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Maximum prefix-sum for a given range
• Difficulty Level : Hard
• Last Updated : 04 Feb, 2021

Given an array of n integers and q queries, each query having a range from l to r. Find the maximum prefix-sum for the range l – r.
Examples:

```Input: a[] = {-1, 2, 3, -5}
q = 2
l = 0, r = 3
l = 1, r = 3

Output: 4
5

Explanation:- The range (0, 3) in the 1st query has
[-1, 2, 3, -5], since it is prefix,
we have to start from -1. Hence, the
max prefix sum will be -1 + 2 + 3 = 4.
The range (1, 3) in the 2nd query has
[2, 3, -5], since it is prefix, we
have to start from 2. Hence, the max
prefix sum will be 2 + 3 = 5.

Input: a[] = {-2, -3, 4, -1, -2, 1, 5, -3}
q = 1
l = 1 r = 7

Output: 4

Explanation:- The range (1, 7) in the 1st query has
[-3, 4, -1, -2, 1, 5, -3], since it is
prefix, we have to start from -3.
Hence, the max prefix sum will be
-3 + 4 - 1 - 2 + 1 + 5 = 4.```

Normal Approach:
A simple solution is to run a loop from l to r and calculate max prefix sum from l to r for every query.
Time Complexity: O(q * n),
Auxiliary Space: O(1)
Efficient Approach:
An efficient approach will be to build a segment tree where each node stores two values(sum and prefix_sum), and do a range query on it to find the max prefix sum. But for building the segment tree we have to think on what to store on the nodes of the tree.
For finding out the maximum prefix sum, we will require two things, one being the sum and the other prefix sum. The merging will return two things, sum of the ranges and the prefix sum that will store the max(prefix.left, prefix.sum + prefix.right) in the segment tree.
If we have a deep look into it, the max prefix sum for any two range combining will either be the prefix sum from left side or the sum of left side+prefix sum of right side, whichever is max is taken into account.
Representation of Segment trees:
1. Leaf Nodes are the elements of the input array.
2. Each internal node represents some merging of the leaf nodes. The merging may be different for different problems. For this problem, merging is sum of leaves under a node.
An array representation of tree is used to represent Segment Trees. For each node at index i, the left child is at index 2*i+1, right child at 2*i+2 and the parent is at (i-1)/2.
Construction of Segment Tree from given array:
We start with a segment arr[0 . . . n-1]. and every time we divide the current segment into two halves(if it has not yet become a segment of length 1), and then call the same procedure on both halves, and for each such segment we store the sum and prefix sum in corresponding node.
We then do a range query on segment tree to find out the max prefix-sum for the given range, and output the max prefix-sum.
Below is the C++ implementation of above approach.

## CPP

 `// CPP program to find``// maximum prefix sum``#include ``using` `namespace` `std;` `// struct two store two values in one node``struct` `Node {``    ``int` `sum;``    ``int` `prefix;``};` `Node tree[4 * 10000];` `// function to build the segment tree``void` `build(``int` `a[], ``int` `index, ``int` `beg, ``int` `end)``{``    ``if` `(beg == end) {` `        ``// If there is one element in array,``        ``// store it in current node of``        ``// segment tree``        ``tree[index].sum = a[beg];``        ``tree[index].prefix = a[beg];``    ``} ``else` `{``        ``int` `mid = (beg + end) / 2;` `        ``// If there are more than one elements,``        ``// then recur for left and right subtrees``        ``build(a, 2 * index + 1, beg, mid);``        ``build(a, 2 * index + 2, mid + 1, end);` `        ``// adds the sum and stores in the index``        ``// position of segment tree``        ``tree[index].sum = tree[2 * index + 1].sum +``                          ``tree[2 * index + 2].sum;` `        ``// stores the max of prefix-sum either``        ``// from right or from left.``        ``tree[index].prefix = max(tree[2 * index + 1].prefix,``                                 ``tree[2 * index + 1].sum +``                                ``tree[2 * index + 2].prefix);``    ``}``}` `// function to do the range query in the segment``// tree for the maximum prefix sum``Node query(``int` `index, ``int` `beg, ``int` `end, ``int` `l, ``int` `r)``{``    ``Node result;``    ``result.sum = result.prefix = -1;` `    ``// If segment of this node is outside the given``    ``// range, then return the minimum value.``    ``if` `(beg > r || end < l)``        ``return` `result;` `    ``// If segment of this node is a part of given``    ``// range, then return the node of the segment``    ``if` `(beg >= l && end <= r)``        ``return` `tree[index];` `    ``int` `mid = (beg + end) / 2;` `    ``// if left segment of this node falls out of``    ``// range, then recur in the right side of``    ``// the tree``    ``if` `(l > mid)``        ``return` `query(2 * index + 2, mid + 1, end,``                                         ``l, r);` `    ``// if right segment of this node falls out of``    ``// range, then recur in the left side of``    ``// the tree``    ``if` `(r <= mid)``        ``return` `query(2 * index + 1, beg, mid,``                                       ``l, r);` `    ``// If a part of this segment overlaps with``    ``// the given range``    ``Node left = query(2 * index + 1, beg, mid,``                                        ``l, r);``    ``Node right = query(2 * index + 2, mid + 1,``                                   ``end, l, r);` `    ``// adds the sum of the left and right``    ``// segment``    ``result.sum = left.sum + right.sum;` `    ``// stores the max of prefix-sum``    ``result.prefix = max(left.prefix, left.sum +``                            ``right.prefix);` `    ``// returns the value``    ``return` `result;``}` `// driver program to test the program``int` `main()``{` `    ``int` `a[] = { -2, -3, 4, -1, -2, 1, 5, -3 };` `    ``// calculates the length of array``    ``int` `n = ``sizeof``(a) / ``sizeof``(a);` `    ``// calls the build function to build``    ``// the segment tree``    ``build(a, 0, 0, n - 1);` `    ``// find the max prefix-sum between``    ``// 3rd and 5th index of array``    ``cout << query(0, 0, n - 1, 3, 5).prefix``         ``<< endl;` `    ``// find the max prefix-sum between``    ``// 0th and 7th index of array``    ``cout << query(0, 0, n - 1, 1, 7).prefix``         ``<< endl;` `    ``return` `0;``}`

## Java

 `// JAVA program to find``// maximum prefix sum``class` `GFG``{` `// two store two values in one node``static` `class` `Node``{``    ``int` `sum;``    ``int` `prefix;``};` `static` `Node []tree = ``new` `Node[``4` `* ``10000``];``static``{``    ``for``(``int` `i = ``0``; i < tree.length; i++)``        ``tree[i] = ``new` `Node();``}` `// function to build the segment tree``static` `void` `build(``int` `a[], ``int` `index, ``int` `beg, ``int` `end)``{``    ``if` `(beg == end)``    ``{` `        ``// If there is one element in array,``        ``// store it in current node of``        ``// segment tree``        ``tree[index].sum = a[beg];``        ``tree[index].prefix = a[beg];``    ``} ``else` `{``        ``int` `mid = (beg + end) / ``2``;` `        ``// If there are more than one elements,``        ``// then recur for left and right subtrees``        ``build(a, ``2` `* index + ``1``, beg, mid);``        ``build(a, ``2` `* index + ``2``, mid + ``1``, end);` `        ``// adds the sum and stores in the index``        ``// position of segment tree``        ``tree[index].sum = tree[``2` `* index + ``1``].sum +``                        ``tree[``2` `* index + ``2``].sum;` `        ``// stores the max of prefix-sum either``        ``// from right or from left.``        ``tree[index].prefix = Math.max(tree[``2` `* index + ``1``].prefix,``                                ``tree[``2` `* index + ``1``].sum +``                                ``tree[``2` `* index + ``2``].prefix);``    ``}``}` `// function to do the range query in the segment``// tree for the maximum prefix sum``static` `Node query(``int` `index, ``int` `beg, ``int` `end, ``int` `l, ``int` `r)``{``    ``Node result = ``new` `Node();``    ``result.sum = result.prefix = -``1``;` `    ``// If segment of this node is outside the given``    ``// range, then return the minimum value.``    ``if` `(beg > r || end < l)``        ``return` `result;` `    ``// If segment of this node is a part of given``    ``// range, then return the node of the segment``    ``if` `(beg >= l && end <= r)``        ``return` `tree[index];` `    ``int` `mid = (beg + end) / ``2``;` `    ``// if left segment of this node falls out of``    ``// range, then recur in the right side of``    ``// the tree``    ``if` `(l > mid)``        ``return` `query(``2` `* index + ``2``, mid + ``1``, end,``                                        ``l, r);` `    ``// if right segment of this node falls out of``    ``// range, then recur in the left side of``    ``// the tree``    ``if` `(r <= mid)``        ``return` `query(``2` `* index + ``1``, beg, mid,``                                    ``l, r);` `    ``// If a part of this segment overlaps with``    ``// the given range``    ``Node left = query(``2` `* index + ``1``, beg, mid,``                                        ``l, r);``    ``Node right = query(``2` `* index + ``2``, mid + ``1``,``                                ``end, l, r);` `    ``// adds the sum of the left and right``    ``// segment``    ``result.sum = left.sum + right.sum;` `    ``// stores the max of prefix-sum``    ``result.prefix = Math.max(left.prefix, left.sum +``                            ``right.prefix);` `    ``// returns the value``    ``return` `result;``}` `// Driver code``public` `static` `void` `main(String[] args)``{` `    ``int` `a[] = { -``2``, -``3``, ``4``, -``1``, -``2``, ``1``, ``5``, -``3` `};` `    ``// calculates the length of array``    ``int` `n = a.length;` `    ``// calls the build function to build``    ``// the segment tree``    ``build(a, ``0``, ``0``, n - ``1``);` `    ``// find the max prefix-sum between``    ``// 3rd and 5th index of array``    ``System.out.print(query(``0``, ``0``, n - ``1``, ``3``, ``5``).prefix``        ``+``"\n"``);` `    ``// find the max prefix-sum between``    ``// 0th and 7th index of array``    ``System.out.print(query(``0``, ``0``, n - ``1``, ``1``, ``7``).prefix``        ``+``"\n"``);``}``}` `// This code is contributed by PrinciRaj1992`

## C#

 `// C# program to find``// maximum prefix sum``using` `System;` `class` `GFG``{` `// two store two values in one node``class` `Node``{``    ``public` `int` `sum;``    ``public` `int` `prefix;``};` `static` `Node []tree = ``new` `Node[4 * 10000];` `// function to build the segment tree``static` `void` `build(``int` `[]a, ``int` `index, ``int` `beg, ``int` `end)``{``    ``if` `(beg == end)``    ``{` `        ``// If there is one element in array,``        ``// store it in current node of``        ``// segment tree``        ``tree[index].sum = a[beg];``        ``tree[index].prefix = a[beg];``    ``} ``else` `{``        ``int` `mid = (beg + end) / 2;` `        ``// If there are more than one elements,``        ``// then recur for left and right subtrees``        ``build(a, 2 * index + 1, beg, mid);``        ``build(a, 2 * index + 2, mid + 1, end);` `        ``// adds the sum and stores in the index``        ``// position of segment tree``        ``tree[index].sum = tree[2 * index + 1].sum +``                        ``tree[2 * index + 2].sum;` `        ``// stores the max of prefix-sum either``        ``// from right or from left.``        ``tree[index].prefix = Math.Max(tree[2 * index + 1].prefix,``                                ``tree[2 * index + 1].sum +``                                ``tree[2 * index + 2].prefix);``    ``}``}` `// function to do the range query in the segment``// tree for the maximum prefix sum``static` `Node query(``int` `index, ``int` `beg, ``int` `end, ``int` `l, ``int` `r)``{``    ``Node result = ``new` `Node();``    ``result.sum = result.prefix = -1;` `    ``// If segment of this node is outside the given``    ``// range, then return the minimum value.``    ``if` `(beg > r || end < l)``        ``return` `result;` `    ``// If segment of this node is a part of given``    ``// range, then return the node of the segment``    ``if` `(beg >= l && end <= r)``        ``return` `tree[index];` `    ``int` `mid = (beg + end) / 2;` `    ``// if left segment of this node falls out of``    ``// range, then recur in the right side of``    ``// the tree``    ``if` `(l > mid)``        ``return` `query(2 * index + 2, mid + 1, end,``                                        ``l, r);` `    ``// if right segment of this node falls out of``    ``// range, then recur in the left side of``    ``// the tree``    ``if` `(r <= mid)``        ``return` `query(2 * index + 1, beg, mid,``                                    ``l, r);` `    ``// If a part of this segment overlaps with``    ``// the given range``    ``Node left = query(2 * index + 1, beg, mid,``                                        ``l, r);``    ``Node right = query(2 * index + 2, mid + 1,``                                ``end, l, r);` `    ``// adds the sum of the left and right``    ``// segment``    ``result.sum = left.sum + right.sum;` `    ``// stores the max of prefix-sum``    ``result.prefix = Math.Max(left.prefix, left.sum +``                            ``right.prefix);` `    ``// returns the value``    ``return` `result;``}` `// Driver code``public` `static` `void` `Main(String[] args)``{` `    ``for``(``int` `i = 0; i < tree.Length; i++)``        ``tree[i] = ``new` `Node();``    ``int` `[]a = { -2, -3, 4, -1, -2, 1, 5, -3 };` `    ``// calculates the length of array``    ``int` `n = a.Length;` `    ``// calls the build function to build``    ``// the segment tree``    ``build(a, 0, 0, n - 1);` `    ``// find the max prefix-sum between``    ``// 3rd and 5th index of array``    ``Console.Write(query(0, 0, n - 1, 3, 5).prefix``        ``+``"\n"``);` `    ``// find the max prefix-sum between``    ``// 0th and 7th index of array``    ``Console.Write(query(0, 0, n - 1, 1, 7).prefix``        ``+``"\n"``);``}``}` `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 program to find``# maximum prefix sum` `# struct two store two values in one node` `# function to build the segment tree``def` `build(a, index, beg, end):``    ``global` `tree` `    ``if` `(beg ``=``=` `end):` `        ``# If there is one element in array,``        ``# store it in current node of``        ``# segment tree``        ``tree[index][``0``] ``=` `a[beg]``        ``tree[index][``1``] ``=` `a[beg]``    ``else``:``        ``mid ``=` `(beg ``+` `end) ``/``/` `2` `        ``# If there are more than one elements,``        ``# then recur for left and right subtrees``        ``build(a, ``2` `*` `index ``+` `1``, beg, mid)``        ``build(a, ``2` `*` `index ``+` `2``, mid ``+` `1``, end)` `        ``# adds the sum and stores in the index``        ``# position of segment tree``        ``tree[index][``0``] ``=` `tree[``2` `*` `index ``+` `1``][``0``] ``+` `tree[``2` `*` `index ``+` `2``][``0``]` `        ``# stores the max of prefix-sum either``        ``# from right or from left.``        ``tree[index][``1``] ``=` `max``(tree[``2` `*` `index ``+` `1``][``1``],tree[``2` `*` `index ``+` `1``][``0``] ``+` `tree[``2` `*` `index ``+` `2``][``1``])` `# function to do the range query in the segment``# tree for the maximum prefix sum``def` `query(index, beg, end, l, r):``    ``global` `tree``    ``result ``=` `[``-``1``, ``-``1``]``    ``# result = result = -1` `    ``# If segment of this node is outside the given``    ``# range, then return the minimum value.``    ``if` `(beg > r ``or` `end < l):``        ``return` `result` `    ``# If segment of this node is a part of given``    ``# range, then return the node of the segment``    ``if` `(beg >``=` `l ``and` `end <``=` `r):``        ``return` `tree[index]` `    ``mid ``=` `(beg ``+` `end) ``/``/` `2` `    ``# if left segment of this node falls out of``    ``# range, then recur in the right side of``    ``# the tree``    ``if` `(l > mid):``        ``return` `query(``2` `*` `index ``+` `2``, mid ``+` `1``, end, l, r)` `    ``# if right segment of this node falls out of``    ``# range, then recur in the left side of``    ``# the tree``    ``if` `(r <``=` `mid):``        ``return` `query(``2` `*` `index ``+` `1``, beg, mid, l, r)` `    ``# If a part of this segment overlaps with``    ``# the given range``    ``left ``=` `query(``2` `*` `index ``+` `1``, beg, mid, l, r)``    ``right ``=` `query(``2` `*` `index ``+` `2``, mid ``+` `1``, end, l, r)` `    ``# adds the sum of the left and right``    ``# segment``    ``result[``0``] ``=` `left[``0``] ``+` `right[``0``]` `    ``# stores the max of prefix-sum``    ``result[``1``] ``=` `max``(left[``1``], left[``0``] ``+` `right[``1``])` `    ``# returns the value``    ``return` `result`  `# driver program to test the program``if` `__name__ ``=``=` `'__main__'``:` `    ``a ``=` `[``-``2``, ``-``3``, ``4``, ``-``1``, ``-``2``, ``1``, ``5``, ``-``3` `]` `    ``tree ``=` `[[``0``,``0``] ``for` `i ``in` `range``(``4` `*` `10000``)]` `    ``# calculates the length of array``    ``n ``=` `len``(a)` `    ``# calls the build function to build``    ``# the segment tree``    ``build(a, ``0``, ``0``, n ``-` `1``)` `    ``# find the max prefix-sum between``    ``# 3rd and 5th index of array``    ``print``(query(``0``, ``0``, n ``-` `1``, ``3``, ``5``)[``1``])` `    ``# find the max prefix-sum between``    ``# 0th and 7th index of array``    ``print``(query(``0``, ``0``, n ``-` `1``, ``1``, ``7``)[``1``])` `    ``# This code is contributed by mohit kumar 29.`

Output:

```-1
4```

Time Complexity:
Time Complexity for tree construction is O(n). There are total 2n-1 nodes, and value of every node is calculated only once in tree construction.
Time complexity to every query is O(log n).
Time complexity for the problem is O(q * log n)