Hello Everyone,
Given an array arr[] of length N, the task is to find the sum of the maximum elements of every possible sub-array of the array.
Examples:
Input : arr[] = {1, 3, 1, 7} Output : 42 Max of all sub-arrays: {1} - 1 {1, 3} - 3 {1, 3, 1} - 3 {1, 3, 1, 7} - 7 {3} - 3 {3, 1} - 3 {3, 1, 7} - 7 {1} - 1 {1, 7} - 7 {7} - 7 1 + 3 + 3 + 7 + 3 + 3 + 7 + 1 + 7 + 7 = 42 Input : arr[] = {1, 1, 1, 1, 1} Output : 15
Approach :
In this article, we will learn how to solve this problem using divide and conquer.
Let’s assume that element at ith index is largest of all. For any sub-array that contains index ‘i’, the element at ‘i’ will always be maximum in the sub-array.
If element at ith index is largest, we can safely say, that element ith index will be largest in (i+1)(N-i) subarrays. So, its total contribution will be arr[i](i+1)*(N-i). Now, we will divide the array in two parts, (0, i-1) and (i+1, N-1) and apply the same algorithms to both of them separately.
So our general recurrence relation will be:
maxSumSubarray(arr, l, r) = arr[i](r-i+1)(i-l+1) + maxSumSubarray(arr, l, i-1) + maxSumSubarray(arr, i+1, r) where i is index of maximum element in range [l, r].
Now, we need a way to efficiently answer rangeMax() queries. Segment tree will be an efficient way to answer this query. We will need to answer this query at most N times. Thus, the time complexity of our divide and conquer algorithm will O(Nlog(N)).
If we have to answer the problem “Sum of minimum of all subarrays” then we will use the segment tree to answer rangeMin() queries. For this, you can go through the article segment tree range minimum.
Below is the implementation code:
// C++ implementation of the above approach
#include <bits/stdc++.h>
#define seg_max 51
using
namespace
std;
// Array to store segment tree.
// In first we will store the maximum
// of a range
// In second, we will store index of
// that range
pair<
int
,
int
> seg_tree[seg_max];
// Size of array declared global
// to maintain simplicity in code
int
n;
// Function to build segment tree
pair<
int
,
int
> buildMaxTree(
int
l,
int
r,
int
i,
int
arr[])
{
// Base case
if
(l == r) {
seg_tree[i] = { arr[l], l };
return
seg_tree[i];
}
// Finding the maximum among left and right child
seg_tree[i] = max(buildMaxTree(l, (l + r) / 2, 2 * i + 1, arr),
buildMaxTree((l + r) / 2 + 1, r, 2 * i + 2, arr));
// Returning the maximum to parent
return
seg_tree[i];
}
// Function to perform range-max query in segment tree
pair<
int
,
int
> rangeMax(
int
l,
int
r,
int
arr[],
int
i = 0,
int
sl = 0,
int
sr = n - 1)
{
// Base cases
if
(sr < l || sl > r)
return
{ INT_MIN, -1 };
if
(sl >= l and sr <= r)
return
seg_tree[i];
// Finding the maximum among left and right child
return
max(rangeMax(l, r, arr, 2 * i + 1, sl, (sl + sr) / 2),
rangeMax(l, r, arr, 2 * i + 2, (sl + sr) / 2 + 1, sr));
}
// Function to find maximum sum subarray
int
maxSumSubarray(
int
arr[],
int
l = 0,
int
r = n - 1)
{
// base case
if
(l > r)
return
0;
// range-max query to determine
// largest in the range.
pair<
int
,
int
> a = rangeMax(l, r, arr);
// divide the array in two parts
return
a.first * (r - a.second + 1) * (a.second - l + 1)
+ maxSumSubarray(arr, l, a.second - 1)
+ maxSumSubarray(arr, a.second + 1, r);
}
// Driver Code
int
main()
{
// Input array
int
arr[] = { 1, 3, 1, 7 };
// Size of array
n =
sizeof
(arr) /
sizeof
(
int
);
// Builind the segment-tree
buildMaxTree(0, n - 1, 0, arr);
cout << maxSumSubarray(arr);
return
0;
}
Output:
42
Time complexity : O(Nlog(N))