## largest sum of contiguous subarray No Larger than k

maximum sum subarray having sum less than or equal to given sum

max sum of rectangle no larger than k

maximum subarray

max sum of rectangle no larger than k huahua

maximum sum subarray of size k - leetcode

maximum size subarray sum equals k

maximum sum of subarray close to k

For example, we have

{2,2,-1}, when k = 0, return -1. when k = 3, return 3.

This is even tricky because we have negative numbers and an additional variable k. k can be any value, negative, don't make any assumption.

I cannot refer to https://en.wikipedia.org/wiki/Maximum_subarray_problem and https://www.youtube.com/watch?v=yCQN096CwWM to solve this problem.

Can any body help me? Better use Java or JavaScript.

Here is a classic algorithm o(n) for the maximum(no variable k):

public int maxSubArray(int[] nums) { int max = nums[0]; int tsum = nums[0]; for(int i=1;i<nums.length;i++){ tsum = Math.max(tsum+nums[i],nums[i]); max = Math.max(max,tsum); } return max; }

This is an o(nlogn) solution referred to https://www.quora.com/Given-an-array-of-integers-A-and-an-integer-k-find-a-subarray-that-contains-the-largest-sum-subject-to-a-constraint-that-the-sum-is-less-than-k

private int maxSumSubArray(int[] a , int k){ int max = Integer.MIN_VALUE; int sumj = 0; TreeSet<Integer> ts = new TreeSet(); ts.add(0); for(int i=0;i<a.length;i++){ sumj += a[i]; Integer gap = ts.ceiling(sumj - k); if(gap != null) max = Math.max(max, sumj - gap); ts.add(sumj); } return max; }

**largest sum of contiguous subarray No Larger than k,** largest sum of contiguous subarray No Larger than k find subarray with max sum less or equal than k // slight variation of "Maximum Subarray" (no less than k � // find subarray with max sum less or equal than k // slight variation of "Maximum Subarray" (no less than k requirement) which can be seen as a special case where k==Integer.MAX_VALUE so we find the max subarray sum less than +infinity which is the maximum subarray sum public int maxSubArray (int [] nums) { long k = Integer.Max_VALUE; // parameter long maxLessThanK = Integer.MIN_VALUE; TreeSet<Long> set = new TreeSet<>(); set. add (0L); // so that subarrays starting from index 0 can be

Here's a naive algorithm that runs in O(n²).

std::array<int, 3> input = {2, 2, -1}; int k = -1; int sum = 0, largestSum = *std::min_element(input.begin(), input.end()) -1; int i = 0, j = 0; int start = 0, end = 0; while (largestSum != k && i != input.size()) { sum += input[j]; if (sum <= k && sum > largestSum) { largestSum = sum; start = i; end = j; } ++j; if (j == input.size()) { ++i; j = i; sum = 0; } }

That's C++ but it shouldn't be hard to write in Java or Javascript.
It basically tries every sum possible (there are `n*(n+1)/2`

) and stops if it finds k.

`largestSum`

must be initialized to a low-enough value. Since the minimum element of the input could equal k, I subtracted 1 to it.
`start`

and `end`

are the first and last indices of the final subarray.

Of course, it could be improved if you had any constraints on the inputs.

**Maximum sum subarray having sum less than or equal to given sum ,** Longest subarray whose elements can be made equal by maximum K increments � Subarray with difference between maximum and minimum element greater� Largest Sum Contiguous Subarray Write an efficient program to find the sum of contiguous subarray within a one-dimensional array of numbers which has the largest sum. Recommended: Please solve it on “ PRACTICE ” first, before moving on to the solution.

I was influenced by the classic solution mentioned in the question. This problem can be simply solved by an o(n^2) solution:

private int maxSumSubArray(int[] a , int k){ int max = Integer.MIN_VALUE; for(int i=0;i<a.length;i++){ int tsum = 0; for(int j=i;j<a.length;j++){ tsum += a[j]; if(tsum <= k) max=Math.max(max,tsum); } } return max; }

**Largest Sum Contiguous Subarray,** Maximum sum contiguous subarray within a one-dimensional array of is updated to 7 because max_ending_here is greater than max_so_far for i=7, a[7] = -3� Because the sum of rectangle [ [0, 1], [-2, 3]] is 2 and 2 is the max number no larger than k (k = 2).

**Maximum Sum of Subarray Close to K,** Given an array, find the maximum sum of subarray close to k but not larger than k. Java Solution The solution to this problem is obvious when we draw. Given an array, find the maximum sum of subarray close to k but not larger than k. Java Solution. The solution to this problem is obvious when we draw the following diagram.

**LeetCode – Max Sum of Rectangle No Larger Than K (Java),** The key to the optimal solution is using a tree set to calculate the maximum sum of subarray close to k. Java Solution 1. public int maxSumSubmatrix(int[]� 363. Max Sum of Rectangle No Larger Than K (Hard) 364. Nested List Weight Sum II (Medium) 366. Find Leaves of Binary Tree (Medium) 368. Largest Divisible Subset (Medium) 369. Plus One Linked List (Medium) 370.

**Improved algorithms for the k maximum-sums ,** m�n, our algorithm for finding the k maximum-sum subarrays is the first one Linear-time algorithms for finding all the non-overlapping maximal segments were given in [7,18]. We first output those elements whose values are greater than ai. Given an array A of n real numbers, the maximum subarray problem is to find a contiguous subarray which has the largest sum. The k-maximum subarrays problem is to find k such subarrays with the largest sums. For the 1-maximum subarray the well known divide-and-conquer algorithm, presented in most textbooks, although suboptimal, is easy to implement and can be made optimal with a simple change that speeds up the combine phase.

##### Comments

- Any complexity requirements ?
- No more requirement, can you solve it?
- What value must be not greater than k? Length of subarray or sum of subarray? Answer for test [1, 2, 3], k = 2 is 5 or 2?
- As the title, max sum of contiguous subarray no bigger than k.