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Solution of divide array into group with min sum group questions
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''' | ||
A conveyor belt has packages that must be shipped from one port to another within D days. | ||
The i-th package on the conveyor belt has a weight of weights[i]. Each day, we load the ship with packages on the conveyor belt (in the order given by weights). We may not load more weight than the maximum weight capacity of the ship. | ||
Return the least weight capacity of the ship that will result in all the packages on the conveyor belt being shipped within D days. | ||
Example 1: | ||
Input: weights = [1,2,3,4,5,6,7,8,9,10], D = 5 | ||
Output: 15 | ||
Explanation: | ||
A ship capacity of 15 is the minimum to ship all the packages in 5 days like this: | ||
1st day: 1, 2, 3, 4, 5 | ||
2nd day: 6, 7 | ||
3rd day: 8 | ||
4th day: 9 | ||
5th day: 10 | ||
Note that the cargo must be shipped in the order given, so using a ship of capacity 14 and splitting the packages into parts like (2, 3, 4, 5), (1, 6, 7), (8), (9), (10) is not allowed. | ||
Example 2: | ||
Input: weights = [3,2,2,4,1,4], D = 3 | ||
Output: 6 | ||
Explanation: | ||
A ship capacity of 6 is the minimum to ship all the packages in 3 days like this: | ||
1st day: 3, 2 | ||
2nd day: 2, 4 | ||
3rd day: 1, 4 | ||
Note: | ||
1 <= D <= weights.length <= 50000 | ||
1 <= weights[i] <= 500 | ||
''' | ||
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class Solution(object): | ||
def shipWithinDays(self, weights, D): | ||
""" | ||
:type weights: List[int] | ||
:type D: int | ||
:rtype: int | ||
""" | ||
high, low = sum(weights)+1, max(weights) | ||
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while(low < high): | ||
mid = (high+low)/2 | ||
temp_left = mid | ||
packet_at_left = D-1 | ||
for weight in weights: | ||
if weight <= mid: | ||
if temp_left < weight: | ||
if packet_at_left == 0: | ||
low = mid+1 | ||
break | ||
packet_at_left -= 1 | ||
temp_left = mid-weight | ||
else: | ||
temp_left -= weight | ||
else: | ||
high = mid | ||
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return low | ||
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class Solution(object): | ||
def shipWithinDays(self, weights, D): | ||
""" | ||
:type weights: List[int] | ||
:type D: int | ||
:rtype: int | ||
""" | ||
left, right = max(weights), sum(weights) | ||
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while left < right: | ||
curr_sum, groups, invalid = 0, 0, True | ||
mid = left + ((right-left) >> 1) | ||
for weight in weights: | ||
if weight > mid: | ||
invalid = False | ||
break | ||
if curr_sum + weight > mid: | ||
groups += 1 | ||
curr_sum = 0 | ||
curr_sum += weight | ||
if invalid and groups < D: | ||
right = mid | ||
else: | ||
left = mid + 1 | ||
return left |
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''' | ||
Given an array which consists of non-negative integers and an integer m, you can split the array into m non-empty continuous subarrays. Write an algorithm to minimize the largest sum among these m subarrays. | ||
Note: | ||
If n is the length of array, assume the following constraints are satisfied: | ||
1 ≤ n ≤ 1000 | ||
1 ≤ m ≤ min(50, n) | ||
Examples: | ||
Input: | ||
nums = [7,2,5,10,8] | ||
m = 2 | ||
Output: | ||
18 | ||
Explanation: | ||
There are four ways to split nums into two subarrays. | ||
The best way is to split it into [7,2,5] and [10,8], | ||
where the largest sum among the two subarrays is only 18. | ||
''' | ||
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class Solution(object): | ||
def splitArray(self, nums, m): | ||
""" | ||
:type nums: List[int] | ||
:type m: int | ||
:rtype: int | ||
""" | ||
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left, right = max(nums), sum(nums) | ||
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while left < right: | ||
mid = left + ((right-left) >> 1) | ||
curr_sum, invalid, groups = 0, True, 0 | ||
for num in nums: | ||
if num > mid: | ||
inalid = False | ||
break | ||
if num + curr_sum > mid: | ||
groups += 1 | ||
curr_sum = 0 | ||
curr_sum += num | ||
if invalid and groups < m: | ||
right = mid | ||
else: | ||
left = mid + 1 | ||
return left | ||
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