I'm posting my code for a LeetCode problem copied here. If you would like to review, please do so. Thank you for your time!
Problem
Given an array which consists of non-negative integers and an integer
m
, you can split the array intom
non-empty continuous subarrays. Write an algorithm to minimize the largest sum among thesem
subarrays.Note:
If
n
is the length of array, assume the following constraints are satisfied:$$ 1 \le n \le 1000 \\ 1 \le m \le \min(50, n) $$
Example:
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.
Code
#include <vector>
class Solution {
public:
static int splitArray(const std::vector<int> &nums, const int m) {
size_t lo = 0;
size_t hi = 0;
for (int num : nums) {
lo = std::max(lo, (size_t) num);
hi += num;
}
while (lo < hi) {
size_t mid = lo + (hi - lo) / 2;
if (partitioning_is_valid(nums, m - 1, mid)) {
hi = mid;
} else {
lo = mid + 1;
}
}
return lo;
}
private:
static bool partitioning_is_valid(const std::vector<int> &nums, int partitions, size_t partition_max_sum) {
size_t accumulate_curr_partition = 0;
for (int num : nums) {
if (num > partition_max_sum) {
return false;
} else if (accumulate_curr_partition + num <= partition_max_sum) {
accumulate_curr_partition += num;
} else {
partitions--;
accumulate_curr_partition = num;
if (partitions < 0) {
return false;
}
}
}
return true;
}
};
Reference
LeetCode has a template for answering questions. There is usually a class named Solution
with one or more public
functions which we are not allowed to rename. For this question, the template is:
class Solution {
public:
int splitArray(vector<int>& nums, int m) {
}
};
lo = *std ::max_element(nums.begin(), nums.end())
andhi = std :: accumulate(nums.begin(), nums.end(), 0)
and remove thefor
loop. Do make sure that to use this functions you need to include<algorithm>
header file which by default is included in the Leet-Code default template. \$\endgroup\$