Maximum sum subarray of size range [L, R]
Last Updated :
08 Mar, 2024
Given an integer array arr[] of size N and two integer L and R. The task is to find the maximum sum subarray of size between L and R (both inclusive).
Example:
Input: arr[] = {1, 2, 2, 1}, L = 1, R = 3
Output: 5
Explanation:
Subarray of size 1 are {1}, {2}, {2}, {1} and maximum sum subarray = 2 for subarray {2}.
Subarray of size 2 are {1, 2}, {2, 2}, {2, 1}, and maximum sum subarray = 4 for subarray {2, 2}.
Subarray of size 3 are {1, 2, 2}, {2, 2, 1}, and maximum sum subarray = 5 for subarray {2, 2, 1}.
Hence the maximum possible sum subarray is 5.
Input: arr[] = {-1, -3, -7, -11}, L = 1, R = 4
Output: -1
Approach:
- Here we will use the concept of sliding window which is discuss in this post.
- First calculate prefix sum of array in array pre[].
- Next iterate over the range L to N -1, and consider all subarray of size L to R.
- Create a multiset for storing prefix sums of subarray length L to R.
- Now to find maximum sum subarray ending at index i just subtract pre[i] and minimum of all values from pre[i - L] to pre[i - R].
- Finally return maximum of all sums.
Below is the implementation of the above approach:
C++
// C++ program to find Maximum sum
// subarray of size between L and R.
#include <bits/stdc++.h>
using namespace std;
// function to find Maximum sum subarray
// of size between L and R
void max_sum_subarray(vector<int> arr,
int L, int R)
{
int n = arr.size();
int pre[n] = { 0 };
// calculating prefix sum
pre[0] = arr[0];
for (int i = 1; i < n; i++) {
pre[i] = pre[i - 1] + arr[i];
}
multiset<int> s1;
// maintain 0 for initial
// values of i upto R
// Once i = R, then
// we need to erase that 0 from
// our multiset as our first
// index of subarray
// cannot be 0 anymore.
s1.insert(0);
int ans = INT_MIN;
ans = max(ans, pre[L - 1]);
// we maintain flag to
// counter if that initial
// 0 was erased from set or not.
int flag = 0;
for (int i = L; i < n; i++) {
// erase 0 from multiset once i=b
if (i - R >= 0) {
if (flag == 0) {
auto it = s1.find(0);
s1.erase(it);
flag = 1;
}
}
// insert pre[i-L]
if (i - L >= 0)
s1.insert(pre[i - L]);
// find minimum value in multiset.
ans = max(ans,
pre[i] - *s1.begin());
// erase pre[i-R]
if (i - R >= 0) {
auto it = s1.find(pre[i - R]);
s1.erase(it);
}
}
cout << ans << endl;
}
// Driver code
int main()
{
int L, R;
L = 1;
R = 3;
vector<int> arr = { 1, 2, 2, 1 };
max_sum_subarray(arr, L, R);
return 0;
}
// Java program to find Maximum sum
// subarray of size between L and R.
import java.util.*;
class GFG {
// function to find Maximum sum subarray
// of size between L and R
static void max_sum_subarray(List<Integer> arr, int L, int R){
int n = arr.size();
int[] pre = new int[n + 1];
// calculating prefix sum
// here pre[0] = 0
for (int i = 1; i <= n; i++) {
pre[i] = pre[i - 1]+arr.get(i - 1);
}
// treemap for storing prefix sums for
// subarray length L to R
TreeMap<Integer, Integer> s1 = new TreeMap<>();
int ans = Integer.MIN_VALUE;
for (int i = L; i <= n; i++) {
// if i > R, erase pre[i - R - 1]
// note that pre[0] = 0
if (i > R) {
// decrement count of pre[i - R - 1]
s1.put(pre[i - R - 1], s1.get(pre[i - R - 1])-1);
// if count is zero, element is not present
// in map so remove it
if (s1.get(pre[i - R - 1]) == 0)
s1.remove(pre[i - R - 1]);
}
// insert pre[i - L]
s1.put(pre[i - L], s1.getOrDefault(pre[i - L], 0)+1);
// find minimum value in treemap.
ans = Math.max(ans, pre[i] - s1.firstKey());
}
System.out.println(ans);
}
// Driver code
public static void main(String[] args){
int L, R;
L = 1;
R = 3;
List<Integer> arr = Arrays.asList(1, 2, 2, 1);
max_sum_subarray(arr, L, R);
}
}
// This code is contributed by Utkarsh Sharma
// C# program to find Maximum sum
// subarray of size between L and R.
using System;
using System.Collections.Generic;
class GFG
{
// function to find Maximum sum subarray
// of size between L and R
static void max_sum_subarray(List<int> arr, int L, int R)
{
int n = arr.Count;
int[] pre = new int[n];
// calculating prefix sum
pre[0] = arr[0];
for (int i = 1; i < n; i++)
{
pre[i] = pre[i - 1] + arr[i];
}
List<int> s1 = new List<int>();
// maintain 0 for initial
// values of i upto R
// Once i = R, then
// we need to erase that 0 from
// our multiset as our first
// index of subarray
// cannot be 0 anymore.
s1.Add(0);
int ans = Int32.MinValue;
ans = Math.Max(ans, pre[L - 1]);
// we maintain flag to
// counter if that initial
// 0 was erased from set or not.
int flag = 0;
for (int i = L; i < n; i++)
{
// erase 0 from multiset once i=b
if (i - R >= 0)
{
if (flag == 0)
{
int it = s1.IndexOf(0);
s1.RemoveAt(it);
flag = 1;
}
}
// insert pre[i-L]
if (i - L >= 0)
s1.Add(pre[i - L]);
// find minimum value in multiset.
ans = Math.Max(ans, pre[i] - s1[0]);
// erase pre[i-R]
if (i - R >= 0)
{
int it = s1.IndexOf(pre[i - R]);
s1.RemoveAt(it);
}
}
Console.WriteLine(ans);
}
// Driver code
static void Main()
{
int L, R;
L = 1;
R = 3;
List<int> arr = new List<int>(){1, 2, 2, 1};
max_sum_subarray(arr, L, R);
}
}
// This code is contributed by divyesh072019
// Javascript program to find Maximum sum
// subarray of size between L and R.
// function to find Maximum sum subarray
// of size between L and R
function max_sum_subarray(arr,L,R)
{
let n = arr.length;
let pre = new Array(n);
// calculating prefix sum
pre[0] = arr[0];
for (let i = 1; i < n; i++)
{
pre[i] = pre[i - 1] + arr[i];
}
let s1 = []
// maintain 0 for initial
// values of i upto R
// Once i = R, then
// we need to erase that 0 from
// our multiset as our first
// index of subarray
// cannot be 0 anymore.
s1.push(0);
let ans = Number.MIN_VALUE;
ans = Math.max(ans, pre[L - 1]);
// we maintain flag to
// counter if that initial
// 0 was erased from set or not.
let flag = 0;
for (let i = L; i < n; i++)
{
// erase 0 from multiset once i=b
if (i - R >= 0)
{
if (flag == 0)
{
let it = s1.indexOf(0);
s1.splice(it,1);
flag = 1;
}
}
// insert pre[i-L]
if (i - L >= 0)
s1.push(pre[i - L]);
// find minimum value in multiset.
ans = Math.max(ans, pre[i] - s1[0]);
// erase pre[i-R]
if (i - R >= 0)
{
let it = s1.indexOf(pre[i - R]);
s1.splice(it,1);
}
}
document.write(ans);
}
// Driver code
let L, R;
L = 1;
R = 3;
let arr = [1, 2, 2, 1];
max_sum_subarray(arr, L, R);
// This code is contributed by avanitrachhadiya2155
def max_sum_subarray(arr, L, R):
n = len(arr)
pre = [0] * n
# calculating prefix sum
pre[0] = arr[0]
for i in range(1, n):
pre[i] = pre[i - 1] + arr[i]
s1 = set()
# maintain 0 for initial
# values of i up to R
# Once i = R, then
# we need to erase that 0 from
# our set as our first
# index of subarray
# cannot be 0 anymore.
s1.add(0)
ans = float('-inf')
ans = max(ans, pre[L - 1])
# we maintain flag to
# counter if that initial
# 0 was erased from set or not.
flag = 0
for i in range(L, n):
# erase 0 from set once i=b
if i - R >= 0:
if flag == 0:
s1.remove(0)
flag = 1
# insert pre[i-L]
if i - L >= 0:
s1.add(pre[i - L])
# find minimum value in set.
ans = max(ans, pre[i] - min(s1))
# erase pre[i-R]
if i - R >= 0:
s1.remove(pre[i - R])
print(ans)
# Driver code
if __name__ == "__main__":
L, R = 1, 3
arr = [1, 2, 2, 1]
max_sum_subarray(arr, L, R)
Time Complexity: O (N * log N)
Auxiliary Space: O (N)
Similar Reads
Maximum sum two non-overlapping subarrays of given size Given an array, we need to find two subarrays with a specific length K such that sum of these subarrays is maximum among all possible choices of subarrays. Examples: Input : arr[] = [2, 5, 1, 2, 7, 3, 0] K = 2 Output : 2 5 7 3 We can choose two arrays of maximum sum as [2, 5] and [7, 3], the sum of
12 min read
Max sum of M non-overlapping subarrays of size K Given an array and two numbers M and K. We need to find the max sum of sums of M subarrays of size K (non-overlapping) in the array. (Order of array remains unchanged). K is the size of subarrays and M is the count of subarray. It may be assumed that size of array is more than m*k. If total array si
15+ min read
Maximum of all subarrays of size K using Segment Tree Given an array arr[] and an integer K, the task is to find the maximum for each and every contiguous subarray of size K. Examples: Input: arr[] = {1, 2, 3, 1, 4, 5, 2, 3, 6}, K = 3 Output: 3 3 4 5 5 5 6 Explanation: Maximum of 1, 2, 3 is 3 Maximum of 2, 3, 1 is 3 Maximum of 3, 1, 4 is 4 Maximum of 1
14 min read
Maximum of all Subarrays of size k using set in C++ STL Given an array of size N and an integer K, the task is to find the maximum for each and every contiguous sub-array of size K and print the sum of all these values in the end. Examples: Input: arr[] = {4, 10, 54, 11, 8, 7, 9}, K = 3 Output: 182 Input: arr[] = {1, 2, 3, 4, 1, 6, 7, 8, 2, 1}, K = 4 Out
3 min read
CSES Solutions - Maximum Subarray Sum Given an array arr[] of N integers, your task is to find the maximum sum of values in a contiguous, nonempty subarray. Examples: Input: N = 8, arr[] = {-1, 3, -2, 5, 3, -5, 2, 2}Output: 9Explanation: The subarray with maximum sum is {3, -2, 5, 3} with sum = 3 - 2 + 5 + 3 = 9. Input: N = 6, arr[] = {
5 min read