- Split Array Largest Sum
Hard
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Given an integer array nums and an integer k, split nums into k non-empty subarrays such that the largest sum of any subarray is minimized.
Return the minimized largest sum of the split.
A subarray is a contiguous part of the array.
Example 1:
Input: nums = [7,2,5,10,8], k = 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], k = 2
Output: 9
Explanation: There are four ways to split nums into two subarrays.
The best way is to split it into [1,2,3] and [4,5], where the largest sum among the two subarrays is only 9.
Constraints:
1 <= nums.length <= 1000
0 <= nums[i] <= 106
1 <= k <= min(50, nums.length)
解法1：Binary Search。
不过感觉这题题目似乎有问题，应该不是刚好划分成k个组，而是<=k个组都可以。
class Solution {
public:
int splitArray(vector<int>& nums, int k) {
int n = nums.size();
int totalSum = 0, maxNum = 0;
for (auto num : nums) {
maxNum = max(maxNum, num);
totalSum += num;
}
int start = maxNum, end = totalSum;
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (numSubArray(nums, mid) <= k) {
end = mid;
} else {
start = mid;
}
}
if (numSubArray(nums, start) <= k) return start;
return end;
}
private:
// return num of subarrays with maximum sum of each subarray
int numSubArray(vector<int>& nums, int maxSum) {
int n = nums.size();
int res = 0, sum = 0;
for (int i = 0; i < n; i++) {
sum += nums[i];
if (sum > maxSum) {
sum = nums[i];
res ++;
}
}
if (sum > 0) res++;
return res;
}
};