Maximum subarray sum possible after removing at most K array elements
Last Updated :
11 Apr, 2023
Given an array arr[] of size N and an integer K, the task is to find the maximum subarray sum by removing at most K elements from the array.
Examples:
Input: arr[] = { -2, 1, 3, -2, 4, -7, 20 }, K = 1
Output: 26
Explanation:
Removing arr[5] from the array modifies arr[] to { -2, 1, 3, -2, 4, 20 }
Subarray with maximum sum is { 1, 3, -2, 4, 20 }.
Therefore, the required output is 26.
Input:arr[] = { -1, 1, -1, -1, 1, 1 }, K=2
Output: 3
Explanation:
Removing arr[2] and arr[3] from the array modifies arr[] to { – 1, 1, 1, 1}
Subarray with maximum sum is { 1, 1, 1 }.
Therefore, the required output is 3.
Approach: The problem can be solved using Dynamic Programming. The idea is to use the concept of Kadane’s algorithm. Follow the steps below to solve the problem:
- Traverse the array arr[] and for every array element following two operations needs to be performed:
- Remove the current array element from the subarray.
- Include the current array element in the subarray.
- Therefore, the recurrence relation to solve this problem is as follows:
mxSubSum(i, j) = max(max(0, arr[i] + mxSubSum(i – 1, j)), mxSubSum(i – 1, j – 1))
i: Stores index of array element
j: Maximum count of elements that can be removed from the subarray
mxSubSum(i, j): Return maximum subarray sum from the subarray { arr[i], arr[N – 1] } by removing K – j array elements.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
#define M 100
int mxSubSum(int i, int* arr,
int j, int dp[][M])
{
if (i == 0) {
return dp[i][j] = max(0, arr[i]);
}
if (dp[i][j] != -1) {
return dp[i][j];
}
int X = max(0, arr[i]
+ mxSubSum(i - 1, arr, j, dp));
if (j == 0) {
return dp[i][j] = X;
}
int Y = mxSubSum(i - 1, arr, j - 1, dp);
return dp[i][j] = max(X, Y);
}
int MaximumSubarraySum(int n, int* arr, int k)
{
int dp[M][M];
memset(dp, -1, sizeof(dp));
mxSubSum(n - 1, arr, k, dp);
int res = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j <= k; j++) {
res = max(res, dp[i][j]);
}
}
if (*max_element(arr, arr + n) < 0) {
res = *max_element(arr, arr + n);
}
return res;
}
int main()
{
int arr[] = { -2, 1, 3, -2, 4, -7, 20 };
int K = 1;
int N = sizeof(arr) / sizeof(arr[0]);
cout << MaximumSubarraySum(N, arr, K) << endl;
return 0;
}
|
Java
import java.util.*;
class GFG{
static final int M = 100;
static int mxSubSum(int i, int []arr,
int j, int dp[][])
{
if (i == 0) {
return dp[i][j] = Math.max(0, arr[i]);
}
if (dp[i][j] != -1) {
return dp[i][j];
}
int X = Math.max(0, arr[i]
+ mxSubSum(i - 1, arr, j, dp));
if (j == 0)
{
return dp[i][j] = X;
}
int Y = mxSubSum(i - 1, arr, j - 1, dp);
return dp[i][j] = Math.max(X, Y);
}
static int MaximumSubarraySum(int n, int []arr, int k)
{
int [][]dp = new int[M][M];
for (int i = 0; i < M; i++)
for (int j = 0; j < M; j++)
dp[i][j] = -1;
mxSubSum(n - 1, arr, k, dp);
int res = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j <= k; j++) {
res = Math.max(res, dp[i][j]);
}
}
if (Arrays.stream(arr).max().getAsInt() < 0) {
res = Arrays.stream(arr).max().getAsInt();
}
return res;
}
public static void main(String[] args)
{
int arr[] = { -2, 1, 3, -2, 4, -7, 20 };
int K = 1;
int N = arr.length;
System.out.print(MaximumSubarraySum(N, arr, K) +"\n");
}
}
|
Python3
M = 100
def mxSubSum(i, arr, j):
global dp
if (i == 0):
dp[i][j] = max(0, arr[i])
return dp[i][j]
if (dp[i][j] != -1):
return dp[i][j]
X = max(0, arr[i] + mxSubSum(i - 1, arr, j))
if (j == 0):
dp[i][j] = X
return X
Y = mxSubSum(i - 1, arr, j - 1)
dp[i][j] = max(X, Y)
return dp[i][j]
def MaximumSubarraySum(n, arr, k):
mxSubSum(n - 1, arr, k)
res = 0
for i in range(n):
for j in range(k + 1):
res = max(res, dp[i][j])
if (max(arr) < 0):
res = max(arr)
return res
if __name__ == '__main__':
dp = [[-1 for i in range(100)] for i in range(100)]
arr = [-2, 1, 3, -2, 4, -7, 20]
K = 1
N = len(arr)
print(MaximumSubarraySum(N, arr, K))
|
C#
using System;
using System.Collections;
class GFG{
static int M = 100;
static int mxSubSum(int i, int []arr,
int j, int [,]dp)
{
if (i == 0)
{
return dp[i, j] = Math.Max(0, arr[i]);
}
if (dp[i, j] != -1)
{
return dp[i, j];
}
int X = Math.Max(0, arr[i] +
mxSubSum(i - 1, arr, j, dp));
if (j == 0)
{
return dp[i, j] = X;
}
int Y = mxSubSum(i - 1, arr, j - 1, dp);
return dp[i, j] = Math.Max(X, Y);
}
static int MaximumSubarraySum(int n, int []arr, int k)
{
int [,]dp = new int[M, M];
for(int i = 0; i < M; i++)
for(int j = 0; j < M; j++)
dp[i, j] = -1;
mxSubSum(n - 1, arr, k, dp);
int res = 0;
for(int i = 0; i < n; i++)
{
for(int j = 0; j <= k; j++)
{
res = Math.Max(res, dp[i, j]);
}
}
Array.Sort(arr);
if (arr[n - 1] < 0)
{
res = arr[n - 1];
}
return res;
}
public static void Main(String[] args)
{
int []arr = { -2, 1, 3, -2, 4, -7, 20 };
int K = 1;
int N = arr.Length;
Console.WriteLine(MaximumSubarraySum(N, arr, K));
}
}
|
Javascript
<script>
var M = 100;
function mxSubSum(i, arr, j, dp)
{
if (i == 0)
{
dp[i][j] = Math.max(0, arr[i]);
return dp[i][j];
}
if (dp[i][j] != -1)
{
return dp[i][j];
}
var X = Math.max(
0, arr[i] + mxSubSum(i - 1, arr, j, dp));
if (j == 0)
{
dp[i][j] = X;
return dp[i][j]
}
var Y = mxSubSum(i - 1, arr, j - 1, dp);
dp[i][j] = Math.max(X, Y);
return dp[i][j]
}
function MaximumSubarraySum(n, arr, k)
{
var dp = Array.from(Array(M), () => Array(M).fill(-1));
mxSubSum(n - 1, arr, k, dp);
var res = 0;
for(var i = 0; i < n; i++)
{
for(var j = 0; j <= k; j++)
{
res = Math.max(res, dp[i][j]);
}
}
if (arr.reduce((a, b) => Math.max(a, b)) < 0)
{
res = arr.reduce((a, b) => Math.max(a, b));
}
return res;
}
var arr = [ -2, 1, 3, -2, 4, -7, 20 ];
var K = 1;
var N = arr.length;
document.write( MaximumSubarraySum(N, arr, K));
</script>
|
Time Complexity: O(N * K)
Auxiliary Space: O(N * K)
Efficient approach : Using DP Tabulation method ( Iterative approach )
The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memorization(top-down) because memorization method needs extra stack space of recursion calls.
Steps to solve this problem :
- Create a vector to store the solution of the subproblems.
- Initialize the table with base cases
- Fill up the table iteratively
- Return the final solution
Implementation :
C++
#include <bits/stdc++.h>
using namespace std;
#define M 100
int MaximumSubarraySum(int n, int* arr, int k)
{
int dp[n][k + 1];
memset(dp, 0, sizeof(dp));
dp[0][0] = max(0, arr[0]);
for (int i = 1; i < n; i++) {
for (int j = 0; j <= k; j++) {
int X = max(0, arr[i] + dp[i - 1][j]);
int Y = (j > 0) ? dp[i - 1][j - 1] : 0;
dp[i][j] = max(X, Y);
}
}
int res = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j <= k; j++) {
res = max(res, dp[i][j]);
}
}
if (*max_element(arr, arr + n) < 0) {
res = *max_element(arr, arr + n);
}
return res;
}
int main()
{
int arr[] = { -2, 1, 3, -2, 4, -7, 20 };
int K = 1;
int N = sizeof(arr) / sizeof(arr[0]);
cout << MaximumSubarraySum(N, arr, K) << endl;
return 0;
}
|
Java
import java.util.Arrays;
class Main
{
static int maximumSubarraySum(int n, int[] arr, int k) {
int[][] dp = new int[n][k + 1];
for (int i = 0; i < n; i++) {
Arrays.fill(dp[i], 0);
}
dp[0][0] = Math.max(0, arr[0]);
for (int i = 1; i < n; i++) {
for (int j = 0; j <= k; j++) {
int X = Math.max(0, arr[i] + dp[i - 1][j]);
int Y = (j > 0) ? dp[i - 1][j - 1] : 0;
dp[i][j] = Math.max(X, Y);
}
}
int res = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j <= k; j++) {
res = Math.max(res, dp[i][j]);
}
}
if (Arrays.stream(arr).max().getAsInt() < 0) {
res = Arrays.stream(arr).max().getAsInt();
}
return res;
}
public static void main(String[] args) {
int[] arr = { -2, 1, 3, -2, 4, -7, 20 };
int K = 1;
int N = arr.length;
System.out.println(maximumSubarraySum(N, arr, K));
}
}
|
Python
def MaximumSubarraySum(n, arr, k):
dp = [[0 for j in range(k+1)] for i in range(n)]
dp[0][0] = max(0, arr[0])
for i in range(1, n):
for j in range(k+1):
X = max(0, arr[i] + dp[i - 1][j])
Y = dp[i - 1][j - 1] if j > 0 else 0
dp[i][j] = max(X, Y)
res = 0
for i in range(n):
for j in range(k+1):
res = max(res, dp[i][j])
if max(arr) < 0:
res = max(arr)
return res
if __name__ == '__main__':
arr = [-2, 1, 3, -2, 4, -7, 20]
K = 1
N = len(arr)
print(MaximumSubarraySum(N, arr, K))
|
C#
using System;
using System.Linq;
class MainClass
{
static int maximumSubarraySum(int n, int[] arr, int k)
{
int[][] dp = new int[n][];
for (int i = 0; i < n; i++)
{
dp[i] = new int[k + 1];
for (int j = 0; j <= k; j++)
{
dp[i][j] = 0;
}
}
dp[0][0] = Math.Max(0, arr[0]);
for (int i = 1; i < n; i++)
{
for (int j = 0; j <= k; j++)
{
int X = Math.Max(0, arr[i] + dp[i - 1][j]);
int Y = (j > 0) ? dp[i - 1][j - 1] : 0;
dp[i][j] = Math.Max(X, Y);
}
}
int res = 0;
for (int i = 0; i < n; i++)
{
for (int j = 0; j <= k; j++)
{
res = Math.Max(res, dp[i][j]);
}
}
if (arr.Max() < 0)
{
res = arr.Max();
}
return res;
}
public static void Main()
{
int[] arr = { -2, 1, 3, -2, 4, -7, 20 };
int K = 1;
int N = arr.Length;
Console.WriteLine(maximumSubarraySum(N, arr, K));
}
}
|
Javascript
function MaximumSubarraySum(n, arr, k) {
let dp = new Array(n);
for (let i = 0; i < n; i++) {
dp[i] = new Array(k + 1).fill(0);
}
dp[0][0] = Math.max(0, arr[0]);
for (let i = 1; i < n; i++) {
for (let j = 0; j <= k; j++) {
let X = Math.max(0, arr[i] + dp[i - 1][j]);
let Y = (j > 0) ? dp[i - 1][j - 1] : 0;
dp[i][j] = Math.max(X, Y);
}
}
let res = 0;
for (let i = 0; i < n; i++) {
for (let j = 0; j <= k; j++) {
res = Math.max(res, dp[i][j]);
}
}
if (Math.max(...arr) < 0) {
res = Math.max(...arr);
}
return res;
}
let arr = [-2, 1, 3, -2, 4, -7, 20];
let K = 1;
let N = arr.length;
console.log(MaximumSubarraySum(N, arr, K));
|
Output:
26
Time Complexity: O(N * K)
Auxiliary Space: O(N * K)
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