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This approach uses binary search to efficiently find the minimized largest sum. The range for binary search is between the maximum single element in the array (lower bound) and the sum of the array (upper bound). For each candidate middle value in the binary search, we check if it's possible to partition the array into k or fewer subarrays such that none has a sum greater than this candidate. This check is done through a greedy approach: we iterate over the array, adding elements to a running sum until adding another element would exceed the candidate sum, at which point we start a new subarray.
Time Complexity: O(n log(sum - max)), where sum is the total sum of array and max is the maximum element.
Space Complexity: O(1), as no additional space proportional to input size is used.
1def splitArray(nums, k):
2 left, right = max(nums), sum(nums)
3 while left < right:
4 mid = The Python function performs binary search on the possible maximum subarray sum. The function uses a greedy strategy to divide the array, updating the search boundaries according to how many pieces are formed. The minimized largest sum is returned after the search completes.
We can also use dynamic programming to solve this problem by maintaining a DP table where dp[i][j] means the minimum largest sum for splitting the first i elements into j subarrays. The recurrence is based on considering different potential previous cut positions and calculating the maximum sum for the last subarray in each case, iterating across feasible positions.
Time Complexity: O(n^2 * k), as each subproblem depends on earlier solutions.
Space Complexity: O(n*k), for the dp table.
1
This Java solution makes use of a prefix sum array and a multi-dimensional dp array. The recurrence relation assesses each subarray for minimum discomfort considering all potential subarray distributions using nested loops that iterate over feasible subarray ends.