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Find all combinations of an array and get top k sum elements

I have an array of numbers say [1,2,3,1,1000] , now I want to get all possible combinations of this array and calculate its sum. Combinations are valid such that two combinations have different subset of elements. Then order all the sum values in descending order and get the top k elements.
Example:
[1,2,3,1,1000]
Combinations:
Duplicates of earlier ones are striked out, for example (3,1) matches the earlier (1,3).
(), (1), (2), (3), (1), (1000), (1,2), (1,3), (1,1), (1,1000), (2,3), (2,1), (2,1000), (3,1), (3,1000), (1,1000), (1,2,3), (1,2,1), (1,2,1000), (1,3,1), (1,3,1000), (1,1,1000), (2,3,1), (2,3,1000), (2,1,1000), (3,1,1000), (1,2,3,1), (1,2,3,1000), (1,2,1,1000), (1,3,1,1000), (2,3,1,1000), (1,2,3,1,1000)
And the corresponding sums:
0, 1, 2, 3, 1, 1000, 3, 4, 2, 1001, 5, 3, 1002, 4, 1003, 1001, 6, 4, 1003, 5, 1004, 1002, 6, 1005, 1003, 1004, 7, 1006, 1004, 1005, 1006, 1007
Getting top k=3, sums = 1007, 1006, 1005
So output is [1007, 1006, 1005].
Constraints:
Array size n = 1 to 105
Array elements -109 to 109
k ranges from 1 to 2000
This is my code, reference taken from here:
static List<Long> printDistSum(int arr[]) {
List<Long> list = new ArrayList<>();
int n = arr.length;
// There are totoal 2^n subsets
long total = (long) Math.pow(2, n);
// Consider all numbers from 0 to 2^n - 1
for (int i = 0; i < total; i++) {
long sum = 0;
// Consider binary representation of
// current i to decide which elements
// to pick.
for (int j = 0; j < n; j++)
if ((i & (1 << j)) != 0)
sum += arr[j];
// Print sum of picked elements.
list.add(sum);
}
return list;
}
This code works for small range of inputs but times out for large range of inputs. How to solve this program.

I probably have solution that should be good enough. It has time complexity O(n * k * log(k)).
First we need to calculate max sum - sum of all positive values.
Next we need to iterate over positive values, from smallest to largest. For each of these values we calculate sums of new combinations (at the start we have one combination with max sum).
New combinations will not contains given value so we need to substract it from sum.
At the end we need to iterate over negative values. These values are not belongs to combinations from previous step so we need to add these values to sums.
In every iteration are needed only k maximum sums. I used the PriorityQueue to store these sums. That class use heap data structure so adding/removing values has logarithmic time.
Code:
private static long[] findSums(int[] array, int k) {
long maxSum = Arrays.stream(array).filter(it -> it >= 0).sum();
int[] positives = Arrays.stream(array).filter(it -> it >= 0).sorted().toArray();
int[] negatives = Arrays.stream(array).filter(it -> it < 0).sorted().toArray();
// sort time complexity is O(n*log(n))
PriorityQueue<Long> sums = new PriorityQueue<>(k); // priority queue is implemented using heap so adding element has time complexity O(log(n))
sums.add(maxSum); // we start with max sum - combination of all positive elements
int previous = Integer.MIN_VALUE;
Long[] previousAddedSums = {};
Long[] sumsToIterate;
// iterate over positive values
for (int i = 0; i < positives.length; i++) {
if (positives[i] == previous) {
sumsToIterate = previousAddedSums;
} else {
sumsToIterate = sums.toArray(new Long[sums.size()]);
}
previousAddedSums = new Long[sumsToIterate.length];
for (int j = 0; j < sumsToIterate.length; j++) {
long newSum = sumsToIterate[j] - positives[i];
// new sum is calculated - value positives[i] is removed from combination (subtracted from sum of that combination)
sums.add(newSum);
previousAddedSums[j] = newSum;
if (sums.size() > k) {
sums.poll(); // only first k maximum sums are needed at the moment
}
}
previous = positives[i];
}
previous = Integer.MAX_VALUE;
// iterate over negative values in reverse order
for (int i = negatives.length - 1; i >= 0; i--) {
if (negatives[i] == previous) {
sumsToIterate = previousAddedSums;
} else {
sumsToIterate = sums.toArray(new Long[sums.size()]);
}
previousAddedSums = new Long[sumsToIterate.length];
for (int j = 0; j < sumsToIterate.length; j++) {
long newSum = sumsToIterate[j] + negatives[i]; // value negatives[i] is added to combination (added to sum of that combination)
sums.add(newSum);
previousAddedSums[j] = newSum;
if (sums.size() > k) {
sums.poll();
}
}
previous = negatives[i];
}
long[] result = new long[sums.size()];
for (int i = sums.size() - 1; i >=0 ; i--) {
result[i] = sums.poll();
}
// get sums from priority queue in proper order
return result;
// this whole method has time complexity O(n * k * log(k))
// k is less than or equal 2000 so it should be good enough ;)
}
Demo: https://ideone.com/yf6POI
Edit: I have fixed my solution. Instead of iterating over distinct values I check if current value is same like previous. In that case I use combinations (sums) created in previous step. This prevents from creating duplicates of combinations.
I'm sorry if I didn't explain this well enough. I don't have experience in describing algorithmic / mathematical things in english.

Pls ignore all previous posts cuz they are all wrong.
Intuitively, we gotta use backtrack to find all desired combos, but it's impossible to backtrack on 10^5 elements.
Constraint 1 <= n <= 10^5 alludes that our algorithm bottlenecked by O(nlogn) sorting
Constraint 1 <= k <= min(2000,2^n) alludes that we can backtrack on k elements since k is less than 11. 2^11=2024/log(2000)=11 -- actually this "2^n" gives away solution :)
My algorithm (nlog(n) + 2^k)
sort the array
Record the highest score combo which is the sum of all positive integers
Find a window in the sorted array of math.min(log(k)--which is less than 11,n) elements -- worst case, this window consists of the
lowest 11 absolute values in the sorted array. Several approaches to
achieve that, since the candidates must be inside 22 elements
window(11 smallest positive values + 11 biggest negative values), we
can use PriorityQueue of size 11 scanning over these 22 elements. or
we can use two pointers to find the sliding window of size 11.
backtrack on this 11 absolute value elements window, find sum of each combo and put them into a size k/k-1 PriorityQueue. (k is for
the case that there's no positive elements)
result is the sum of all positive integers plus (sum deducted by each of k-1 elements in PriorityQueue).

I was also asked the same question yesterday but sadly I was not able to solve it yesterday. I have tried solving it today and think I have the answer today.
First of all I don't think that different subsets mean different costs in a set i.e in array of [1,2,3,1] both subsets are valid => [1,2,3] and [2,3,1] as they both use different 1's. Now here is my solution keeping this in mind. But if you really want to keep distinct elements in set then you can simply remove the multiple elements and do partial_sort then.
Logic
Store sum of all +ve nos. in a variable, say maxsum.
Convert the negative nos. to their absolute values.
Get lowest min(k-1, n) elements in sorted order.
Find all their combinations and subtract them from the maxsum.
While finding all their combinations we only need lowest k-1 combos. So we have to find a way to keep the number of combinations to that. For that use a sorted data structure and limit its size to k and then for every element in the sorted array iterate through the combos and add those combos to the sorted data structure if the end element of that data structure is greater. Also pop the end element after that.
For taking care of the above point I am using 2 vectors since the order already remains sorted.
The proposed solution has time complexity of O(n*log(k) + k^2).
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long int ll;
template <class T>
void print(vector<T> topSumm)
{
for (ll itr : topSumm)
cout << itr << '\t';
cout << '\n';
}
vector<ll> mergeSortedArrays(vector<ll> &minns, vector<ll> &temp)
{
vector<ll> ans(minns.size() + temp.size());
int i{0}, j{0}, k{0};
while (i < minns.size() && j < temp.size())
{
if (temp[j] < minns[i])
ans[k++] = temp[j++];
else
ans[k++] = minns[i++];
}
while (i < minns.size())
ans[k++] = minns[i++];
while (j < temp.size())
ans[k++] = temp[j++];
return ans;
}
vector<ll> topKSum(vector<int> &arr, int k)
{
int n{(int)arr.size()};
ll maxSumm{0};
for (int i{0}; i < n; ++i)
{
if (arr[i] > 0)
maxSumm += arr[i];
else
arr[i] = -arr[i];
}
int nk{min(k - 1, n)};
partial_sort(arr.begin(), arr.begin() + nk, arr.end());
vector<ll> minns{0, maxSumm};
ll summ{};
bool breakOuter{false};
for (int i{0}; i < nk; ++i)
{
vector<ll> temp;
for (ll nums : minns)
{
summ = nums + arr[i];
if (minns.size() + temp.size() < k)
temp.push_back(summ);
else
{
if (minns.back() > summ)
{
minns.pop_back();
temp.push_back(summ);
}
else
{
if (nums == 0)
breakOuter = true;
break;
}
}
}
if (breakOuter)
break;
minns = mergeSortedArrays(minns, temp);
}
vector<ll> ans(k);
int i{0};
for (ll nums : minns)
ans[i++] = maxSumm - nums;
return ans;
}
int main()
{
int t;
cin >> t;
while (t--)
{
int n, k;
cin >> n >> k;
vector<int> arr(n);
ll maxSumm{0};
for (int i{0}; i < n; ++i)
cin >> arr[i];
vector<ll> topSums = topKSum(arr, k);
print<ll>(topSums);
}
return 0;
}

Total all matching integer elements in an array

I was looking over some basic questions that might be asked in an interview. Using basic for loops (No hash maps etc)I want to sum up all the matching elements in an array.For example, 6 matching elements will result in (Matched: 6) and a total of 36 in the example below.An example of this could be rolling dice, and the score is the total of all the dice that match.
public static void main(String[] args) {
int arr[] = {6,6,6,6,6,6};
int matched = 1;
int value = 0;
int total = 0;
for(int i=0;i<arr.length;i++){
for(int j=(i+1);j<arr.length;j++){
if(ar[i] == ar[j]){
matched++;
value = ar[i];
break;
}
}
total = (matched * value);
} // End for loop
System.out.println("Matched:"+(matched)+"");
System.out.println("Total:"+total);
}
But, if the array was for example...
int arr[] = {6,1,1,6,6,6};
The output I get will be (Matched:5) and an a total of 30.Can could I store the matching pair of 1's and add them to the total using as little basic code as possible?

Interpreting the question as "provide the total number of values that occur more than once and their sum", and taking into account that nothing "fancy" (sic) such as a map can be used:
int[] array = {6, 1, 1, 6, 6, 6};
int sum = 0;
int matches = 0;
for (int i = 0; i < array.length; i++) {
for (int j = 0; j < array.length; j++) {
if (i != j && array[i] == array[j]) {
sum += array[i];
matches++;
break;
}
}
}
System.out.println(matches); // 6
System.out.println(sum); // 26

If you are allowed to use arrays - if not too fancy - then you could use 3 loops:
The first finds minimum and maximum elements
The second determines the occurrence of each item
The third calculates the totals
In Java this would look like this:
public static void main(String[] args) {
int[] array = {6, 3, 3, 6, 6, 6, 4, 4, 20};
int min = array[0];
int max = array[0];
// Find min max
for (int i : array) {
if (i < min) {
min = i;
}
if (i > max) {
max = i;
}
}
// Use array to sum up all elements
int[] holderArray = new int[max - min + 1];
for (int i : array) {
holderArray[i - min]++;
}
// Calculate occurrence and sums
for(int i = 0; i < holderArray.length; i++) {
if(holderArray[i] > 0) {
System.out.printf("number: %2d; count: %2d; sum: %2d%n", i + min, holderArray[i], (i + min) * holderArray[i]);
}
}
}
This prints out:
number: 3; count: 2; sum: 6
number: 4; count: 2; sum: 8
number: 6; count: 4; sum: 24
number: 20; count: 1; sum: 20

I don't completely understand your question, but based from what I understood, you want to get the sum of all matched numbers if its greater than 1? In that case there is a O(n) solution that you could use.
Create an empty Map then iterate over the array, and if the current number is not within the map add it to the map with value 1, if the current element is already existing in the map then just increment it's value (val++), at the end you will have a map and for each key (each distinct number) you will have the number of matched numbers from the array as the value. All you need to do is iterate over the pairs of key,val multiply each key*val then sum up the multiplied values and you get your correct total. And if you need just the number of matched, you can in another variable sum up just the vals.
So for lets say array [1,1,5,1,5,2,2,5,1], your map will something like:
{1:4, 2:2, 5:3},
and your totals:
total matches: 9
total: 23
I hope this helps!

Here is a code fragment I just wrote. This method takes in an array of integers and creates a map of each unique integer and it's count in the list. In the second part of the method, the HashMap object countOfNumbersMap is iterated and the sum of each element is printed.
private void findSumOfMatchingNumbers(int array[]) {
HashMap<Integer, Integer> countOfNumbersMap = new HashMap<>();
// Setting up the map of unique integers and it's count
for (int i = 0 ; i < array.length ; i++) {
if (countOfNumbersMap.containsKey(array[i])) {
int currentCount = countOfNumbersMap.get(array[i]);
countOfNumbersMap.put(array[i], currentCount + 1);
} else {
countOfNumbersMap.put(array[i], 1);
}
}
for (Integer integer : countOfNumbersMap.keySet()) {
int sum = countOfNumbersMap.get(integer) * integer;
System.out.println(String.format("Number = %d, Sum = %d", integer, sum));
}
}
The worst case runtime of the program is O(n).

If the range of your integers is limited, the easiest way to do this would be to create a histogram (i.e. create an array, where under index i you store the number of occurrences of the number i).
From that, it's easy to find elements that occur more than once, and sum them up. This solution has a complexity of O(n+k), where k is the range of your integers.
Another solution is to sort the array,then the matching numbers will be next to each other, and it's easy to count them. This has O(nlogn) complexity.
If these methods are not allowed, here is a solution in O(n^2) that only uses for loops:
Sum up the whole array
With a double loop, find all elements that are unique, subtract them from the sum, and count their number.
The remaining sum is the sum of all elements occurring more than once, and the count of unique elements subtracted from the length of the array gives the number of matching element.

How to find the largest difference in change in an array? - java

Suppose I declare an array:
int[] arr = {10,2,7,11,3};
The largest (positive) change in this array would be 9 as 11 - 2 = 9.
How would I write a method that find the largest change in code with the smaller integer occurring earlier?
Thank you,

I rewrote the answer since I misunderstood the question.
The simplest but almost certainly not the most efficient way to do this is to check every change and comparing it to the previous one. If it is bigger, discard the previous one and remember this one instead.
int change = arr[1] - arr[0]; //give it an initial value, if we find a bigger change we will replace it
for(int i = 0; i < arr.length - 1; i++) {
for(int j = i + 1; i < arr.length; j++) {
if(arr[j]-arr[i] > change) {
change = arr[j]-arr[i];
}
}
}
This will still give an answer even if there are no positive changes. If you do not want that, you can modify it. It is trivial.
Keep in mind that arr.length - 1 is important in the outer loop.

Looks like you want to find the smallest number and the largest number in the list and comparing it with the largest number on the list.
List first item as minimum.
Compare to next item and if greater, assign that number as lesser.
List first item as largest.
Compare to next item and if smaller, assign that number as greater.
At the end of the list, you will have the least and the greatest number. The difference will be the difference between the smallest and the largest.
Hope that helps.

Interesting question! You could brute-force this pretty easily, as I'm still thinking for a more creative solution.
int [] arr = {5, 4, 3, 2, 1};
int biggestDifference = arr[1]-arr[0];
for (int i = 0; i < arr.length - 1; i++) {
for (int j = i + 1; j < arr.length; j++) {
if ((arr[j] - arr[i]) > biggestDifference) {
biggestDifference = arr[j] - arr[i];
}
}
}

Sorting the array would ensure that the smallest number will always be at the front and largest at the back.
public static void main(String []args){
int[] arr = {10,2,7,11,3};
int diff = findBiggestDiff(arr);
System.out.println(diff);
}
public static int findBiggestDiff(int[] arr){
Arrays.sort(arr);
int diff = arr[arr.length-1] - arr[0];
return diff;
}

Explanation on Integer.MAX_VALUE and Integer.MIN_VALUE to find min and max value in an array

I don't seem to understand how Integer.MAX_VALUE and Integer.MIN_VALUE help in finding the min and max value in an array.
I understand how this method (pseudocode below) works when finding the min and max values:
max = A[0], min = A[0]
for each i in A
if A[i] > max then max = A[i]
if A[i] < min then min = A[i]
But as for this method, I don't understand the purpose of Integer.MAX_VALUE and Integer.MIN_VALUE:
import java.util.Scanner;
class MyClass {
public static void main(String[] args) {
int[] numbers; // declaring the data type of numbers
numbers = new int[3]; //assigning the number of values numbers will contain
int smallest = Integer.MAX_VALUE, largest = Integer.MIN_VALUE;
Scanner input = new Scanner(System.in);
System.out.println("Please enter 3 numbers");
for(int counter = 0; counter<numbers.length;counter++) {
numbers[counter] = input.nextInt();
}
for(int i = 0; i<numbers.length; i++) {
if(numbers[i]<smallest)
smallest = numbers[i];
else if(numbers[i]>largest)
largest = numbers[i];
}
System.out.println("Largest is "+largest);
System.out.println("Smallest is "+smallest);
}
}
System.out.println(Integer.MAX_VALUE) gives 2147483647
System.out.println(Integer.MIN_VALUE) gives -2147483648
So what purpose do Integer.MIN_VALUE and Integer.MIN_VALUE serve in the comparisons?

but as for this method, I don't understand the purpose of Integer.MAX_VALUE and Integer.MIN_VALUE.
By starting out with smallest set to Integer.MAX_VALUE and largest set to Integer.MIN_VALUE, they don't have to worry later about the special case where smallest and largest don't have a value yet. If the data I'm looking through has a 10 as the first value, then numbers[i]<smallest will be true (because 10 is < Integer.MAX_VALUE) and we'll update smallest to be 10. Similarly, numbers[i]>largest will be true because 10 is > Integer.MIN_VALUE and we'll update largest. And so on.
Of course, when doing this, you must ensure that you have at least one value in the data you're looking at. Otherwise, you end up with apocryphal numbers in smallest and largest.
Note the point Onome Sotu makes in the comments:
...if the first item in the array is larger than the rest, then the largest item will always be Integer.MIN_VALUE because of the else-if statement.
Which is true; here's a simpler example demonstrating the problem (live copy):
public class Example
{
public static void main(String[] args) throws Exception {
int[] values = {5, 1, 2};
int smallest = Integer.MAX_VALUE;
int largest = Integer.MIN_VALUE;
for (int value : values) {
if (value < smallest) {
smallest = value;
} else if (value > largest) {
largest = value;
}
}
System.out.println(smallest + ", " + largest); // 1, 2 -- WRONG
}
}
To fix it, either:
Don't use else, or
Start with smallest and largest equal to the first element, and then loop the remaining elements, keeping the else if.
Here's an example of that second one (live copy):
public class Example
{
public static void main(String[] args) throws Exception {
int[] values = {5, 1, 2};
int smallest = values[0];
int largest = values[0];
for (int n = 1; n < values.length; ++n) {
int value = values[n];
if (value < smallest) {
smallest = value;
} else if (value > largest) {
largest = value;
}
}
System.out.println(smallest + ", " + largest); // 1, 5
}
}

Instead of initializing the variables with arbitrary values (for example int smallest = 9999, largest = 0) it is safer to initialize the variables with the largest and smallest values representable by that number type (that is int smallest = Integer.MAX_VALUE, largest = Integer.MIN_VALUE).
Since your integer array cannot contain a value larger than Integer.MAX_VALUE and smaller than Integer.MIN_VALUE your code works across all edge cases.

By initializing the min/max values to their extreme opposite, you avoid any edge cases of values in the input: Either one of min/max is in fact one of those values (in the case where the input consists of only one of those values), or the correct min/max will be found.
It should be noted that primitive types must have a value. If you used Objects (ie Integer), you could initialize value to null and handle that special case for the first comparison, but that creates extra (needless) code. However, by using these values, the loop code doesn't need to worry about the edge case of the first comparison.
Another alternative is to set both initial values to the first value of the input array (never a problem - see below) and iterate from the 2nd element onward, since this is the only correct state of min/max after one iteration. You could iterate from the 1st element too - it would make no difference, other than doing one extra (needless) iteration over the first element.
The only sane way of dealing with inout of size zero is simple: throw an IllegalArgumentException, because min/max is undefined in this case.

Find the largest sum including at most two consecutive elements from an array

I've been playing around a bit with the algorithms for getting the largest sum with no two adjacent elements in an array but I was thinking:
If we have an array with n elements and we want to find the largest sum so that 3 elements never touch. That's to say if we have the array a = [2, 5, 3, 7, 8, 1] we can pick 2 and 5 but not 2, 5 and 3 because then we have 3 in a row. The larget sum with these rules for this array would be: 22 (2 and 5, 7 and 8. 2+5+7+8=22)
I'm not sure how I would implement this, any ideas?
Edit:
I've only come so far as to think about what might be good to do:
Let's just stick to the same array:
int[] a = {2, 5, 3, 7, 8, 1};
int{} b = new int[n}; //an array to store results in
int n = a.length;
// base case
b[1] = a[1];
// go through each element:
for(int i = 1; i < n; i++)
{
/* find each possible way of going to the next element
use Math.max to take the "better" option to store in the array b*/
}
return b[n]; // return the last (biggest) element.
This is just a thought I got in my head, hasn't reached longer than this.

Algorithm for Maximum sum such that no two elements are adjacent:
Loop for all elements in arr[] and maintain two sums incl and excl where incl = Max sum including the previous element and excl = Max sum excluding the previous element.
Max sum excluding the current element will be max(incl, excl) and max sum including the current element will be excl + current element (Note that only excl is considered because elements cannot be adjacent).
At the end of the loop return max of incl and excl.
Implementation:
#include<stdio.h>
/*Function to return max sum such that no two elements
are adjacent */
int FindMaxSum(int arr[], int n)
{
int incl = arr[0];
int excl = 0;
int excl_new;
int i;
for (i = 1; i < n; i++)
{
/* current max excluding i */
excl_new = (incl > excl)? incl: excl;
/* current max including i */
incl = excl + arr[i];
excl = excl_new;
}
/* return max of incl and excl */
return ((incl > excl)? incl : excl);
}
/* Driver program to test above function */
int main()
{
int arr[] = {5, 5, 10, 100, 10, 5};
printf("%d \n", FindMaxSum(arr, 6));
getchar();
return 0;
}
Time Complexity: O(n)
Space Complexity: O(1)
Edit 1:
If you understand the above code, we can easily do this problem by maintaining the count of already adjacent numbers for previous position.
Here is a working implementation to the required question
//We could assume we store optimal result upto i in array sum
//but we need only sum[i-3] to sum[i-1] to calculate sum[i]
//so in this code, I have instead maintained 3 ints
//So that space complexity to O(1) remains
#include<stdio.h>
int max(int a,int b)
{
if(a>b)
return 1;
else
return 0;
}
/*Function to return max sum such that no three elements
are adjacent */
int FindMaxSum(int arr[], int n)
{
int a1 = arr[0]+arr[1];//equivalent to sum[i-1]
int a2 =arr[0];//equivalent to sum[i-2]
int a3 = 0;//equivalent to sum [i-3]
int count=2;
int crr = 0;//current maximum, equivalent to sum[i]
int i;
int temp;
for (i = 2; i < n; i++)
{
if(count==2)//two elements were consecutive for sum[i-1]
{
temp=max(a2+arr[i],a1);
if(temp==1)
{
crr= a2+arr[i];
count = 1;
}
else
{
crr=a1;
count = 0;
}
//below is the case if we sould have rejected arr[i-2]
// to include arr[i-1],arr[i]
if(crr<(a3+arr[i-1]+arr[i]))
{
count=2;
crr=a3+arr[i-1]+arr[i];
}
}
else//case when we have count<2, obviously add the number
{
crr=a1+arr[i];
count++;
}
a3=a2;
a2=a1;
a1=crr;
}
return crr;
}
/* Driver program to test above function */
int main()
{
int arr[] = {2, 5, 3, 7, 8, 1};
printf("%d \n", FindMaxSum(arr, 6));
return 0;
}
Time Complexity: O(n)
Space Complexity: O(1)

adi's solution can be easily generalized to allow up to n adjacent elements to be included in the sum. The trick is to maintain an array of n + 1 elements, where the k-th element in the array (0 ≤ k ≤ n) gives the maximum sum assuming that the k previous inputs are included in the sum and the k+1-th isn't:
/**
* Find maximum sum of elements in the input array, with at most n adjacent
* elements included in the sum.
*/
public static int maxSum (int input[], int n) {
int sums[] = new int[n+1]; // new int[] fills the array with zeros
int max = 0;
for (int x: input) {
int newMax = max;
// update sums[k] for k > 0 by adding x to the old sums[k-1]
// (loop from top down to avoid overwriting sums[k-1] too soon)
for (int k = n; k > 0; k--) {
sums[k] = sums[k-1] + x;
if (sums[k] > newMax) newMax = sums[k];
}
sums[0] = max; // update sums[0] to best sum possible if x is excluded
max = newMax; // update maximum sum possible so far
}
return max;
}
Like adi's solution, this one also runs in linear time (to be exact, O(mn), where m is the length of the input and n is the maximum number of adjacent elements allowed in the sum) and uses a constant amount of memory independent of the input length (O(n)). In fact, it could even be easily modified to process input streams whose length is not known in advance.

I would imagine putting the array into a binary tree in that order. That way you can keep track of which element is next to each other. Then just simply do an if (node is not directly linked to each other) to sum the nodes which are not next to each other. You can potentially do it with recursion and return the maximum number, makes things easier to code. Hope it helps.

For a set with n entries, there are 2^n ways to partition it. So to generate all possible sets, just loop from 0:2^n-1 and pick the elements from the array with those entries set to 1 (bear with me; I'm getting to your question):
max = 0;
for (i = 0; i < 1<<n; ++i) {
sum = 0;
for (j = 0; j < n; ++j) {
if (i & (1<<j)) { sum += array[j]; }
}
if (sum > max) { /* store max and store i */ }
}
This will find the maximum way to sum the entries of an array. Now, the issue you want is that you don't want to allow all values of i - specifically those that contain 3 consecutive 1's. This can be done by testing if the number 7 (b111) is available at any bit-shift:
for (i = 0; i < 1<<n; ++i) {
for (j = 0; j < n-2; ++j) {
if ((i & (7 << j)) == (7 << j)) { /* skip this i */ }
}
...