrLen -= aNum
ans += 1
if arrLen <= target:
return ans
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
from collections import Counter
class Solution:
def minSetSize(self, arr: List[int]) -> int:
half=len(arr)/2
count=0
dic=Counter(arr)
for i in dic.most_common():
half-=i[1]
count+=1
if half<=0:
return count
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
length=len(arr)
bar=length//2
d={}
for i in range(len(arr)):
d[arr[i]]=d.get(arr[i],0)+1
res=0
l=list(sorted(d.values(),reverse=True))
for value in l:
length-=value
res+=1
if length<=bar:
return res
return res
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
from collections import Counter
class Solution:
def minSetSize(self, arr: List[int]) -> int:
counts = Counter(arr)
elements = 0
count = len(arr) / 2
for (key, value_count) in sorted(list(counts.items()), key=lambda x: x[1], reverse=True):
if count <= 0:
break
elements += 1
count -= value_count
return elements
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
objective = len(arr)//2
res = ret = 0
c = collections.Counter(arr)
for k, v in c.most_common():
res += v
ret += 1
if res >= objective:
break
return ret
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
from collections import Counter
class Solution:
def minSetSize(self, arr: List[int]) -> int:
c = Counter(arr)
a, res = 0, 0
for n, c in c.most_common():
a += c
res += 1
if a >= len(arr) // 2:
break
return res
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
n = len(arr)
count = Counter(arr)
res = 0
s = 0
for key, rep in count.most_common():
s += 1
res += rep
if res >=n//2 :
return s
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
cnter = collections.Counter(arr).most_common()
counting = 0
for i, blep in enumerate(cnter):
if counting + blep[1] >= len(arr)/2:
return i + 1
else:
counting += blep[1]
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
c=Counter(arr)
c=list(c.values())
c.sort(reverse=True)
k=len(c)//2
s=0
count=0
for i in range(len(c)):
s+=c[i]
count+=1
if s>=len(arr)//2:
return count
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
from collections import Counter
count = 0
ans = 0
c = Counter(arr)
target = len(arr)/2
for k,v in sorted(list(c.items()), key=lambda x: x[1], reverse=True):
count+=v
ans+=1
if count >= target: return ans
return 0
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
c=collections.Counter(arr)
n=len(arr)
res=0
count=0
for c,v in c.most_common():
count+=v
res+=1
if count>=n//2:
break
return res
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
d = {}
for k in arr:
if k in d:
d[k] += 1
else:
d[k] = 1
t = []
for k, v in d.items():
t.append(v)
t.sort(reverse=True)
r = 0
ii = 0
while r < len(arr)//2 + len(arr)%1 :
r += t[ii]
ii+=1
return ii
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
num = {}
for i in arr:
if i in num:
num[i] += 1
else:
num[i] = 1
sort_nums = sorted(num.items(), key=lambda x: x[1], reverse=True)
total = 0
ind = 0
for i in sort_nums:
if total < len(arr) / 2:
total += i[1]
ind += 1
else:
break
return ind
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
ans=0
d={}
l=[]
res=0
for i in arr:
if i in d:
d[i]+=1
else:
d[i]=1
for i in d:
l.append(d[i])
l.sort(reverse=True)
print(l)
for i in range(len(l)):
res+=l[i]
if res>=(len(arr)//2):
return i+1
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
A = {}
for x in arr:
A[x] = A.get(x,0) + 1
p = 0
count = 0
for x in sorted(list(A.items()), key = lambda x: -x[1]):
p += x[1]
count += 1
if p >= len(arr) // 2:
return count
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
dic = {}
for num in arr:
dic[num] = dic.get(num, 0) + 1
array = [v for k, v in sorted(list(dic.items()), key=lambda item: item[1], reverse=True)]
target = len(arr)/2
current, i = 0, 0
for num in array:
i += 1
current += num
if current >= target:
return i
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
d = dict()
for x in arr:
d.setdefault(x, 0)
d[x] += 1
c = sorted(d.values(), reverse=True)
for i in range(1, len(c)):
c[i] += c[i-1]
for i, x in enumerate(c):
if x >= len(arr) // 2:
return i + 1
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
n = len(arr)
counter = collections.Counter(arr)
cur = 0
ans = 0
for num in sorted(list(counter.values()) ,reverse = True):
cur += num
ans += 1
if cur >= n//2:
return ans
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
mem = {}
for i in arr:
if i in mem:
mem[i]+=1
else:
mem[i]=1
sorted_array = sorted(list(mem.values()), reverse=True)
counter = 0
res = 0
print(sorted_array)
for i in sorted_array:
res +=i
counter += 1
print(res)
if res>=len(arr)/2:
return counter
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
d={}
for i in arr:
if i in list(d.keys()):
d[i]+=1
else:
d[i]=1
s= len(arr)
ans=0
for j in sorted(list(d.items()),key=lambda x:x[1],reverse=True):
s-=j[1]
ans+=1
if s<=(len(arr)//2):
return ans
else:
continue
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
tmp_dict = {}
for e in arr:
if e not in tmp_dict:
tmp_dict[e] = 1
else:
tmp_dict[e] += 1
tmp = [(key,tmp_dict[key]) for key in tmp_dict]
tmp.sort(key=lambda x:x[1],reverse=True)
res = []
cur = 0
i = 0
while cur<len(arr)//2:
res.append(tmp[i][0])
cur += tmp[i][1]
i += 1
return len(res)
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
from collections import Counter
class Solution:
def minSetSize(self, arr) -> int:
counterArr = [0] * (len(arr) + 1)
for value in Counter(arr).values():
counterArr[value] += 1
steps = 0
total = 0
print(counterArr)
for i in reversed(range(len(arr) + 1)):
num = counterArr[i]
for j in range(num):
total += i
steps += 1
if total >= len(arr) // 2:
return steps
return steps
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
occ = {}
for num in arr:
if num in occ:
occ[num] += 1
else:
occ[num] = 1
occ_lst = list(occ.items())
occ_lst.sort(key=lambda x: x[1], reverse=True)
total_elems_removed = 0
removal_set_size = 0
for i, (_, num_occur) in enumerate(occ_lst):
total_elems_removed += num_occur
if total_elems_removed >= math.ceil(len(arr) / 2.0):
return i + 1
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
h = {}
for i in arr:
if not i in h:
h[i] = 0
h[i] += 1
v = sorted(list(h.values()),reverse = True)
m = len(arr)
t = 0
p = 0
i = 0
print(v)
while t < m//2:
t += v[i]
i+=1
p+=1
return p
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
d = {}
n = len(arr)
for a in arr:
d[a] = d.get(a,0) + 1
val = sorted(list(d.items()), key = lambda x :x[1])[::-1]
count =0
for i,(_,val_i) in enumerate(val):
count += val_i
print(count)
if count>=n//2:
return i +1
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
dict_freq = {}
for i in range(len(arr)):
if arr[i] in dict_freq:
dict_freq[arr[i]] += 1
else:
dict_freq[arr[i]] = 1
target_size = len(arr)//2
sort_dict = dict(sorted(list(dict_freq.items()), key=lambda x: x[1], reverse=True))
print(sort_dict)
for i in dict_freq:
if dict_freq[i] >= target_size:
return 1
else:
break
cnt = 0
count = 0
for i in sort_dict:
cnt = cnt + sort_dict[i]
count = count+1
if cnt >= target_size:
return count
return 8
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
from collections import Counter
class Solution:
def minSetSize(self, arr: List[int]) -> int:
countOfArray = Counter(arr)
listOfValues = list(countOfArray.values())
listOfValues.sort(reverse = True)
lenOfArray = len(arr)
halfTheLength = lenOfArray//2
ret = 0
for val in listOfValues:
lenOfArray = lenOfArray - val
ret += 1
if lenOfArray <= halfTheLength :
break
return ret
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
if len(set(arr)) == 1:
return 1
maps = {x:0 for x in set(arr)}
for x in arr:
maps[x]+=1
maps = sorted(list(maps.items()), key=lambda x: x[1], reverse=True)
mincount = len(arr)//2
ans = 0
currLength,sub = len(arr),9
# print(maps)
for x in range(len(maps)):
key,val = maps[x]
if currLength > mincount:
currLength -= val
ans += 1
else:
return ans
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
arr_len = len(arr)
# arr_int = list(set(arr))
int_dict={}
val=[]
for i in arr:
int_dict[i] = int_dict.setdefault(i,0) + 1
for i in int_dict:
val.append(int_dict[i])
val.sort()
count_num=0
loc=0
for i in val[::-1]:
count_num+=i
loc+=1
if count_num>=arr_len/2:
break
return loc
# if len(set(arr)) == arr_len:
# return int(arr_len+0.5)
# else:
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
c = Counter(arr)
result = list()
count = 0
for k,v in sorted(c.items(),key = lambda item: item[1],reverse=True):
if count < len(arr)//2:
result.append(k)
count += v
return len(result)
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
d = {}
n = len(arr)
for a in arr:
d[a] = d.get(a,0) + 1
val = sorted(list(d.items()), key = lambda x :x[1])[::-1]
count =0
for i,(_,val_i) in enumerate(val):
count += val_i
if count>=n//2:
return i +1
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def floor_counter(self,n):
if n%2==0:
return n/2
else:
return (n+1)/2
def minSetSize(self, arr: List[int]) -> int:
ceiling = self.floor_counter(len(arr))
freq = {}
for elem in arr:
if elem in freq:
freq[elem] += 1
else:
freq[elem] = 1
sorted_freq = sorted(freq.items(), key=lambda k:k[1], reverse=True)
sum = 0
type_counter = []
for i in range(len(sorted_freq)):
sum +=(sorted_freq[i][1])
print(sum)
type_counter.append(sorted_freq[i][0])
if sum>=ceiling:
return len(type_counter)
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
from collections import defaultdict
class Solution:
def minSetSize(self, arr: List[int]) -> int:
occurances = defaultdict(int)
for num in arr:
occurances[num] += 1
dictlist = []
for key, value in occurances.items():
temp = [key,value]
dictlist.append(temp)
dictlist = sorted(dictlist, key=lambda x: x[1])
count = 0
removals =0
while (count < len(arr) / 2):
count += dictlist[-1][1]
dictlist.pop()
removals +=1
return removals
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
n = len(arr)
arr_map = Counter(arr)
arr_tup = [(item, arr_map[item]) for item in arr_map]
#print(arr_tup)
arr_tup = sorted(arr_tup, key= lambda x : -x[1]) #Decrementing based on frequency
#print(arr_tup)
count = 0
total = 0
for item in arr_tup:
total += item[1] #Counting max freq
print((item[0]))
count += 1
if total >= n/2:
return count
''' n = len(arr)
freq = {}
for i in arr:
if i in freq:
freq[i] += 1
else:
freq[i] = 1
new_set = set()
max_freq = max(freq, key = freq.get)
del(freq[max_freq])
new_set.add(max_freq)
arr = [val for val in arr if val != max_freq]
while True:
if len(arr) <= n//2:
return len(new_set)
else:
max_freq = max(freq, key = freq.get)
del(freq[max_freq])
new_set.add(max_freq)
arr = [val for val in arr if val != max_freq]'''
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
from collections import *
class Solution:
def minSetSize(self, arr: List[int]) -> int:
d = Counter()
for num in arr:
d[num] += 1
counts = list(d.values())
counts.sort(reverse=True)
total = 0
for (i,count) in enumerate(counts):
total += count
if total >= len(arr)//2:
return i+1
return None # should never reach here
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
import collections
class Solution:
def minSetSize(self, arr: List[int]) -> int:
minSizeToRemove = len(arr) //2
# print(minSizeToRemove)
counter = collections.Counter(arr)
# print(counter)
sortedCounters = sorted(counter.items(), key=lambda x: x[1], reverse=True)
sum = 0
minSet = set()
for i in sortedCounters:
sum+=i[1]
minSet.add(i[0])
# print(i[0], i[1], sum, minSet)
if sum >= minSizeToRemove:
break
return len(minSet)
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
# if not arr:
# return 0
# ceil = len(arr) / 2.0
# count = collections.Counter(arr)
# sort = count.most_common()
# sum = 0
# for idx,val in enumerate(sort):
# sum += val[1]
# if sum >= ceil:
# return idx+1
# return idx+1
if not arr:
return 0
counter = collections.Counter(arr)
max_val = max(counter.values())
buckets = [0] * (max_val+1)
for count in list(counter.values()):
buckets[count] += 1
set_size = 0
arr_num_to_remove = len(arr) // 2
bucket = max_val
while arr_num_to_remove > 0:
max_need_from_bucket = math.ceil(arr_num_to_remove / bucket)
set_size_increase = min(buckets[bucket],max_need_from_bucket)
set_size += set_size_increase
arr_num_to_remove -= set_size_increase * bucket
bucket -= 1
return set_size
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
numberDict = {}
for i in range(len(arr)):
if arr[i] not in numberDict:
numberDict[arr[i]] = 1
else:
numberDict[arr[i]] += 1
sortedOrders = sorted(list(numberDict.items()), key=lambda x: x[1], reverse=True)
total = 0
setSize = 0
for i in sortedOrders:
if total >= (len(arr)//2):
return setSize
total += i[1]
setSize += 1
return setSize
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
# total_count = len(arr)
# values = set(arr)
# values = list(values)
# possible_combinations = []
# for i in values:
# possible_combinations.append([i,arr.count(i)])
# result = [seq for i in range(len([item[1] for item in possible_combinations]), 0, -1) for seq in itertools.combinations([item[1] for item in possible_combinations], i) if sum(seq) >= total_count//2]
# return min(len(i) for i in result)
total_count = 0
arr_mc = collections.Counter(arr).most_common()
print(arr_mc)
for i in range(len(arr_mc)):
total_count += arr_mc[i][1]
if total_count >= len(arr)/2:
return i + 1
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
from collections import Counter
class Solution:
def minSetSize(self, arr: List[int]) -> int:
sz = len(arr)
d = sorted(list(Counter(arr).items()), key=lambda x: -x[1])
num = 0
numElems = 0
for e in d:
numElems += e[1]
if numElems < sz//2:
num += 1
else:
return num+1
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
d = {}
n = len(arr)
for num in arr:
if num not in d:
d[num] = 0
d[num] += 1
num_freq = list(sorted([v for k,v in d.items()]))
tot = 0
m = len(num_freq)
i = 0
while(tot < n/2 and i<m):
tot += num_freq[m-i-1]
i+=1
print(i)
print(num_freq)
return i
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def check(self, arr, to_remove):
return len(tuple(filter(lambda x: not to_remove.contains(x), arr))) <= len(arr) / 2
def minSetSize(self, arr: List[int]) -> int:
all_nums = set(arr)
freq = {n:0 for n in all_nums}
for num in arr:
freq[num] += 1
nums_by_count = sorted(freq.items(), key=lambda x: x[1], reverse=True)
remove_count = 0
result = 0
half = len(arr) / 2
for item in nums_by_count:
num, count = item
remove_count += count
result += 1
if remove_count >= half:
return result
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
dict_={}
for num in arr:
if num not in dict_:
dict_[num]=1
else:
dict_[num]+=1
sort_list = sorted(dict_.keys(),key=lambda x: dict_[x],reverse=True)
n = len(arr)
size = 0
for num in sort_list:
n-=dict_[num]
size+=1
if n<len(arr)//2+1:
return size
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
# get length of array
length = len(arr)
# build dict to count how many times each int appears
counts = {}
for num in arr:
if num not in counts:
counts[num] =1
else:
counts[num] += 1
# print(counts)
# get values from dict, sort in descending order
descending = sorted(counts.values(), reverse = True)
# print(descending)
# initialize 2 variables: count and total
count = 0
total = 0
# loop over descending list of counts
for num in descending:
# add each number to our total
total += num
# increment count by 1
count += 1
# if our total is half or more, return count
if total >= length/2:
return count
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
from collections import Counter
c = Counter (arr)
ls = []
for i, c in c.items ():
ls .append ((c, i))
ls.sort (reverse=True)
#print (ls)
l = 0
r = len (ls)-1
res = 0
items = 0
for i in range (len(ls)):
res += 1
items += ls[i][0]
#print (items)
if items >= (len(arr)//2):
break
return res
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
arr.sort()
freq=[]
k=0
while k<len(arr):
j=k+1
fr=1
while j<len(arr):
if arr[j]==arr[k]:
fr+=1
else:
break
j+=1
k+=1
freq.append(fr)
k+=1
freq.sort()
freq.reverse()
k=0
no=0
count=0
while k<len(freq) and count<len(arr)/2:
if freq[0]>=len(arr)/2:
no+=1
break
count=count+freq[k]
no+=1
k+=1
return no
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
# get counts of each element
counts = {}
for i, elem in enumerate(arr):
counts[elem] = 1 if elem not in counts else counts[elem]+1
# keep getting max and checking if the length is less than half upon removal until it is
size = len(arr)
count = 0
vals = sorted(counts.values(),reverse=True)
while size > len(arr)/2:
maxElem = vals[count]
size -= maxElem
count += 1
return count
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
import math
from collections import Counter
class Solution:
def minSetSize(self, arr: List[int]) -> int:
curr = len(arr)
target = math.ceil(len(arr) / 2)
ans = 0
counts = Counter(arr)
for count in sorted(list(counts.values()), reverse=True):
curr -= count
ans += 1
if curr <= target:
return ans
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
# get the most freq occurring ints
d = {}
for a in arr:
if a in d:
d[a] += 1
else:
d[a] = 1
q = []
for k in d:
heappush(q, (-d[k], k))
size = len(arr)
half = len(arr) / 2
numPop = 0
while size > half:
poped = heappop(q)
size += poped[0]
numPop += 1
return numPop
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
targetLen = len(arr) / 2
elementsRemoved = depth = 0
data = collections.Counter(arr)
for key, val in data.most_common():
elementsRemoved += val
depth += 1
if elementsRemoved >= targetLen:
return depth
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
half = len(arr) // 2
d = {}
for num in arr:
if num not in d:
d[num] = 0
d[num] += 1
lst = sorted(list(d.values()), reverse=True)
accum = 0
res = 0
for item in lst:
if accum < half:
accum += item
res += 1
return res
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
d = collections.defaultdict(int)
for i in arr:
d[i] += 1
res = len(arr)
orig = len(arr)
count = 0
freq = [(d[i],i) for i in d]
freq.sort(reverse=True)
count = 0
for i in freq:
res -= i[0]
count+= 1
if res <= orig//2:
return count
return -1
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
from collections import defaultdict
class Solution:
def minSetSize(self, arr: List[int]) -> int:
count_map = defaultdict(int)
for num in arr:
count_map[num] += 1
res_arr = []
for num, count in count_map.items():
res_arr.append((count, num))
res_arr = sorted(res_arr)[::-1]
count = 0
index = 0
while count < len(arr) // 2:
count += res_arr[index][0]
index += 1
return index
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
halfLength = len(arr)//2
valDict = {}
for val in arr:
if val in valDict:
valDict[val] += 1
else:
valDict[val] = 1
setSize = 0
lengthValsRemoved = 0
for val in sorted(valDict.values(), reverse=True):
setSize += 1
lengthValsRemoved += val
if lengthValsRemoved >= halfLength:
break
return setSize
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
from collections import Counter
class Solution:
def minSetSize(self, arr: List[int]) -> int:
n = len(arr)
counts = Counter(arr)
sorted_counts = sorted(counts.items(), key=lambda x: (-x[1], x[0]))
count = 0
ans = 0
for sc in sorted_counts:
count += sc[1]
ans += 1
if count >= n//2:
return ans
return ans
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
frequency = defaultdict(int)
for num in arr:
frequency[num] += 1
arrLength = len(arr)
pairs = []
for key,val in frequency.items():
pairs.append((val,key))
pairs.sort(reverse = True)
cumSum = 0
for counter,pair in enumerate(pairs):
frequency,key = pair
cumSum += frequency
if cumSum >= arrLength/2:
return counter+1
|
Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.
Return the minimum size of the set so that at least half of the integers of the array are removed.
Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.
Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.
Example 3:
Input: arr = [1,9]
Output: 1
Example 4:
Input: arr = [1000,1000,3,7]
Output: 1
Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5
|
class Solution:
def minSetSize(self, arr: List[int]) -> int:
countOfArray = Counter(arr)
listOfValues = list(countOfArray.values())
listOfValues.sort(reverse = True)
lenOfArray = len(arr)
halfTheLength = lenOfArray//2
ret = 0
for val in listOfValues:
lenOfArray = lenOfArray - val
ret += 1
if lenOfArray <= halfTheLength :
break
return ret
|
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