description stringlengths 171 4k | code stringlengths 94 3.98k | normalized_code stringlengths 57 4.99k |
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Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, array: List[int], hour: int) -> int:
if len(array) == hour:
return max(array)
low = 1
high = max(array)
ans = high
while low <= high:
mid = (low + high) // 2
h = hour
for i in range(len(array)):
if array[i] <= mid:
h -= 1
elif array[i] % mid == 0:
h -= array[i] // mid
else:
h -= array[i] // mid + 1
if h < 0:
break
if h >= 0:
ans = mid
high = mid - 1
else:
low = mid + 1
return ans | CLASS_DEF FUNC_DEF VAR VAR VAR IF FUNC_CALL VAR VAR VAR RETURN FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR NUMBER IF BIN_OP VAR VAR VAR NUMBER VAR BIN_OP VAR VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR NUMBER IF VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
def condition(k) -> bool:
count = 0
for v in piles:
count += int(math.ceil(v / k))
if count > H:
return False
return True
left, right = 1, max(piles)
while left < right:
mid = left + (right - left) // 2
if condition(mid):
right = mid
else:
left = mid + 1
return left | CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR VAR IF VAR VAR RETURN NUMBER RETURN NUMBER VAR ASSIGN VAR VAR NUMBER FUNC_CALL VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
l = 1
r = max(piles)
def findTime(count):
hr = 0
for p in piles:
hr += (p - 1) // count + 1
if hr > H:
return False
return True
while l < r:
mid = l + (r - l) // 2
if findTime(mid):
r = mid
else:
l = mid + 1
return l | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER IF VAR VAR RETURN NUMBER RETURN NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
lo = sum(piles) // H
hi = max(piles)
if len(piles) == 1:
return ceil(piles[0] / H)
while lo <= hi:
mid = (lo + hi) // 2
count = sum(map(lambda x: ceil(x / mid), piles))
if count > H:
lo = mid + 1
else:
hi = mid - 1
return lo | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER RETURN FUNC_CALL VAR BIN_OP VAR NUMBER VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR VAR VAR IF VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
def sop(A, H, K):
if K == 0:
return False
count = 0
for i in A:
if i % K == 0:
count += i // K
else:
count += i // K + 1
if count <= H:
return True
return False
l = sum(piles) // H
r = max(piles)
print((l, r))
print(sop(piles, H, l))
if len(piles) == 1:
return l + 1
m = (l + r) // 2
while l <= r:
if sop(piles, H, l) == False and sop(piles, H, l + 1) == True:
return l + 1
if sop(piles, H, r - 1) == False and sop(piles, H, r) == True:
return r
if sop(piles, H, m) == False:
l = m
m = (l + r) // 2
else:
r = m
m = (l + r) // 2 | CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF IF VAR NUMBER RETURN NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF BIN_OP VAR VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR RETURN NUMBER RETURN NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR IF FUNC_CALL VAR VAR NUMBER RETURN BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER WHILE VAR VAR IF FUNC_CALL VAR VAR VAR VAR NUMBER FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER NUMBER RETURN BIN_OP VAR NUMBER IF FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR VAR VAR VAR NUMBER RETURN VAR IF FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
maxPile = 0
for i in piles:
if i > maxPile:
maxPile = i
low = 1
high = maxPile
minVal = maxPile
while low <= high:
middle = (low + high) // 2
numHours = 0
for j in piles:
numHours += j // middle
if j % middle > 0:
numHours += 1
if numHours <= H:
minVal = middle
high = middle - 1
else:
low = middle + 1
return minVal | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR BIN_OP VAR VAR IF BIN_OP VAR VAR NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
l = 1
r = max(piles)
while l < r:
m = l + (r - l) // 2
h = 0
for i in piles:
h += math.ceil(i / m)
if h > H:
l = m + 1
else:
r = m
return l | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR FUNC_CALL VAR BIN_OP VAR VAR IF VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
def count(k):
hours = 0
for i in range(len(piles)):
hours += (piles[i] + k - 1) // k
return hours
l = 1
r = 1000000000
while l < r:
m = l + (r - l) // 2
if count(m) > H:
l = m + 1
else:
r = m
return l | CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER VAR RETURN VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
lo = 1
hi = max(piles)
ans = 0
while hi >= lo:
mid = (lo + hi) // 2
if self.check(mid, piles, H):
ans = mid
hi = mid - 1
else:
lo = mid + 1
return ans
def check(self, mid, piles, H):
i = 0
count = 0
new = piles[:]
while i < len(new):
if count > H:
return False
if mid >= new[i]:
count += 1
i += 1
else:
count += new[i] // mid + 1
i += 1
if count > H:
return False
return True | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR WHILE VAR FUNC_CALL VAR VAR IF VAR VAR RETURN NUMBER IF VAR VAR VAR VAR NUMBER VAR NUMBER VAR BIN_OP BIN_OP VAR VAR VAR NUMBER VAR NUMBER IF VAR VAR RETURN NUMBER RETURN NUMBER |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
left = 1
right = max(piles)
def can(k):
return sum((p - 1) // k + 1 for p in piles) <= H
while left < right:
mid = (left + right) // 2
if can(mid):
right = mid
else:
left = mid + 1
return right | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_DEF RETURN FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER VAR VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
mini = 1
maxi = max(piles)
ans = 0
while mini <= maxi:
mid = mini + (maxi - mini) // 2
value = sum((p - 1) // mid + 1 for p in piles)
if value <= H:
ans = mid
maxi = mid - 1
else:
mini = mid + 1
return ans | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
N = len(piles)
start = 1
end = max(piles)
while start + 1 < end:
mid = start + (end - start) // 2
if self._can_complete(piles, mid, H):
end = mid
else:
start = mid
if self._can_complete(piles, start, H):
return start
else:
return end
def _can_complete(self, piles, K, H):
i = 0
N = len(piles)
c = 0
for d in piles:
c += math.ceil(d / K)
return c <= H | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR WHILE BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR IF FUNC_CALL VAR VAR VAR VAR RETURN VAR RETURN VAR VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR VAR VAR FUNC_CALL VAR BIN_OP VAR VAR RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
def cnt(k, piles):
ans = 0
for p in piles:
ans += math.ceil(p / float(k))
return int(ans)
piles.sort()
l = 1
h = piles[-1]
mid = (l + h) // 2
while l <= h:
s = cnt(mid, piles)
if s <= H:
h = mid - 1
else:
l = mid + 1
mid = (l + h) // 2
return l | CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR VAR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER WHILE VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
l, h = 1, max(piles)
def canFinish(k):
hoursTaken = sum([ceil(p / k) for p in piles])
return hoursTaken <= H
while l < h:
mid = l + (h - l) // 2
if canFinish(mid):
h = mid
else:
l = mid + 1
return l | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR VAR NUMBER FUNC_CALL VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR RETURN VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
def can(K):
cur = H
for a in piles:
cur -= a // K
cur -= a % K != 0
return cur >= 0
l, r = 1, sum([(a + 1) for a in piles])
while l < r:
mid = (l + r) // 2
if can(mid):
r = mid
else:
l = mid + 1
return l | CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF ASSIGN VAR VAR FOR VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR NUMBER RETURN VAR NUMBER ASSIGN VAR VAR NUMBER FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
l, r = 1, max(piles)
while l < r:
m = (l + r) // 2
if sum((p + m - 1) // m for p in piles) > H:
l = m + 1
else:
r = m
return l | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR VAR NUMBER FUNC_CALL VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
if not piles:
return 0
l, r = 1, max(piles)
def canDo(pace):
ans = 0
for v in piles:
ans -= -v // pace
return ans <= H
while l < r:
m = (l + r) // 2
if canDo(m):
r = m
else:
l = m + 1
return l | CLASS_DEF FUNC_DEF VAR VAR VAR IF VAR RETURN NUMBER ASSIGN VAR VAR NUMBER FUNC_CALL VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR VAR BIN_OP VAR VAR RETURN VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def canEat(self, piles, bananas, H):
cur_H = 0
for pile in piles:
if pile % bananas == 0:
cur_H += pile // bananas
else:
cur_H += pile // bananas + 1
if cur_H > H:
return False
return True
def minEatingSpeed(self, piles: List[int], H: int) -> int:
min_speed = 1
max_speed = max(piles)
while min_speed < max_speed:
mid = (min_speed + max_speed) // 2
if self.canEat(piles, mid, H):
max_speed = mid
else:
min_speed = mid + 1
return min_speed | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR IF BIN_OP VAR VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR RETURN NUMBER RETURN NUMBER FUNC_DEF VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
def possible(K):
count = 0
for banana in piles:
if banana % K == 0:
count += banana / K
else:
count += banana // K + 1
if count <= H:
return True
else:
return False
left, right = 1, max(piles)
while left <= right:
mid = (left + right) // 2
if possible(mid):
right = mid - 1
else:
left = mid + 1
return left | CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR IF BIN_OP VAR VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR RETURN NUMBER RETURN NUMBER ASSIGN VAR VAR NUMBER FUNC_CALL VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
def time_needed(K):
return sum(p // K + (1 if p % K else 0) for p in piles)
lower = max(1, sum(piles) // H)
upper = max(piles)
while lower < upper:
if time_needed(lower) <= H:
return lower
lower += 1
mid = (lower + upper) // 2
if time_needed(mid) <= H:
upper = mid
else:
lower = mid + 1
return lower | CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF RETURN FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR NUMBER NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR VAR IF FUNC_CALL VAR VAR VAR RETURN VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
def possible(k):
return sum((num - 1) // k + 1 for num in piles) <= H
left, right = 1, max(piles)
while left < right:
mid = (left + right) // 2
if not possible(mid):
left = mid + 1
else:
right = mid
return left | CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF RETURN FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER VAR VAR VAR ASSIGN VAR VAR NUMBER FUNC_CALL VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
low, high = 1, max(piles)
while low < high:
K = low + (high - low) // 2
h = sum([((pile - 1) // K + 1) for pile in piles])
if h <= H:
high = K
else:
low = K + 1
return low | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR VAR NUMBER FUNC_CALL VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
if H == len(piles):
return max(piles)
l = 1
r = max(piles) + 1
while l < r:
m = l + (r - l) // 2
if self.minHours(piles, m) <= H:
r = m
elif self.minHours(piles, m) > H:
l = m + 1
return l
def minHours(self, piles, k):
h = 0
for p in piles:
if p % k == 0:
h += p // k
else:
h = h + p // k + 1
return h | CLASS_DEF FUNC_DEF VAR VAR VAR IF VAR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR IF FUNC_CALL VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR IF BIN_OP VAR VAR NUMBER VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP VAR VAR NUMBER RETURN VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
l = max(1, sum(piles) // H)
r = max(piles) // (H // len(piles)) + 1
def hours(rate):
count = 0
for x in piles:
count += (x + rate - 1) // rate
return count
while l <= r:
mid = (l + r) // 2
h = hours(mid)
if h > H:
l = mid + 1
else:
ans = mid
r = mid - 1
return ans | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP FUNC_CALL VAR VAR BIN_OP VAR FUNC_CALL VAR VAR NUMBER FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER VAR RETURN VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
if H == len(piles):
return max(piles)
elif sum(piles) <= H:
return 1
cntr = collections.Counter(piles)
lst = set()
for key in cntr.keys():
lst.add((key, cntr[key]))
s = 2
e = max(piles) - 1
while s <= e:
K = int((s + e) / 2)
total = 0
for pile, n in lst:
q, r = divmod(pile, K)
if r:
total += (q + 1) * n
else:
total += q * n
if total > H:
break
if total <= H:
e = K - 1
else:
s = K + 1
return s | CLASS_DEF FUNC_DEF VAR VAR VAR IF VAR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR RETURN NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER WHILE VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR IF VAR VAR BIN_OP BIN_OP VAR NUMBER VAR VAR BIN_OP VAR VAR IF VAR VAR IF VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
def eatingPossible(cap, piles, H):
h = 0
for pile in piles:
h += pile // cap
if pile % cap != 0:
h += 1
if h > H:
return False
return True
piles.sort()
left = 1
right = max(piles)
while left < right:
mid = left + (right - left) // 2
if eatingPossible(mid, piles, H):
right = mid
else:
left = mid + 1
return left | CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR VAR BIN_OP VAR VAR IF BIN_OP VAR VAR NUMBER VAR NUMBER IF VAR VAR RETURN NUMBER RETURN NUMBER EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
def useDays(pile, speed):
days = pile // speed
return days if pile % speed == 0 else days + 1
def CanDone(piles, speed, H):
numDays = 0
for num in piles:
numDays += useDays(num, speed)
return numDays <= H
maxEat = max(piles)
left, right = 1, maxEat + 1
while left < right:
mid = left + (right - left >> 1)
if CanDone(piles, mid, H):
right = mid
else:
left = mid + 1
return left | CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF ASSIGN VAR BIN_OP VAR VAR RETURN BIN_OP VAR VAR NUMBER VAR BIN_OP VAR NUMBER FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR VAR FUNC_CALL VAR VAR VAR RETURN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER BIN_OP VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
left = 1
right = max(piles)
while left < right:
mid = (left + right) // 2
if self.get_total_days(piles, mid) > H:
left = mid + 1
else:
right = mid
return left
def get_total_days(self, piles: List[int], num_eat: int) -> int:
if num_eat == 0:
return 0
total_days = 0
for pile in piles:
total_days += math.ceil(float(pile) / num_eat)
return total_days | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR RETURN VAR VAR FUNC_DEF VAR VAR VAR IF VAR NUMBER RETURN NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def hours(self, piles, k):
cnt = 0
for p in piles:
if p <= k:
cnt = cnt + 1
else:
cnt = cnt + p // k + 1
return cnt
def minEatingSpeed(self, piles: List[int], H: int) -> int:
if len(piles) == 1:
return 1 + piles[0] // H
l = 0
r = max(piles)
while l <= r:
mid = (l + r) // 2
hrs = self.hours(piles, mid)
if hrs <= H:
ans = mid
r = mid - 1
else:
l = mid + 1
return ans | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP VAR VAR NUMBER RETURN VAR FUNC_DEF VAR VAR VAR IF FUNC_CALL VAR VAR NUMBER RETURN BIN_OP NUMBER BIN_OP VAR NUMBER VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
def time(val):
t = len(piles)
for pile in piles:
t += (pile - 1) // val
return t
s = sum(piles)
l, r = math.ceil(s / H), math.ceil(s / (H - len(piles) + 1))
while l <= r:
mid = l + (r - l) // 2
t = time(mid)
if t <= H:
r = mid - 1
else:
l = mid + 1
return l | CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR FOR VAR VAR VAR BIN_OP BIN_OP VAR NUMBER VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR BIN_OP VAR VAR FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP VAR FUNC_CALL VAR VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
s, e = 1, max(piles)
minK = float("inf")
while s <= e:
m = (s + e) // 2
if self.eat(piles, m) > H:
s = m + 1
else:
minK = min(minK, m)
e = m - 1
return minK
def eat(self, piles, K):
ret = 0
for p in piles:
ret += math.ceil(p / K)
return ret | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR STRING WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR VAR FUNC_CALL VAR BIN_OP VAR VAR RETURN VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
def possible(K):
return sum(ceil(pile / K) for pile in piles) <= H
total = sum(piles)
lo, hi = ceil(total / H), ceil(total / (H - len(piles) + 1))
while lo < hi:
mid = lo + (hi - lo) // 2
if possible(mid):
hi = mid
else:
lo = mid + 1
return lo | CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR BIN_OP VAR VAR FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP VAR FUNC_CALL VAR VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
def ifeatK(k):
hour = 0
for pile in piles:
hour += math.ceil(pile / k)
return hour
start = 1
end = sum(piles)
while start + 1 < end:
mid = (start + end) // 2
if ifeatK(mid) < H:
end = mid
elif ifeatK(mid) > H:
start = mid
else:
end = mid
if ifeatK(start) > H:
return end
if ifeatK(start) == H:
return start
if ifeatK(end) == H:
return end
return start | CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR VAR FUNC_CALL VAR BIN_OP VAR VAR RETURN VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR WHILE BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR ASSIGN VAR VAR IF FUNC_CALL VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR IF FUNC_CALL VAR VAR VAR RETURN VAR IF FUNC_CALL VAR VAR VAR RETURN VAR IF FUNC_CALL VAR VAR VAR RETURN VAR RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
k = sum(piles) // H
if k == 0:
return 1
while True:
h = 0
prev = k
for p in piles:
print(p, k)
if p > 1:
h += (p - 1) // k + 1
else:
h += 1
if h > H:
break
if h <= H:
return k
k += 1 | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR IF VAR NUMBER RETURN NUMBER WHILE NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR VAR IF VAR NUMBER VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER IF VAR VAR IF VAR VAR RETURN VAR VAR NUMBER VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
def is_valid(speed):
curr = 0
for n in piles:
q, r = divmod(n, speed)
if r:
q += 1
curr += q
if curr > H:
return False
return True
l, r = 1, max(piles)
res = -1
while l <= r:
m = l + (r - l) // 2
if is_valid(m):
res = m
r = m - 1
else:
l = m + 1
return res | CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR IF VAR VAR NUMBER VAR VAR IF VAR VAR RETURN NUMBER RETURN NUMBER ASSIGN VAR VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles, H):
start = sum(piles) // H + 1
while start <= max(piles):
suplus = sum(map(lambda x: (start - x % start) % start, piles))
if suplus + sum(piles) <= start * H:
return start
else:
start += 1
return max(piles) | CLASS_DEF FUNC_DEF ASSIGN VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR NUMBER WHILE VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP BIN_OP VAR BIN_OP VAR VAR VAR VAR IF BIN_OP VAR FUNC_CALL VAR VAR BIN_OP VAR VAR RETURN VAR VAR NUMBER RETURN FUNC_CALL VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
if H == len(piles):
return max(piles)
else:
if sum(piles) % H == 0:
tempMin = sum(piles) // H
else:
tempMin = sum(piles) // H + 1
tempMax = max(piles)
while tempMin != tempMax:
tempMid = (tempMax - tempMin) // 2 + tempMin
if self.ifOk(piles, H, tempMid):
tempMax = tempMid
else:
tempMin = tempMid + 1
return tempMin
def ifOk(self, piles, H, K):
hourNeed = 0
for pile in piles:
if pile % K == 0:
hourNeed = hourNeed + pile // K
else:
hourNeed = hourNeed + pile // K + 1
if hourNeed > H:
return False
return True | CLASS_DEF FUNC_DEF VAR VAR VAR IF VAR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER VAR IF FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR IF BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP VAR VAR NUMBER IF VAR VAR RETURN NUMBER RETURN NUMBER |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
if len(piles) == 1:
return piles[0] // H if piles[0] % H == 0 else piles[0] // H + 1
l = sum(piles) // H
h = max(piles) * len(piles) // H + 1
def feasible(k):
return sum((p - 1) // k + 1 for p in piles) <= H
while l < h:
mid = (l + h) // 2
if feasible(mid):
h = mid
else:
l = mid + 1
return l | CLASS_DEF FUNC_DEF VAR VAR VAR IF FUNC_CALL VAR VAR NUMBER RETURN BIN_OP VAR NUMBER VAR NUMBER BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR NUMBER FUNC_DEF RETURN FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER VAR VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
n = len(piles)
if not n:
return -1
def binary_search(l, r):
nonlocal piles, H
while l < r:
mid = (l + r) // 2
if num_hours(mid) <= H:
r = mid
else:
l = mid + 1
return l
def num_hours(bananas_per_hour):
nonlocal piles
res = 0
for i in piles:
if bananas_per_hour >= i:
res += 1
continue
res += i // bananas_per_hour
res += 1 if i % bananas_per_hour else 0
return res
return binary_search(1, max(piles))
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11] | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR RETURN NUMBER FUNC_DEF WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR NUMBER NUMBER RETURN VAR RETURN FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR VAR EXPR LIST NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def time_taken(self, piles, n):
return sum(v // n + (1 if v % n else 0) for v in piles)
def minEatingSpeed(self, piles: List[int], H: int) -> int:
lo = max(1, sum(piles) // H)
hi = max(piles)
t_hi = self.time_taken(piles, hi)
while lo < hi:
mid = (lo + hi) // 2
t_lo = self.time_taken(piles, lo)
t_mid = self.time_taken(piles, mid)
if t_lo <= H:
return lo
elif t_mid <= H:
hi, t_hi = mid, t_mid
lo += 1
else:
lo = mid + 1
return lo | CLASS_DEF FUNC_DEF RETURN FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR NUMBER NUMBER VAR VAR FUNC_DEF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR RETURN VAR IF VAR VAR ASSIGN VAR VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
lo, hi = 1, max(piles)
ans = hi
while lo <= hi:
k = (lo + hi) // 2
time = 0
for i in piles:
t = int(ceil(i / k))
if not t:
t += 1
time += t
if time <= H:
hi = k - 1
ans = k
else:
lo = k + 1
return ans | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR VAR IF VAR VAR NUMBER VAR VAR IF VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
if len(piles) == 1:
div, rem = divmod(piles[0], H)
if rem == 0:
return div
else:
return div + 1
left = 0
right = sum(piles)
while left < right:
mid = (left + right) // 2
if self.CalcTime(piles, mid, H):
right = mid
else:
left = mid + 1
return left
def CalcTime(self, piles, K, H):
h = 0
for p in piles:
div, rem = divmod(p, K)
h += div
if rem > 0:
h += 1
if h > H:
return False
return h <= H | CLASS_DEF FUNC_DEF VAR VAR VAR IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR NUMBER VAR IF VAR NUMBER RETURN VAR RETURN BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR VAR IF VAR NUMBER VAR NUMBER IF VAR VAR RETURN NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def get_hours(self, piles, k):
ans = 0
for x in piles:
if x % k == 0:
ans += x // k
else:
ans += int(x // k) + 1
return ans
def minEatingSpeed(self, piles: List[int], H: int) -> int:
n, max_val = len(piles), max(piles)
if H < n:
return 1 << 31
start, end = 1, max_val
while start <= end:
mid = (end - start) // 2 + start
h = self.get_hours(piles, mid)
if h <= H:
end = mid - 1
else:
start = mid + 1
return start if start <= max_val else start - 1 | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR IF BIN_OP VAR VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP FUNC_CALL VAR BIN_OP VAR VAR NUMBER RETURN VAR FUNC_DEF VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR VAR RETURN BIN_OP NUMBER NUMBER ASSIGN VAR VAR NUMBER VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR VAR BIN_OP VAR NUMBER VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
lenth = len(piles)
s = (min(piles) - 1) * lenth // H + 1
m = (max(piles) - 1) * lenth // H + 1
if int(m) < m:
m = int(m) + 1
if s == m:
return s
while s < m:
mid = (s + m) // 2
count = sum([((i - 1) // mid + 1) for i in piles])
if count > H:
s = mid + 1
else:
m = mid
return s | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR RETURN VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER VAR VAR IF VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
def can_finish(k):
h = 0
for p in piles:
d, r = divmod(p, k)
if r:
d += 1
h += d
return h <= H
l = 1
r = max(piles)
while l < r:
m = (l + r) // 2
if can_finish(m):
r = m
else:
l = m + 1
return l | CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR IF VAR VAR NUMBER VAR VAR RETURN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
max_k = 0
for value in piles:
max_k = max(max_k, value)
def possible(k):
hours = 0
for item in piles:
hours += math.ceil(item / k)
if hours > H:
return False
if hours <= H:
return True
return False
def binary_search(low, high):
mid = low + (high - low) // 2
if possible(low):
return low
elif possible(mid):
return binary_search(low, mid)
else:
return binary_search(mid + 1, high)
def binary_search_iter(low, high):
pass
return binary_search(1, max_k) | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR VAR FUNC_CALL VAR BIN_OP VAR VAR IF VAR VAR RETURN NUMBER IF VAR VAR RETURN NUMBER RETURN NUMBER FUNC_DEF ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR RETURN VAR IF FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR VAR RETURN FUNC_CALL VAR BIN_OP VAR NUMBER VAR FUNC_DEF RETURN FUNC_CALL VAR NUMBER VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
lo, hi = 1, 1000000001
ans = -1
N = len(piles)
def poss(val):
cnt = 0
for n in range(N):
cnt += math.ceil(piles[n] / val)
if cnt <= H:
return True
return False
while lo <= hi:
mid = lo + (hi - lo) // 2
if poss(mid):
ans = mid
hi = mid - 1
else:
lo = mid + 1
return ans | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR BIN_OP VAR VAR VAR IF VAR VAR RETURN NUMBER RETURN NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
ceiling = lambda x, y: (x + y - 1) // y
def timeTaken(v):
totalTime = 0
for p in piles:
totalTime += ceiling(p, v)
return totalTime
def binarySearch(s, e, up_bound):
if s == e:
return s
m = (s + e) // 2
if timeTaken(m) <= up_bound:
return binarySearch(s, m, up_bound)
else:
return binarySearch(m + 1, e, up_bound)
pmax = 1
for p in piles:
pmax = max(pmax, p)
return binarySearch(1, pmax, H) | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR VAR FUNC_CALL VAR VAR VAR RETURN VAR FUNC_DEF IF VAR VAR RETURN VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR RETURN FUNC_CALL VAR VAR VAR VAR RETURN FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN FUNC_CALL VAR NUMBER VAR VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
def possible(K):
return sum(math.ceil(p / K) for p in piles) <= H
N = len(piles)
if H == N:
return max(piles)
sm = sum(piles)
lo, hi = int(sm / H) + 1, math.ceil(sm / (H - N))
while lo < hi:
mi = (lo + hi) // 2
if not possible(mi):
lo = mi + 1
else:
hi = mi
return hi | CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR VAR RETURN FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP FUNC_CALL VAR BIN_OP VAR VAR NUMBER FUNC_CALL VAR BIN_OP VAR BIN_OP VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
def is_ok(piles, H, mid):
sum = 0
for i in range(len(piles)):
if piles[i] <= mid:
sum += 1
else:
sum += math.ceil(piles[i] / mid)
return sum <= H
ok = sum(piles)
ng = 0
while abs(ok - ng) > 1:
mid = (ok + ng) // 2
if is_ok(piles, H, mid):
ok = mid
else:
ng = mid
return ok | CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR VAR VAR RETURN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER WHILE FUNC_CALL VAR BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
right = 10**9
left = 1
piles.sort()
while left + 1 < right:
mid = (left + right) // 2
if self.canEatAll(piles, mid, H):
right = mid
else:
left = mid
return left if self.canEatAll(piles, left, H) else right
def canEatAll(self, piles, speed, H):
res = 0
for pile in piles:
res += (pile - 1) // speed + 1
return res <= H | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR WHILE BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR RETURN FUNC_CALL VAR VAR VAR VAR VAR VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
n = len(piles)
piles.sort()
def finish(k):
return sum((p - 1) // k + 1 for p in piles) <= H
r = 0
while r < n:
if finish(piles[r]):
break
else:
r += 1
start = 1 if r == 0 else piles[0]
end = piles[r]
while start < end:
mid = (start + end) // 2
if finish(mid):
end = mid
else:
start = mid + 1
return start | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER VAR VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR IF FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
def simulate(k):
hours = 0
for pile in piles:
hours += ceil(pile / k)
return hours
lower = 1
upper = 1
for pile in piles:
upper = max(upper, pile)
while lower < upper:
midpoint = (upper + lower) // 2
if simulate(midpoint) > H:
lower = midpoint + 1
else:
upper = midpoint
return upper | CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR VAR FUNC_CALL VAR BIN_OP VAR VAR RETURN VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
bananas = sum(piles)
K = bananas // H + (bananas % H != 0)
while True:
hours_needed = 0
for pile in piles:
hours_needed += pile // K
if pile % K != 0:
hours_needed += 1
if hours_needed <= H:
return K
K += 1 | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR NUMBER WHILE NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR BIN_OP VAR VAR IF BIN_OP VAR VAR NUMBER VAR NUMBER IF VAR VAR RETURN VAR VAR NUMBER VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
def is_feasible(rate, H, piles):
count = 0
for i in range(len(piles)):
count += piles[i] // rate
if piles[i] % rate != 0:
count += 1
if count > H:
return False
return True
l = 1
h = sum(piles)
ans = float("inf")
while l <= h:
mid = (l + h) // 2
if is_feasible(mid, H, piles):
ans = min(ans, mid)
h = mid - 1
else:
l = mid + 1
return ans | CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR IF BIN_OP VAR VAR VAR NUMBER VAR NUMBER IF VAR VAR RETURN NUMBER RETURN NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR STRING WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
piles.sort()
l, r = 1, max(piles)
while l <= r:
m = (l + r) // 2
if sum([((x - 1) // m + 1) for x in piles]) > H:
l = m + 1
elif sum([math.ceil(x / m) for x in piles]) < H:
r = m - 1
else:
return m
return l | CLASS_DEF FUNC_DEF VAR VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR VAR NUMBER FUNC_CALL VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
low = 1
high = max(piles)
def possible(k):
time = 0
for pile in piles:
time += (pile - 1) // k + 1
return time <= H
while low <= high:
mid = low + (high - low) // 2
outcome = possible(mid)
if possible(mid):
high = mid - 1
else:
low = mid + 1
return low | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER RETURN VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
def eats(piles, h, k):
hours = 0
for pile in piles:
hours += (pile - 1) // k + 1
return hours <= h
lo = 1
hi = 10**9
while lo < hi:
mid = lo + (hi - lo) // 2
if eats(piles, H, mid):
hi = mid
else:
lo = mid + 1
return lo | CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER RETURN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
n = len(piles)
if n >= H:
return max(piles)
def check(m):
ret = 0
for p in piles:
ret += p // m + (p % m != 0)
return ret
upper = max(piles)
lo, hi = 1, upper
while lo <= hi:
mid = (lo + hi) // 2
if check(mid) <= H:
hi = mid - 1
else:
lo = mid + 1
return lo | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR VAR RETURN FUNC_CALL VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
self.piles = piles
l, r = 1, max(piles)
while l <= r:
K = (l + r) // 2
if self.eat_time(K) <= H:
r = K - 1
else:
l = K + 1
return l
def eat_time(self, K):
cnt = 0
for banana in self.piles:
q, mod = divmod(banana, K)
cnt += q
if mod != 0:
cnt += 1
return cnt | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR NUMBER FUNC_CALL VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR VAR IF VAR NUMBER VAR NUMBER RETURN VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
lo, hi = 1, max(piles)
while lo < hi:
mid = (lo + hi) // 2
cur_cap, num_hrs = 0, 0
for p in piles:
cur_cap = p
if cur_cap <= mid:
num_hrs += 1
else:
num_hrs += cur_cap // mid + 1
if num_hrs > H:
lo = mid + 1
else:
hi = mid
return lo | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR VAR NUMBER FUNC_CALL VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER FOR VAR VAR ASSIGN VAR VAR IF VAR VAR VAR NUMBER VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR RETURN VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
n = len(piles)
if not n:
return -1
def binary_search(l, r):
nonlocal piles, H
while l < r:
mid = (l + r) // 2
if num_hours(mid) <= H:
r = mid
else:
l = mid + 1
return l
def num_hours(k):
nonlocal n, piles
res = 0
for i in range(n):
num = piles[i]
if k >= num:
res += 1
continue
res += num // k
res += 1 if num % k else 0
return res
return binary_search(1, max(piles)) | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR RETURN NUMBER FUNC_DEF WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF VAR VAR VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR NUMBER NUMBER RETURN VAR RETURN FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
def helper(speed):
h = 0
for p in piles:
h += math.ceil(p / speed)
return h
if not piles or len(piles) == 0:
return 0
start, end = 1, max(piles)
while start + 1 < end:
mid = start + (end - start) // 2
h = helper(mid)
if h <= H:
end = mid
else:
start = mid
return start if helper(start) <= H else end | CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR VAR FUNC_CALL VAR BIN_OP VAR VAR RETURN VAR IF VAR FUNC_CALL VAR VAR NUMBER RETURN NUMBER ASSIGN VAR VAR NUMBER FUNC_CALL VAR VAR WHILE BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR RETURN FUNC_CALL VAR VAR VAR VAR VAR VAR |
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
Constraints:
1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9 | class Solution:
def minEatingSpeed(self, piles: List[int], H: int) -> int:
def valid(mid):
if mid == 0:
return False
hours = 0
for i in piles:
if i < mid:
hours += 1
else:
hours += math.ceil(i / mid)
return hours <= H
if not piles:
return 0
piles = sorted(piles)
s, e = 0, sum(piles)
ans = -1
while s <= e:
mid = (s + e) // 2
if valid(mid):
ans = mid
e = mid - 1
else:
s = mid + 1
return ans | CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF IF VAR NUMBER RETURN NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR VAR RETURN VAR VAR IF VAR RETURN NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Dr. Evil is interested in math and functions, so he gave Mahmoud and Ehab array a of length n and array b of length m. He introduced a function f(j) which is defined for integers j, which satisfy 0 ≤ j ≤ m - n. Suppose, c_{i} = a_{i} - b_{i} + j. Then f(j) = |c_1 - c_2 + c_3 - c_4... c_{n}|. More formally, $f(j) =|\sum_{i = 1}^{n}(- 1)^{i - 1} *(a_{i} - b_{i + j})|$.
Dr. Evil wants Mahmoud and Ehab to calculate the minimum value of this function over all valid j. They found it a bit easy, so Dr. Evil made their task harder. He will give them q update queries. During each update they should add an integer x_{i} to all elements in a in range [l_{i};r_{i}] i.e. they should add x_{i} to a_{l}_{i}, a_{l}_{i} + 1, ... , a_{r}_{i} and then they should calculate the minimum value of f(j) for all valid j.
Please help Mahmoud and Ehab.
-----Input-----
The first line contains three integers n, m and q (1 ≤ n ≤ m ≤ 10^5, 1 ≤ q ≤ 10^5) — number of elements in a, number of elements in b and number of queries, respectively.
The second line contains n integers a_1, a_2, ..., a_{n}. ( - 10^9 ≤ a_{i} ≤ 10^9) — elements of a.
The third line contains m integers b_1, b_2, ..., b_{m}. ( - 10^9 ≤ b_{i} ≤ 10^9) — elements of b.
Then q lines follow describing the queries. Each of them contains three integers l_{i} r_{i} x_{i} (1 ≤ l_{i} ≤ r_{i} ≤ n, - 10^9 ≤ x ≤ 10^9) — range to be updated and added value.
-----Output-----
The first line should contain the minimum value of the function f before any update.
Then output q lines, the i-th of them should contain the minimum value of the function f after performing the i-th update .
-----Example-----
Input
5 6 3
1 2 3 4 5
1 2 3 4 5 6
1 1 10
1 1 -9
1 5 -1
Output
0
9
0
0
-----Note-----
For the first example before any updates it's optimal to choose j = 0, f(0) = |(1 - 1) - (2 - 2) + (3 - 3) - (4 - 4) + (5 - 5)| = |0| = 0.
After the first update a becomes {11, 2, 3, 4, 5} and it's optimal to choose j = 1, f(1) = |(11 - 2) - (2 - 3) + (3 - 4) - (4 - 5) + (5 - 6) = |9| = 9.
After the second update a becomes {2, 2, 3, 4, 5} and it's optimal to choose j = 1, f(1) = |(2 - 2) - (2 - 3) + (3 - 4) - (4 - 5) + (5 - 6)| = |0| = 0.
After the third update a becomes {1, 1, 2, 3, 4} and it's optimal to choose j = 0, f(0) = |(1 - 1) - (1 - 2) + (2 - 3) - (3 - 4) + (4 - 5)| = |0| = 0. | def read():
return list(map(int, input().split(" ")))
n, m, q = read()
aa = read()
bb = read()
reqs = [read() for _ in range(q)]
asum = 0
bsum = 0
for i, (a, b) in enumerate(zip(aa, bb)):
asum += a if i % 2 == 0 else -a
bsum += b if i % 2 == 0 else -b
bpos = [bsum]
for i in range(len(aa), len(bb)):
b = bb[i]
rempos = i - len(aa)
bsum += b if i % 2 == 0 else -b
bsum -= bb[rempos] if rempos % 2 == 0 else -bb[rempos]
bpos += [bsum if rempos % 2 == 1 else -bsum]
bpos = sorted(set(bpos))
def closest(arr, value):
l = 0
r = len(arr)
while l + 1 < r:
m = (l + r) // 2
if arr[m] <= value:
l = m
else:
r = m
res = arr[l]
if l + 1 < len(arr) and abs(arr[l + 1] - value) < abs(arr[l] - value):
res = arr[l + 1]
return res
print(abs(asum - closest(bpos, asum)))
for req in reqs:
l, r, x = req
l -= 1
if (r - l) % 2 != 0:
asum += x if l % 2 == 0 else -x
print(abs(asum - closest(bpos, asum))) | FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR VAR VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER NUMBER VAR VAR VAR BIN_OP VAR NUMBER NUMBER VAR VAR ASSIGN VAR LIST VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR BIN_OP VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER NUMBER VAR VAR VAR BIN_OP VAR NUMBER NUMBER VAR VAR VAR VAR VAR LIST BIN_OP VAR NUMBER NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR WHILE BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR VAR IF BIN_OP VAR NUMBER FUNC_CALL VAR VAR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER RETURN VAR EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR VAR VAR FOR VAR VAR ASSIGN VAR VAR VAR VAR VAR NUMBER IF BIN_OP BIN_OP VAR VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER NUMBER VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR VAR VAR |
Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
(i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).
Find the minimum element.
You may assume no duplicate exists in the array.
Example 1:
Input: [3,4,5,1,2]
Output: 1
Example 2:
Input: [4,5,6,7,0,1,2]
Output: 0 | class Solution:
def findMin(self, nums):
if len(nums) == 0:
return -1
if len(nums) == 2:
return min(nums)
low = 0
high = len(nums) - 1
mid = (low + high) // 2
while low < high:
if nums[mid] < nums[high]:
high = mid
elif nums[mid] > nums[high]:
low = mid + 1
mid = (low + high) // 2
return nums[low] | CLASS_DEF FUNC_DEF IF FUNC_CALL VAR VAR NUMBER RETURN NUMBER IF FUNC_CALL VAR VAR NUMBER RETURN FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER WHILE VAR VAR IF VAR VAR VAR VAR ASSIGN VAR VAR IF VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER RETURN VAR VAR |
Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
(i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).
Find the minimum element.
You may assume no duplicate exists in the array.
Example 1:
Input: [3,4,5,1,2]
Output: 1
Example 2:
Input: [4,5,6,7,0,1,2]
Output: 0 | class Solution:
def findMin(self, nums):
if not nums or len(nums) == 0:
return 0
start, end = 0, len(nums) - 1
target = nums[-1]
while start + 1 < end:
mid = (start + end) // 2
if nums[mid] < target:
end = mid
else:
start = mid
if nums[start] < target:
return nums[start]
return nums[end] | CLASS_DEF FUNC_DEF IF VAR FUNC_CALL VAR VAR NUMBER RETURN NUMBER ASSIGN VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER WHILE BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR IF VAR VAR VAR RETURN VAR VAR RETURN VAR VAR |
Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
(i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).
Find the minimum element.
You may assume no duplicate exists in the array.
Example 1:
Input: [3,4,5,1,2]
Output: 1
Example 2:
Input: [4,5,6,7,0,1,2]
Output: 0 | class Solution:
def findMin(self, nums):
if len(nums) == 0:
return None
if nums[-1] > nums[0]:
return nums[0]
boundary = nums[0]
return self.binary_search_min(nums, boundary)
def binary_search_min(self, nums, maximum):
left_boundary = 0
right_boundary = len(nums) - 1
while True:
current_index = left_boundary + (right_boundary - left_boundary) // 2
value = nums[current_index]
if left_boundary == right_boundary:
return value
if value >= maximum:
left_boundary = current_index + 1
else:
right_boundary = current_index | CLASS_DEF FUNC_DEF IF FUNC_CALL VAR VAR NUMBER RETURN NONE IF VAR NUMBER VAR NUMBER RETURN VAR NUMBER ASSIGN VAR VAR NUMBER RETURN FUNC_CALL VAR VAR VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER WHILE NUMBER ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR VAR VAR IF VAR VAR RETURN VAR IF VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR |
Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
(i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).
Find the minimum element.
You may assume no duplicate exists in the array.
Example 1:
Input: [3,4,5,1,2]
Output: 1
Example 2:
Input: [4,5,6,7,0,1,2]
Output: 0 | class Solution:
def findMin(self, nums):
if len(nums) == 1:
return nums[0]
for i in range(1, len(nums)):
if nums[i - 1] > nums[i]:
return nums[i]
return nums[0] | CLASS_DEF FUNC_DEF IF FUNC_CALL VAR VAR NUMBER RETURN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR BIN_OP VAR NUMBER VAR VAR RETURN VAR VAR RETURN VAR NUMBER |
Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
(i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).
Find the minimum element.
You may assume no duplicate exists in the array.
Example 1:
Input: [3,4,5,1,2]
Output: 1
Example 2:
Input: [4,5,6,7,0,1,2]
Output: 0 | class Solution:
def findMin(self, nums):
low, high = 0, len(nums) - 1
while low < high:
mid = (low + high) // 2
if nums[mid] <= nums[high]:
high = mid
else:
low = mid + 1
return nums[low] | CLASS_DEF FUNC_DEF ASSIGN VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
(i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).
Find the minimum element.
You may assume no duplicate exists in the array.
Example 1:
Input: [3,4,5,1,2]
Output: 1
Example 2:
Input: [4,5,6,7,0,1,2]
Output: 0 | class Solution:
def findMin(self, nums):
if len(nums) == 1:
return nums[0]
l = 0
r = len(nums) - 1
while l < r:
m = (l + r) // 2
v = nums[m]
if nums[m - 1] < v:
if nums[r] > v:
r = m - 1
else:
l = m + 1
else:
l = m + 1
if l == r:
if m < l:
if nums[m] < nums[l]:
return nums[m]
else:
return nums[l]
if l < m:
if nums[l - 1] < nums[m]:
return nums[l - 1]
return nums[l]
elif r < l:
return nums[r]
else:
return nums[m]
def findMin_PASSED(self, nums):
if len(nums) == 1:
return nums[0]
retval = float("inf")
l, r = 0, len(nums) - 1
while l < r:
m = (l + r) // 2
v = nums[m]
if v < nums[m + 1]:
if v < nums[r]:
r = m
elif v > nums[r]:
l = m + 1
elif v > nums[m + 1]:
l = m + 1
if nums[m] < nums[m + 1]:
return nums[m]
else:
return nums[m + 1] | CLASS_DEF FUNC_DEF IF FUNC_CALL VAR VAR NUMBER RETURN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR VAR VAR IF VAR BIN_OP VAR NUMBER VAR IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR IF VAR VAR IF VAR VAR VAR VAR RETURN VAR VAR RETURN VAR VAR IF VAR VAR IF VAR BIN_OP VAR NUMBER VAR VAR RETURN VAR BIN_OP VAR NUMBER RETURN VAR VAR IF VAR VAR RETURN VAR VAR RETURN VAR VAR FUNC_DEF IF FUNC_CALL VAR VAR NUMBER RETURN VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR VAR VAR IF VAR VAR BIN_OP VAR NUMBER IF VAR VAR VAR ASSIGN VAR VAR IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER RETURN VAR VAR RETURN VAR BIN_OP VAR NUMBER |
Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
(i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).
Find the minimum element.
You may assume no duplicate exists in the array.
Example 1:
Input: [3,4,5,1,2]
Output: 1
Example 2:
Input: [4,5,6,7,0,1,2]
Output: 0 | class Solution:
def findMin(self, nums):
left, right = 0, len(nums) - 1
mid = 0
while left < right:
mid = left + (right - left) // 2
if nums[right] < nums[mid]:
left = mid + 1
elif nums[right] > nums[mid]:
if nums[mid - 1] > nums[mid]:
return nums[mid]
else:
right = mid - 1
return nums[left] | CLASS_DEF FUNC_DEF ASSIGN VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR VAR VAR IF VAR BIN_OP VAR NUMBER VAR VAR RETURN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
(i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).
Find the minimum element.
You may assume no duplicate exists in the array.
Example 1:
Input: [3,4,5,1,2]
Output: 1
Example 2:
Input: [4,5,6,7,0,1,2]
Output: 0 | class Solution:
def findMin(self, nums):
if len(nums) <= 3:
return min(nums)
lo = 0
hi = len(nums) - 1
mid = (hi + lo) // 2
if nums[mid] < nums[mid - 1] and nums[mid] < nums[mid + 1]:
return nums[mid]
if nums[mid] > nums[lo] and nums[mid] > nums[hi]:
return self.findMin(nums[mid:])
else:
return self.findMin(nums[: mid + 1]) | CLASS_DEF FUNC_DEF IF FUNC_CALL VAR VAR NUMBER RETURN FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER RETURN VAR VAR IF VAR VAR VAR VAR VAR VAR VAR VAR RETURN FUNC_CALL VAR VAR VAR RETURN FUNC_CALL VAR VAR BIN_OP VAR NUMBER |
Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times.
Note: The algorithm should run in linear time and in O(1) space.
Example 1:
Input: [3,2,3]
Output: [3]
Example 2:
Input: [1,1,1,3,3,2,2,2]
Output: [1,2] | class Solution:
def majorityElement(self, nums):
if nums == []:
return []
dct = {}
for el in nums:
if el in dct:
dct[el] += 1
else:
dct[el] = 1
fin = []
for key in dct:
if dct[key] > len(nums) // 3:
fin.append(key)
return fin | CLASS_DEF FUNC_DEF IF VAR LIST RETURN LIST ASSIGN VAR DICT FOR VAR VAR IF VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR LIST FOR VAR VAR IF VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR RETURN VAR |
Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times.
Note: The algorithm should run in linear time and in O(1) space.
Example 1:
Input: [3,2,3]
Output: [3]
Example 2:
Input: [1,1,1,3,3,2,2,2]
Output: [1,2] | class Solution:
def majorityElement(self, nums):
length = len(nums)
ans = []
for i in set(nums):
if nums.count(i) > length / 3:
ans.append(i)
return ans | CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR RETURN VAR |
Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times.
Note: The algorithm should run in linear time and in O(1) space.
Example 1:
Input: [3,2,3]
Output: [3]
Example 2:
Input: [1,1,1,3,3,2,2,2]
Output: [1,2] | class Solution:
def majorityElement(self, nums):
ans = []
if not nums:
return []
a1 = a2 = None
c1 = c2 = 0
for num in nums:
if a1 == num:
c1 += 1
elif a2 == num:
c2 += 1
elif c1 == 0:
a1, c1 = num, 1
elif c2 == 0:
a2, c2 = num, 1
else:
c1 -= 1
c2 -= 1
c1 = c2 = 0
for num in nums:
if num == a1:
c1 += 1
elif num == a2:
c2 += 1
for a, c in ((a1, c1), (a2, c2)):
if c > len(nums) // 3:
ans.append(a)
return ans | CLASS_DEF FUNC_DEF ASSIGN VAR LIST IF VAR RETURN LIST ASSIGN VAR VAR NONE ASSIGN VAR VAR NUMBER FOR VAR VAR IF VAR VAR VAR NUMBER IF VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR IF VAR VAR VAR NUMBER IF VAR VAR VAR NUMBER FOR VAR VAR VAR VAR VAR VAR IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR RETURN VAR |
Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times.
Note: The algorithm should run in linear time and in O(1) space.
Example 1:
Input: [3,2,3]
Output: [3]
Example 2:
Input: [1,1,1,3,3,2,2,2]
Output: [1,2] | class Solution:
def majorityElement(self, nums):
d = dict()
results = []
for num in nums:
d[num] = d.get(num, 0) + 1
if d[num] > len(nums) // 3:
if num not in results:
results.append(num)
return results | CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR VAR ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER IF VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR |
Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times.
Note: The algorithm should run in linear time and in O(1) space.
Example 1:
Input: [3,2,3]
Output: [3]
Example 2:
Input: [1,1,1,3,3,2,2,2]
Output: [1,2] | class Solution:
def majorityElement(self, nums):
if len(nums) <= 2:
return list(set(nums))
can, can2 = 0, 0
num, num2 = 0, 1
for i in range(len(nums)):
if nums[i] == can:
num += 1
elif nums[i] == can2:
num2 += 2
elif num == 0:
num = 1
can = nums[i]
elif num2 == 0:
num2 = 1
can2 = nums[i]
else:
num, num2 = num - 1, num2 - 1
res = []
if nums.count(can) > int(len(nums) / 3):
res.append(can)
if nums.count(can2) > int(len(nums) / 3) and can2 != can:
res.append(can2)
return res | CLASS_DEF FUNC_DEF IF FUNC_CALL VAR VAR NUMBER RETURN FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR NUMBER IF VAR VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR LIST IF FUNC_CALL VAR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR |
Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times.
Note: The algorithm should run in linear time and in O(1) space.
Example 1:
Input: [3,2,3]
Output: [3]
Example 2:
Input: [1,1,1,3,3,2,2,2]
Output: [1,2] | class Solution:
def majorityElement(self, nums):
n1 = n2 = None
c1 = c2 = 0
for i in nums:
if n1 == i:
c1 += 1
elif n2 == i:
c2 += 1
elif c1 == 0:
n1, c1 = i, 1
elif c2 == 0:
n2, c2 = i, 1
else:
c1, c2 = c1 - 1, c2 - 1
return [n for n in [n1, n2] if n is not None and nums.count(n) > len(nums) // 3] | CLASS_DEF FUNC_DEF ASSIGN VAR VAR NONE ASSIGN VAR VAR NUMBER FOR VAR VAR IF VAR VAR VAR NUMBER IF VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER RETURN VAR VAR LIST VAR VAR VAR NONE FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER |
Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times.
Note: The algorithm should run in linear time and in O(1) space.
Example 1:
Input: [3,2,3]
Output: [3]
Example 2:
Input: [1,1,1,3,3,2,2,2]
Output: [1,2] | class Solution:
def majorityElement(self, nums):
N = len(nums)
N3 = N / 3
nums.sort()
result = []
if N < 1:
return nums
_cnt = 0
_num = nums[0]
for n in nums:
if n != _num:
if _cnt > N3:
result.append(_num)
_cnt = 1
_num = n
else:
_cnt += 1
else:
if _cnt > N3:
result.append(_num)
return result | CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR LIST IF VAR NUMBER RETURN VAR ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR IF VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR |
Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times.
Note: The algorithm should run in linear time and in O(1) space.
Example 1:
Input: [3,2,3]
Output: [3]
Example 2:
Input: [1,1,1,3,3,2,2,2]
Output: [1,2] | class Solution:
def majorityElement(self, nums):
ret = []
least = len(nums) / 3
h = {}
for n in nums:
if n in list(h.keys()):
h[n] += 1
else:
h[n] = 1
if h[n] > least and n not in ret:
ret.append(n)
return ret | CLASS_DEF FUNC_DEF ASSIGN VAR LIST ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR DICT FOR VAR VAR IF VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER IF VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR |
Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times.
Note: The algorithm should run in linear time and in O(1) space.
Example 1:
Input: [3,2,3]
Output: [3]
Example 2:
Input: [1,1,1,3,3,2,2,2]
Output: [1,2] | class Solution:
def majorityElement(self, nums):
dic = {}
res = []
n = len(nums)
for i in range(n):
dic[nums[i]] = dic.get(nums[i], 0) + 1
print(dic)
for num, cnt in dic.items():
if cnt > n / 3:
res.append(num)
return res | CLASS_DEF FUNC_DEF ASSIGN VAR DICT ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR FOR VAR VAR FUNC_CALL VAR IF VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR RETURN VAR |
Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times.
Note: The algorithm should run in linear time and in O(1) space.
Example 1:
Input: [3,2,3]
Output: [3]
Example 2:
Input: [1,1,1,3,3,2,2,2]
Output: [1,2] | class Solution:
def majorityElement(self, nums):
ctr = collections.Counter()
for n in nums:
ctr[n] += 1
if len(ctr) == 3:
ctr -= collections.Counter(set(ctr))
return [n for n in ctr if nums.count(n) > len(nums) // 3] | CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR FOR VAR VAR VAR VAR NUMBER IF FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR RETURN VAR VAR VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER |
Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times.
Note: The algorithm should run in linear time and in O(1) space.
Example 1:
Input: [3,2,3]
Output: [3]
Example 2:
Input: [1,1,1,3,3,2,2,2]
Output: [1,2] | class Solution:
def majorityElement(self, nums):
if not nums:
return []
l = len(nums)
candidate1 = candidate2 = 0
count1 = count2 = 0
for i in nums:
if candidate1 == i:
count1 += 1
elif candidate2 == i:
count2 += 1
elif count1 == 0:
candidate1 = i
count1 = 1
elif count2 == 0:
candidate2 = i
count2 = 1
else:
count1 -= 1
count2 -= 1
c1 = nums.count(candidate1)
c2 = nums.count(candidate2)
res = []
if c1 > l / 3:
res.append(candidate1)
if c2 > l / 3 and candidate1 != candidate2:
res.append(candidate2)
return res | CLASS_DEF FUNC_DEF IF VAR RETURN LIST ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR IF VAR VAR VAR NUMBER IF VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST IF VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR |
Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times.
Note: The algorithm should run in linear time and in O(1) space.
Example 1:
Input: [3,2,3]
Output: [3]
Example 2:
Input: [1,1,1,3,3,2,2,2]
Output: [1,2] | class Solution:
def majorityElement(self, nums):
d = {}
for i in nums:
if i in d:
d[i] += 1
else:
d[i] = 1
l = []
for i in list(d.keys()):
if d[i] > len(nums) // 3:
l.append(i)
return l | CLASS_DEF FUNC_DEF ASSIGN VAR DICT FOR VAR VAR IF VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR IF VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR RETURN VAR |
Given an array Arr[] of size N. For every element in the array, the task is to find the index of the farthest element in the array to the right which is smaller than the current element. If no such number exists then print -1.
Note: 0 based indexing.
Example 1:
Input:
N=5
Arr[] = {3, 1, 5, 2, 4}
Output:
3 -1 4 -1 -1
Explanation:
Arr[3] is the farthest smallest element
to the right of Arr[0].
Arr[4] is the farthest smallest element
to the right of Arr[2].
And for the rest of the elements, there is
no smaller element to their right.
Example 2:
Input:
N=5
Arr[] = {1, 2, 3, 4, 0}
Output:
4 4 4 4 -1
Your Task:
You don't need to read input or print anything. Your task is to complete the function farNumber() which takes the N (number of elements in Array Arr) ,Arr[], and returns the array of farthest element to the right for every element of the array.
Expected Time Complexity: O(N*logN)
Expected Auxiliary Space: O(N)
Constraints:
1 ≤ N ≤ 1e5
0 ≤ Arr[i] ≤ 1e9 | class Solution:
def farNumber(self, n, a):
l = []
for i in range(n):
flag = 0
for j in range(n - 1, i, -1):
if a[j] < a[i]:
flag = 1
l.append(j)
break
if flag == 0:
l.append(-1)
return l | CLASS_DEF FUNC_DEF ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR VAR VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER RETURN VAR |
Given an array Arr[] of size N. For every element in the array, the task is to find the index of the farthest element in the array to the right which is smaller than the current element. If no such number exists then print -1.
Note: 0 based indexing.
Example 1:
Input:
N=5
Arr[] = {3, 1, 5, 2, 4}
Output:
3 -1 4 -1 -1
Explanation:
Arr[3] is the farthest smallest element
to the right of Arr[0].
Arr[4] is the farthest smallest element
to the right of Arr[2].
And for the rest of the elements, there is
no smaller element to their right.
Example 2:
Input:
N=5
Arr[] = {1, 2, 3, 4, 0}
Output:
4 4 4 4 -1
Your Task:
You don't need to read input or print anything. Your task is to complete the function farNumber() which takes the N (number of elements in Array Arr) ,Arr[], and returns the array of farthest element to the right for every element of the array.
Expected Time Complexity: O(N*logN)
Expected Auxiliary Space: O(N)
Constraints:
1 ≤ N ≤ 1e5
0 ≤ Arr[i] ≤ 1e9 | class Solution:
def farNumber(self, N, Arr):
l = []
x = 0
m = 0
for i in range(N):
x = N - 1
m = 0
while x > i:
if Arr[i] > Arr[x]:
l.append(x)
m = 1
break
x = x - 1
if m == 0:
l.append(-1)
return l | CLASS_DEF FUNC_DEF ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER RETURN VAR |
Given an array Arr[] of size N. For every element in the array, the task is to find the index of the farthest element in the array to the right which is smaller than the current element. If no such number exists then print -1.
Note: 0 based indexing.
Example 1:
Input:
N=5
Arr[] = {3, 1, 5, 2, 4}
Output:
3 -1 4 -1 -1
Explanation:
Arr[3] is the farthest smallest element
to the right of Arr[0].
Arr[4] is the farthest smallest element
to the right of Arr[2].
And for the rest of the elements, there is
no smaller element to their right.
Example 2:
Input:
N=5
Arr[] = {1, 2, 3, 4, 0}
Output:
4 4 4 4 -1
Your Task:
You don't need to read input or print anything. Your task is to complete the function farNumber() which takes the N (number of elements in Array Arr) ,Arr[], and returns the array of farthest element to the right for every element of the array.
Expected Time Complexity: O(N*logN)
Expected Auxiliary Space: O(N)
Constraints:
1 ≤ N ≤ 1e5
0 ≤ Arr[i] ≤ 1e9 | class Solution:
def farNumber(self, N, Arr):
for i in range(len(Arr)):
flag = 0
for j in reversed(range(len(Arr))):
if Arr[i] > Arr[j] and i < j:
Arr[i] = j
flag = 1
break
if flag == 0:
Arr[i] = -1
return Arr | CLASS_DEF FUNC_DEF FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR NUMBER RETURN VAR |
Given an array Arr[] of size N. For every element in the array, the task is to find the index of the farthest element in the array to the right which is smaller than the current element. If no such number exists then print -1.
Note: 0 based indexing.
Example 1:
Input:
N=5
Arr[] = {3, 1, 5, 2, 4}
Output:
3 -1 4 -1 -1
Explanation:
Arr[3] is the farthest smallest element
to the right of Arr[0].
Arr[4] is the farthest smallest element
to the right of Arr[2].
And for the rest of the elements, there is
no smaller element to their right.
Example 2:
Input:
N=5
Arr[] = {1, 2, 3, 4, 0}
Output:
4 4 4 4 -1
Your Task:
You don't need to read input or print anything. Your task is to complete the function farNumber() which takes the N (number of elements in Array Arr) ,Arr[], and returns the array of farthest element to the right for every element of the array.
Expected Time Complexity: O(N*logN)
Expected Auxiliary Space: O(N)
Constraints:
1 ≤ N ≤ 1e5
0 ≤ Arr[i] ≤ 1e9 | class Solution:
def farNumber(self, n, arr):
arr2 = sorted(enumerate(arr), key=lambda x: [x[1], x[0]])
list1 = [0] * n
list1[arr2[0][0]] = -1
index_max = arr2[0][0]
for i in range(1, n):
if arr2[i][0] > index_max:
list1[arr2[i][0]] = -1
index_max = arr2[i][0]
else:
list1[arr2[i][0]] = index_max
return list1 | CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR LIST VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR NUMBER VAR ASSIGN VAR VAR VAR NUMBER NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER VAR RETURN VAR |
Given an array Arr[] of size N. For every element in the array, the task is to find the index of the farthest element in the array to the right which is smaller than the current element. If no such number exists then print -1.
Note: 0 based indexing.
Example 1:
Input:
N=5
Arr[] = {3, 1, 5, 2, 4}
Output:
3 -1 4 -1 -1
Explanation:
Arr[3] is the farthest smallest element
to the right of Arr[0].
Arr[4] is the farthest smallest element
to the right of Arr[2].
And for the rest of the elements, there is
no smaller element to their right.
Example 2:
Input:
N=5
Arr[] = {1, 2, 3, 4, 0}
Output:
4 4 4 4 -1
Your Task:
You don't need to read input or print anything. Your task is to complete the function farNumber() which takes the N (number of elements in Array Arr) ,Arr[], and returns the array of farthest element to the right for every element of the array.
Expected Time Complexity: O(N*logN)
Expected Auxiliary Space: O(N)
Constraints:
1 ≤ N ≤ 1e5
0 ≤ Arr[i] ≤ 1e9 | class Solution:
def farNumber(self, N, Arr):
ans = []
for i in range(0, N - 1):
sm = -1
se = Arr[i]
for j in range(N - 1, i, -1):
if Arr[j] < se:
sm = j
break
ans.append(sm)
ans.append(-1)
return ans | CLASS_DEF FUNC_DEF ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER RETURN VAR |
Given an array Arr[] of size N. For every element in the array, the task is to find the index of the farthest element in the array to the right which is smaller than the current element. If no such number exists then print -1.
Note: 0 based indexing.
Example 1:
Input:
N=5
Arr[] = {3, 1, 5, 2, 4}
Output:
3 -1 4 -1 -1
Explanation:
Arr[3] is the farthest smallest element
to the right of Arr[0].
Arr[4] is the farthest smallest element
to the right of Arr[2].
And for the rest of the elements, there is
no smaller element to their right.
Example 2:
Input:
N=5
Arr[] = {1, 2, 3, 4, 0}
Output:
4 4 4 4 -1
Your Task:
You don't need to read input or print anything. Your task is to complete the function farNumber() which takes the N (number of elements in Array Arr) ,Arr[], and returns the array of farthest element to the right for every element of the array.
Expected Time Complexity: O(N*logN)
Expected Auxiliary Space: O(N)
Constraints:
1 ≤ N ≤ 1e5
0 ≤ Arr[i] ≤ 1e9 | class Solution:
def farNumber(self, n, a):
pref = [0] * n
pref[n - 1] = a[n - 1]
for i in range(n - 2, -1, -1):
pref[i] = min(a[i], pref[i + 1])
res = []
for i in range(n):
low = i + 1
high = n - 1
ans = -1
while low <= high:
mid = (low + high) // 2
if a[i] <= pref[mid]:
high = mid - 1
else:
ans = mid
low = mid + 1
res.append(ans)
return res | CLASS_DEF FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR RETURN VAR |
Given an array Arr[] of size N. For every element in the array, the task is to find the index of the farthest element in the array to the right which is smaller than the current element. If no such number exists then print -1.
Note: 0 based indexing.
Example 1:
Input:
N=5
Arr[] = {3, 1, 5, 2, 4}
Output:
3 -1 4 -1 -1
Explanation:
Arr[3] is the farthest smallest element
to the right of Arr[0].
Arr[4] is the farthest smallest element
to the right of Arr[2].
And for the rest of the elements, there is
no smaller element to their right.
Example 2:
Input:
N=5
Arr[] = {1, 2, 3, 4, 0}
Output:
4 4 4 4 -1
Your Task:
You don't need to read input or print anything. Your task is to complete the function farNumber() which takes the N (number of elements in Array Arr) ,Arr[], and returns the array of farthest element to the right for every element of the array.
Expected Time Complexity: O(N*logN)
Expected Auxiliary Space: O(N)
Constraints:
1 ≤ N ≤ 1e5
0 ≤ Arr[i] ≤ 1e9 | class Solution:
def farNumber(self, N, Arr):
lista = [-1] * N
for i in range(N):
cn = 0
j = N - 1
flag = 0
while j > i and flag != 1:
if Arr[j] < Arr[i] and j >= cn:
cn = j
lista[i] = cn
flag = 1
else:
j = j - 1
return lista | CLASS_DEF FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR VAR NUMBER IF VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR |
Given an array Arr[] of size N. For every element in the array, the task is to find the index of the farthest element in the array to the right which is smaller than the current element. If no such number exists then print -1.
Note: 0 based indexing.
Example 1:
Input:
N=5
Arr[] = {3, 1, 5, 2, 4}
Output:
3 -1 4 -1 -1
Explanation:
Arr[3] is the farthest smallest element
to the right of Arr[0].
Arr[4] is the farthest smallest element
to the right of Arr[2].
And for the rest of the elements, there is
no smaller element to their right.
Example 2:
Input:
N=5
Arr[] = {1, 2, 3, 4, 0}
Output:
4 4 4 4 -1
Your Task:
You don't need to read input or print anything. Your task is to complete the function farNumber() which takes the N (number of elements in Array Arr) ,Arr[], and returns the array of farthest element to the right for every element of the array.
Expected Time Complexity: O(N*logN)
Expected Auxiliary Space: O(N)
Constraints:
1 ≤ N ≤ 1e5
0 ≤ Arr[i] ≤ 1e9 | class Solution:
def findLower(self, searchArray, ele, N, i):
n = len(searchArray)
l = i + 1
r = n - 1
val = -1
while l <= r:
m = (l + r) // 2
if searchArray[m] < ele:
val = m
l = m + 1
else:
r = m - 1
return val
def farNumber(self, N, arr):
minArray, res = [0] * N, [-1] * N
res[-1] = -1
minArray[-1], minEle = arr[-1], arr[-1]
for i in range(N - 2, -1, -1):
if arr[i] < minEle:
minEle = arr[i]
res[i] = -1
else:
res[i] = self.findLower(minArray, arr[i], N, i)
minArray[i] = minEle
return res | CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR FUNC_DEF ASSIGN VAR VAR BIN_OP LIST NUMBER VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER NUMBER ASSIGN VAR NUMBER VAR VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR RETURN VAR |
Given an array Arr[] of size N. For every element in the array, the task is to find the index of the farthest element in the array to the right which is smaller than the current element. If no such number exists then print -1.
Note: 0 based indexing.
Example 1:
Input:
N=5
Arr[] = {3, 1, 5, 2, 4}
Output:
3 -1 4 -1 -1
Explanation:
Arr[3] is the farthest smallest element
to the right of Arr[0].
Arr[4] is the farthest smallest element
to the right of Arr[2].
And for the rest of the elements, there is
no smaller element to their right.
Example 2:
Input:
N=5
Arr[] = {1, 2, 3, 4, 0}
Output:
4 4 4 4 -1
Your Task:
You don't need to read input or print anything. Your task is to complete the function farNumber() which takes the N (number of elements in Array Arr) ,Arr[], and returns the array of farthest element to the right for every element of the array.
Expected Time Complexity: O(N*logN)
Expected Auxiliary Space: O(N)
Constraints:
1 ≤ N ≤ 1e5
0 ≤ Arr[i] ≤ 1e9 | class Solution:
def SuffixArrayOfMin(self, arr):
suffix_arr = [0] * len(arr)
suffix_arr[-1] = arr[-1]
for i in range(len(arr) - 1, 0, -1):
suffix_arr[i - 1] = min(arr[i - 1], suffix_arr[i])
return suffix_arr
def BinarySearchInSuffixArray(self, arr):
result = []
suffix_arr = self.SuffixArrayOfMin(arr)
for i in range(len(arr)):
start = i + 1
end = len(arr) - 1
ans = -1
while start <= end:
mid = (start + end) // 2
if suffix_arr[mid] < arr[i]:
ans = mid
start = mid + 1
else:
end = mid - 1
result.append(ans)
return result
def farNumber(self, N, Arr):
return self.BinarySearchInSuffixArray(Arr) | CLASS_DEF FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR RETURN VAR FUNC_DEF RETURN FUNC_CALL VAR VAR |
Given an array Arr[] of size N. For every element in the array, the task is to find the index of the farthest element in the array to the right which is smaller than the current element. If no such number exists then print -1.
Note: 0 based indexing.
Example 1:
Input:
N=5
Arr[] = {3, 1, 5, 2, 4}
Output:
3 -1 4 -1 -1
Explanation:
Arr[3] is the farthest smallest element
to the right of Arr[0].
Arr[4] is the farthest smallest element
to the right of Arr[2].
And for the rest of the elements, there is
no smaller element to their right.
Example 2:
Input:
N=5
Arr[] = {1, 2, 3, 4, 0}
Output:
4 4 4 4 -1
Your Task:
You don't need to read input or print anything. Your task is to complete the function farNumber() which takes the N (number of elements in Array Arr) ,Arr[], and returns the array of farthest element to the right for every element of the array.
Expected Time Complexity: O(N*logN)
Expected Auxiliary Space: O(N)
Constraints:
1 ≤ N ≤ 1e5
0 ≤ Arr[i] ≤ 1e9 | class Solution:
def farNumber(self, N, Arr):
a = []
for i in range(N):
for j in range(N - 1, i, -1):
if Arr[j] < Arr[i]:
a.append(j)
break
else:
a.append(-1)
return a | CLASS_DEF FUNC_DEF ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER RETURN VAR |
Given an array Arr[] of size N. For every element in the array, the task is to find the index of the farthest element in the array to the right which is smaller than the current element. If no such number exists then print -1.
Note: 0 based indexing.
Example 1:
Input:
N=5
Arr[] = {3, 1, 5, 2, 4}
Output:
3 -1 4 -1 -1
Explanation:
Arr[3] is the farthest smallest element
to the right of Arr[0].
Arr[4] is the farthest smallest element
to the right of Arr[2].
And for the rest of the elements, there is
no smaller element to their right.
Example 2:
Input:
N=5
Arr[] = {1, 2, 3, 4, 0}
Output:
4 4 4 4 -1
Your Task:
You don't need to read input or print anything. Your task is to complete the function farNumber() which takes the N (number of elements in Array Arr) ,Arr[], and returns the array of farthest element to the right for every element of the array.
Expected Time Complexity: O(N*logN)
Expected Auxiliary Space: O(N)
Constraints:
1 ≤ N ≤ 1e5
0 ≤ Arr[i] ≤ 1e9 | class Solution:
def farNumber(self, N, Arr):
new = [-1] * N
suf = [float("inf")] * N
suf[N - 1] = Arr[N - 1]
p = N - 1
mins = float("inf")
while p >= 0:
mins = min(mins, Arr[p])
suf[p] = mins
p -= 1
for i in range(N - 2, -1, -1):
l = i + 1
r = N - 1
while l <= r:
m = (l + r) // 2
if suf[m] < Arr[i]:
new[i] = m
l = m + 1
else:
r = m - 1
return new | CLASS_DEF FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR BIN_OP LIST FUNC_CALL VAR STRING VAR ASSIGN VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING WHILE VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR |
Given an array Arr[] of size N. For every element in the array, the task is to find the index of the farthest element in the array to the right which is smaller than the current element. If no such number exists then print -1.
Note: 0 based indexing.
Example 1:
Input:
N=5
Arr[] = {3, 1, 5, 2, 4}
Output:
3 -1 4 -1 -1
Explanation:
Arr[3] is the farthest smallest element
to the right of Arr[0].
Arr[4] is the farthest smallest element
to the right of Arr[2].
And for the rest of the elements, there is
no smaller element to their right.
Example 2:
Input:
N=5
Arr[] = {1, 2, 3, 4, 0}
Output:
4 4 4 4 -1
Your Task:
You don't need to read input or print anything. Your task is to complete the function farNumber() which takes the N (number of elements in Array Arr) ,Arr[], and returns the array of farthest element to the right for every element of the array.
Expected Time Complexity: O(N*logN)
Expected Auxiliary Space: O(N)
Constraints:
1 ≤ N ≤ 1e5
0 ≤ Arr[i] ≤ 1e9 | class Solution:
def farNumber(self, N, Arr):
lst = [-1] * N
for i in range(N):
for j in range(N - 1, i, -1):
if Arr[j] < Arr[i]:
lst[i] = j
break
return lst | CLASS_DEF FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR VAR VAR VAR ASSIGN VAR VAR VAR RETURN VAR |
Given an array Arr[] of size N. For every element in the array, the task is to find the index of the farthest element in the array to the right which is smaller than the current element. If no such number exists then print -1.
Note: 0 based indexing.
Example 1:
Input:
N=5
Arr[] = {3, 1, 5, 2, 4}
Output:
3 -1 4 -1 -1
Explanation:
Arr[3] is the farthest smallest element
to the right of Arr[0].
Arr[4] is the farthest smallest element
to the right of Arr[2].
And for the rest of the elements, there is
no smaller element to their right.
Example 2:
Input:
N=5
Arr[] = {1, 2, 3, 4, 0}
Output:
4 4 4 4 -1
Your Task:
You don't need to read input or print anything. Your task is to complete the function farNumber() which takes the N (number of elements in Array Arr) ,Arr[], and returns the array of farthest element to the right for every element of the array.
Expected Time Complexity: O(N*logN)
Expected Auxiliary Space: O(N)
Constraints:
1 ≤ N ≤ 1e5
0 ≤ Arr[i] ≤ 1e9 | class Solution:
def farNumber(self, n, arr):
ans = []
for i in range(n):
check = -1
for j in range(n - 1, i - 1, -1):
if arr[j] < arr[i]:
ans.append(j)
check = 1
break
if check < 0:
ans.append(-1)
return ans | CLASS_DEF FUNC_DEF ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER NUMBER IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER RETURN VAR |
Given an array Arr[] of size N. For every element in the array, the task is to find the index of the farthest element in the array to the right which is smaller than the current element. If no such number exists then print -1.
Note: 0 based indexing.
Example 1:
Input:
N=5
Arr[] = {3, 1, 5, 2, 4}
Output:
3 -1 4 -1 -1
Explanation:
Arr[3] is the farthest smallest element
to the right of Arr[0].
Arr[4] is the farthest smallest element
to the right of Arr[2].
And for the rest of the elements, there is
no smaller element to their right.
Example 2:
Input:
N=5
Arr[] = {1, 2, 3, 4, 0}
Output:
4 4 4 4 -1
Your Task:
You don't need to read input or print anything. Your task is to complete the function farNumber() which takes the N (number of elements in Array Arr) ,Arr[], and returns the array of farthest element to the right for every element of the array.
Expected Time Complexity: O(N*logN)
Expected Auxiliary Space: O(N)
Constraints:
1 ≤ N ≤ 1e5
0 ≤ Arr[i] ≤ 1e9 | class Solution:
def farNumber(self, N, Arr):
r = 0
l = N - 1
while r < N:
if r == l:
Arr[r] = -1
l = N - 1
r = r + 1
elif Arr[r] > Arr[l]:
Arr[r] = l
l = N - 1
r = r + 1
else:
l = l - 1
return Arr | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR IF VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR |
Given an array Arr[] of size N. For every element in the array, the task is to find the index of the farthest element in the array to the right which is smaller than the current element. If no such number exists then print -1.
Note: 0 based indexing.
Example 1:
Input:
N=5
Arr[] = {3, 1, 5, 2, 4}
Output:
3 -1 4 -1 -1
Explanation:
Arr[3] is the farthest smallest element
to the right of Arr[0].
Arr[4] is the farthest smallest element
to the right of Arr[2].
And for the rest of the elements, there is
no smaller element to their right.
Example 2:
Input:
N=5
Arr[] = {1, 2, 3, 4, 0}
Output:
4 4 4 4 -1
Your Task:
You don't need to read input or print anything. Your task is to complete the function farNumber() which takes the N (number of elements in Array Arr) ,Arr[], and returns the array of farthest element to the right for every element of the array.
Expected Time Complexity: O(N*logN)
Expected Auxiliary Space: O(N)
Constraints:
1 ≤ N ≤ 1e5
0 ≤ Arr[i] ≤ 1e9 | class Solution:
def farNumber(self, n, a):
l = []
res = [-1] * n
for i in range(n):
l.append([a[i], i])
l.sort()
mind = None
for i in range(n):
if mind == None or mind < l[i][1]:
mind = l[i][1]
else:
res[l[i][1]] = mind
return res | CLASS_DEF FUNC_DEF ASSIGN VAR LIST ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR LIST VAR VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR NONE FOR VAR FUNC_CALL VAR VAR IF VAR NONE VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER VAR RETURN VAR |
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