
Problem Statement
In the given problem, we have an integer array nums
and an integer k
. The task is to split the array nums
into k
non-empty contiguous subarrays. The objective is to minimize the largest sum among these k
subarrays. In simpler terms, we need to find a way to divide the array such that the sum of the largest subarray (in terms of sum, not length) is as small as possible. The result should be this minimized value of the largest subarray's sum.
Examples
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)
Approach and Intuition
The problem essentially tests the ability to balance partitions of the array such that no single partition is overwhelmingly large compared to others when viewed in terms of their sums. Here's how we approach it:
Example Insights
Input:
nums = [7,2,5,10,8], k = 2
Output:
18
A successful split for minimizing the largest sum for this case is
[7,2,5]
and[10,8]
, resulting in the sums of14
and18
, respectively. Here,18
is the largest sum which is minimized compared to other possible splits.Input:
nums = [1,2,3,4,5], k = 2
Output:
9
A practical split here is
[1,2,3]
and[4,5]
, with the sums being6
and9
. Again,9
is minimized for the largest sum splits.
Logical Breakdown
- Identify the nature of the problem: It calls for partitioning, which hints at dynamic programming or binary search to help decide if a potential maximum sum can be a valid split.
- Establish the search space: The minimum possible value for the largest subarray sum is the maximum single element in
nums
, and the maximum possible value is the sum of all elements innums
. - Utilize binary search within this bounded search space to efficiently hone in on the minimum of these maximum sum values. At each mid-value (potential maximum sum), check if it's possible to create
k
subarrays without exceeding this sum. - Implementation would involve iterating and keeping a running total until adding another element exceeds the mid-value. If during this process, more than
k
subarrays are needed, the mid is too low and should be increased.
Constraints Insights
- The array length of up to
1000
and the values of elements up to10^6
suggest that a direct combinatorial approach would be inefficient. Hence, efficiency in traversal and partition checks must be considered. k
can only be as large as50
or the length of the array, whichever is smaller, putting a natural limit on the number of partitions we need to consider.
This method ensures the solution not only works efficiently but also adheres to the varying sizes and conditions dictated by different inputs within the constraints provided.
Solutions
- C++
class Solution {
public:
int computeMinSubarrays(vector<int>& arr, int limit) {
int sum = 0;
int required = 0;
for (int num : arr) {
if (sum + num <= limit) {
sum += num;
} else {
sum = num;
required++;
}
}
return required + 1;
}
int splitArray(vector<int>& arr, int m) {
int totalSum = 0;
int highestValue = INT_MIN;
for (int num : arr) {
totalSum += num;
highestValue = max(highestValue, num);
}
int low = highestValue;
int high = totalSum;
int bestSmallestMax = 0;
while (low <= high) {
int mid = low + (high - low) / 2;
if (computeMinSubarrays(arr, mid) <= m) {
high = mid - 1;
bestSmallestMax = mid;
} else {
low = mid + 1;
}
}
return bestSmallestMax;
}
};
The provided solution in C++ addresses the problem of determining the smallest possible largest sum of m
subarrays into which a given array arr
can be divided.
Firstly, the program utilizes two main functions:
computeMinSubarrays
accepts an integer arrayarr
and a limit value. It iterates over the array to determine the minimum number of subarrays needed without exceeding the given limit. This is achieved by aggregating elements until the addition of a new element would exceed the limit, at which point a new subarray is started.splitArray
is the main function that utilizes binary search to find the optimal solution. It initializes with the largest individual element of the array aslow
and the sum of all elements ashigh
, ensuring all possible sums are bracketed between these two extremes.The binary search repeatedly calculates the mid-value between
low
andhigh
. For each mid-value, which represents a potential maximum subarray sum,computeMinSubarrays
is called to check how many subarrays are required if that mid-value is the maximum sum allowed per subarray.If the number of subarrays required is less than or equal to
m
, the mid-value is a valid candidate for the answer, andhigh
is adjusted to narrow the search range.Conversely, if the mid-value requires more than
m
subarrays, it means the mid-value is too small, andlow
is adjusted to search for a larger possible value.The binary search continues until
low
surpasseshigh
, at which point the optimal smallest maximum sum of subarrays is found inbestSmallestMax
.
This approach leverages the binary search algorithm for efficient determination of the required value, making the search logarithmic relative to the sum range, combined with linear checks via the computeMinSubarrays
function, resulting in an efficient and effective solution.
- Java
class Solution {
private int countSubarrays(int[] array, int limit) {
int total = 0;
int count = 0;
for (int value : array) {
if (total + value <= limit) {
total += value;
} else {
total = value;
count++;
}
}
return count + 1;
}
public int splitArray(int[] array, int divisions) {
int totalSum = 0;
int highest = Integer.MIN_VALUE;
for (int value : array) {
totalSum += value;
highest = Math.max(highest, value);
}
int lowerBound = highest;
int upperBound = totalSum;
int optimalMax = 0;
while (lowerBound <= upperBound) {
int mid = lowerBound + (upperBound - lowerBound) / 2;
if (countSubarrays(array, mid) <= divisions) {
upperBound = mid - 1;
optimalMax = mid;
} else {
lowerBound = mid + 1;
}
}
return optimalMax;
}
}
The Java solution provided resolves the problem named "Split Array Largest Sum". This program determines the smallest sum among the largest sums of the subarrays when an array is split into a specified number of divisions. Here's a breakdown of how the solution achieves this:
Method
countSubarrays
:- This method calculates how many subarrays are required for a given maximum subarray sum (referred to as
limit
). - It iterates through the array, summing up elements until adding another would surpass the limit. At this point, it starts a new subarray, thus tracking the number of subarrays needed for the specified limit.
- This method calculates how many subarrays are required for a given maximum subarray sum (referred to as
Method
splitArray
:- Uses binary search to find the minimum maximum subarray sum.
- It first calculates the
totalSum
of all array elements and finds the largest single element value,highest
, as potential bounds for the binary search. lowerBound
is set tohighest
andupperBound
tototalSum
.- The binary search is adjusted iteratively by recalculating the middle sum (
mid
) and usingcountSubarrays
to check if the givenmid
can be a feasible largest sum with the allowed number of divisions. - The search space is narrowed based on whether the number of divisions needed is less than or equal to the allowed divisions, adjusting
lowerBound
andupperBound
accordingly.
Finally, the method returns the optimal largest sum for the required divisions. This approach ensures efficiency by leveraging the binary search method to minimize the maximum sum required to split the array within the given constraints.
- Python
class Solution:
def splitArray(self, numbers: List[int], partitions: int) -> int:
def subarrays_needed(max_sum_limit: int) -> int:
total_sum = 0
required_splits = 0
for num in numbers:
if total_sum + num <= max_sum_limit:
total_sum += num
else:
total_sum = num
required_splits += 1
return required_splits + 1
left = max(numbers)
right = sum(numbers)
while left <= right:
mid_value = (left + right) // 2
if subarrays_needed(mid_value) <= partitions:
right = mid_value - 1
minimum_sum = mid_value
else:
left = mid_value + 1
return minimum_sum
The provided code addresses the problem of splitting an array into a specified number of partitions such that the maximum sum of any partition is minimized. This scenario is tackled using a binary search approach over potential sums, implementing it in Python.
- Start by defining the function
splitArray
with parameters for the list of integers and the number of desired partitions. - Introduce an internal helper function
subarrays_needed
that calculates the number of subarrays required given a maximum sum limit. This function iterates through the numbers, building subarrays and counting them based on the given sum limit. - Establish the initial bounds for binary search with:
left
set as the maximum value in the numbers list, ensuring that the maximum sum cannot be lower than any single element.right
as the sum of all elements, representing the scenario where only one partition is used.
- Execute the binary search:
- Calculate the middle point between the current left and right bounds.
- Use the
subarrays_needed
function to determine the number of partitions required for the mid value as a maximum sum. - If the required partitions count is within the allowed number, adjust the
right
boundary to search for a potentially smaller maximum sum. Store this as the current best known minimum sum. - If the partition count exceeds the limit, adjust the
left
boundary to allow for a larger sum limit and continue the search.
- At the end of the binary search, the
minimum_sum
variable holds the minimized largest sum possible under the given constraints, which is returned as the function's result.
This approach efficiently narrows down the potential maximum sums, ensuring an optimal solution through the binary search methodology and careful partition counting.
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