Description
https://leetcode.com/problems/split-array-largest-sum/
Given an array nums 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.
Example 1:
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.
Example 2:
Input: nums = [1,2,3,4,5], m = 2 Output: 9
Example 3:
Input: nums = [1,4,4], m = 3 Output: 4
Constraints:
1 <= nums.length <= 10000 <= nums[i] <= 1061 <= m <= min(50, nums.length)
Explanation
One approach to solving the problem is to use binary search. We can use binary search to search for a largest subarray sum which can make the array split into m parts. The lower bound of the largest subarray sum is the maximum number of the array when m equals the length of the array. The higher bound is the sum of the array when m equals 1. We can base on how many pieces split to adjust the largest subarray sum value.
Python Solution
class Solution:
def splitArray(self, nums: List[int], m: int) -> int:
"""
:type nums: List[int]
:type m: int
:rtype: int
"""
n = len(nums)
start = max(nums)
end = sum(nums)
while start + 1< end:
mid = start + (end - start) // 2
if self.split_into_pieces(nums, mid) > m:
start = mid + 1
else:
end = mid
if self.split_into_pieces(nums, start) <= m:
return start
return end
def split_into_pieces(self, nums, largest_sum):
previous_sum = 0
pieces = 1
for num in nums:
if previous_sum + num > largest_sum:
previous_sum = num
pieces += 1
else:
previous_sum += num
return pieces- Time Complexity: O(Nlog(N))
- Space Complexity: O(1)
class Solution {
public int firstMissingPositive(int[] A) {
int m = 1;
HashSet set = new HashSet();
for(int num: nums){
if(num>m){
set.add(num);
}else if(num==m){
m++;
while(set.contains(m)){
set.remove(m);
m++;
}
}
}
return m;
}
}