# Maximum sub-array sum after dividing array into sub-arrays based on the given queries

Given an array **arr[]** and an integer **k**, we can cut this array at **k** different positions where **k[]** stores the positions of all the cuts required. The task is to print maximum sum among all the cuts after every cut made.

Every cut is of the form of an integer **x** where **x** denotes a cut between **arr[x]** and **arr[x + 1]**.

**Examples:**

Input:arr[] = {4, 5, 6, 7, 8}, k[] = {0, 2, 3, 1}

Output:

26

15

11

8

First cut ->{4} and {5, 6, 7, 8}. Maximum possible sum is 5 + 6 + 7 + 8 = 26

Second cut ->{4}, {5, 6} and {7, 8}. Maximum sum = 15

Third cut ->{4}, {5, 6}, {7} and {8}. Maximum sum = 11

Fourth cut ->{4}, {5}, {6}, {7} and {8}. Maximum sum = 8

Input:arr[] = {1, 2, 3}, k[] = {1}

Output:

3

**Naive approach:** Store the resulting pieces of the array in an ArrayList and after every cut compute linearly the maximum possible sum. But this method would require **O(n*k)** time to answer all the queries.

**Efficient approach:** We can represent each resulting piece of the array as a Piece object with data members **start** (start index of this piece), **end** (end index of this piece) and **value** (sum value of this piece). We can store these pieces in a TreeSet and **sort them by their sum values**. Therefore, after every cut we can get the Piece with largest sum value in O(log(n)).

- We have to make a prefix sum array of the array values to get the sum between two indices in constant time.
- We have to maintain another TreeSet with start indexes of all the current pieces so that we can find the exact piece to cut. For example, for a single piece:
- {1, 8} -> start = 1, end = 2, value = 9 and {6, 3, 9} -> start = 3, end = 5, value = 18.
- In order to cut index 4, we need to cut the second piece into two pieces as {6, 3} ans {9}. So we get the start index of which piece to cut from this TreeSet.

Below is the implementation of the above approach:

`// Java implementation of the approach ` `import` `java.io.IOException; ` `import` `java.io.InputStream; ` `import` `java.util.*; ` ` ` `// Comparator to sort the Pieces ` `// based on their sum values ` `class` `MyComp ` `implements` `Comparator<Piece> { ` ` ` `public` `int` `compare(Piece p1, Piece p2) ` ` ` `{ ` ` ` `if` `(p2.val != p1.val) ` ` ` `return` `p2.val - p1.val; ` ` ` `if` `(p1.start != p2.start) ` ` ` `return` `p2.start - p1.start; ` ` ` `return` `0` `; ` ` ` `} ` `} ` `class` `Piece { ` ` ` `int` `start; ` ` ` `int` `end; ` ` ` `int` `val; ` ` ` ` ` `// Constructor to initialize each Piece ` ` ` `Piece(` `int` `s, ` `int` `e, ` `int` `v) ` ` ` `{ ` ` ` `start = s; ` ` ` `end = e; ` ` ` `val = v; ` ` ` `} ` `} ` ` ` `class` `GFG { ` ` ` ` ` `// Function to perform the given queries on the array ` ` ` `static` `void` `solve(` `int` `n, ` `int` `k, ` `int` `cuts[], ` `int` `A[]) ` ` ` `{ ` ` ` ` ` `// Prefix sum array ` ` ` `int` `sum[] = ` `new` `int` `[n]; ` ` ` `sum[` `0` `] = A[` `0` `]; ` ` ` `for` `(` `int` `i = ` `1` `; i < n; i++) ` ` ` `sum[i] = sum[i - ` `1` `] + A[i]; ` ` ` ` ` `// TreeSet storing all the starts ` ` ` `TreeSet<Integer> t = ` `new` `TreeSet<>(); ` ` ` ` ` `// TreeSet storing the actual pieces ` ` ` `TreeSet<Piece> pq = ` `new` `TreeSet<>(` `new` `MyComp()); ` ` ` `Piece temp[] = ` `new` `Piece[n]; ` ` ` `temp[` `0` `] = ` `new` `Piece(` `0` `, n - ` `1` `, sum[n - ` `1` `]); ` ` ` ` ` `// Added the whole array or Piece of array ` ` ` `// as there is no cuts yet ` ` ` `pq.add(temp[` `0` `]); ` ` ` `t.add(` `0` `); ` ` ` ` ` `for` `(` `int` `i = ` `0` `; i < k; i++) { ` ` ` ` ` `// curr is the piece to be cut ` ` ` `int` `curr = t.floor(cuts[i]); ` ` ` `pq.remove(temp[curr]); ` ` ` `int` `end = temp[curr].end; ` ` ` ` ` `// When a piece with start = s and end = e ` ` ` `// is cut at index i, two pieces are created with ` ` ` `// start = s, end = i and start = i + 1 and end = e ` ` ` `// We remove the previous piece and add ` ` ` `// this one to the TreeSet ` ` ` `temp[curr] ` ` ` `= ` `new` `Piece(curr, cuts[i], ` ` ` `sum[cuts[i]] ` ` ` `- (curr == ` `0` `? ` `0` `: sum[curr - ` `1` `])); ` ` ` `pq.add(temp[curr]); ` ` ` ` ` `temp[cuts[i] + ` `1` `] ` ` ` `= ` `new` `Piece(cuts[i] + ` `1` `, ` ` ` `end, sum[end] - sum[cuts[i]]); ` ` ` `pq.add(temp[cuts[i] + ` `1` `]); ` ` ` ` ` `t.add(curr); ` ` ` `t.add(cuts[i] + ` `1` `); ` ` ` ` ` `System.out.println(pq.first().val); ` ` ` `} ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `main(String[] args) ` ` ` `{ ` ` ` ` ` `int` `A[] = { ` `4` `, ` `5` `, ` `6` `, ` `7` `, ` `8` `}; ` ` ` `int` `n = A.length; ` ` ` `int` `cuts[] = { ` `0` `, ` `2` `, ` `3` `, ` `1` `}; ` ` ` `int` `k = cuts.length; ` ` ` ` ` `solve(n, k, cuts, A); ` ` ` `} ` `} ` |

*chevron_right*

*filter_none*

**Output:**

26 15 11 8

Time Complexity O(n + k Log n)

## Recommended Posts:

- Split array to three subarrays such that sum of first and third subarray is equal and maximum
- Maximum subarray size, such that all subarrays of that size have sum less than k
- Maximum subarray sum in array formed by repeating the given array k times
- Maximum subarray sum by flipping signs of at most K array elements
- Maximum sum after repeatedly dividing N by a divisor
- Range queries to count 1s in a subarray after flip operations
- Maximum of all Subarrays of size k using set in C++ STL
- Maximum sum two non-overlapping subarrays of given size
- Partition into two subarrays of lengths k and (N - k) such that the difference of sums is maximum
- Queries to add, remove and return the difference of maximum and minimum.
- Find the Initial Array from given array after range sum queries
- Queries to find maximum product pair in range with updates
- Maximum subarray sum in O(n) using prefix sum
- Maximum subarray sum modulo m
- Maximum sum subarray such that start and end values are same

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.