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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;
}
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.
I'm working on a a game where I have to generate an int array of 4 elements randomly. My problem is that the mean of all the array elements always have to be a whole number.
Example :
array 1 {4 , 2 , 3, 7} , the mean of the array is 28,75 which is not what I'm looking for,
array 2 {3 , 7 , 6 , 4} , the mean is 20 which is good
Now I could make a loop where I check if the mean of the randomly generated numbers is a whole number but that doesn't seems like an efficient way to do that.
The game I'm working for is mean sum for those who know it.
If the mean is a whole number, then the sum must be divisible by 4.
int[] n = new int[4];
Pick four numbers, and calculate their sum:
int sum = 0;
for (int i = 0; i < 4; ++i) {
sum += (n[i] = random.nextInt());
}
Calculate the remainder of sum / 4:
int r = sum % 4;
So you now need to adjust the sum so that sum % 4 == 0. You can either:
subtract r from any of the elements of the array:
n[random.nextInt(4)] -= r;
or add 4 - r to any element:
n[random.nextInt(4)] += 4 - r;
Ideone demo
Pick a target mean m and random integers n1, n2.
Your array is [m-n1, m+n1, m-n2, m+n2]. Haven't thought about what the properties of this distribution would be, but it should work.
I believe the following function does what you want, given arguments of how many values you want to generate (n) and what's an upper limit for the sum of the values (max).
private static Random r = new Random();
public static int[] makeSet(int n, int max) {
// The next line guarantees the result is divisible by n
int currentMax = n * (1 + r.nextInt(max / n));
Set<Integer> s = new HashSet<Integer>();
// Generate a set of unique values between 0 and the currentMax,
// containing those bounds
s.add(0);
s.add(currentMax);
do {
s.add(r.nextInt(currentMax));
} while(s.size() <= n);
Integer[] values = new Integer[n + 1];
/*
* Convert to array, sort the results, and find successive
* differences. By construction, those differences WILL sum
* to the currentMax, which IS divisible by the number of
* values generated by differencing!
*/
s.toArray(values);
Arrays.sort(values);
int[] results = new int[n];
for(int i = 0; i < n; ++i) {
results[i] = values[i+1] - values[i];
}
return results;
}
Intro: As far as I could search, this question wasn't asked in SO yet.
This is an interview question.
I am not even specifically looking for a code solution, any algorithm/pseudocode will work.
The problem: Given an integer array int[] A and its size N, find 2 non-subsequent (can't be adjacent in the array) elements with minimal sum. Also the answer must not contain the first or last elements (index 0 and n-1). Also the solution should be in O(n) time and space complexity.
E.g. when A = [5, 2, 4, 6, 3, 7] the answer is 5, since 2+3=5.
When A = [1, 2, 3, 3, 2, 1] the answer is 4, since 2+2=4 and you can't choose either of the 1's since the are at the ends of the array.
Attempt: At first I thought that one of the numbers in the solution must be the smallest one in the array (besides the first and last), but this was refuted quickly with the counter-example A = [4, 2, 1, 2, 4] -> 4 (2+2)
Then I thought that if I find the 2 smallest numbers (besides the first and last) in the array, the solution will be those two. This obviously quickly failed because I can't choose 2 adjacent numbers, and if I have to choose non-adjacent numbers then this is the very definition of the question :).
Finally I thought, well, I will just find the 3 smallest numbers (besides the first and last) in the array, and the solution will have to be two of those, since two of those have to not be adjacent to each other.
This also failed due to A = [2, 2, 1, 2, 4, 2, 6] -> 2+1=3 , which seems to work because I will find 2, 1, 2, but assuming I am finding the 2, 1, 2 in indexes 1, 2, 3 this won't necessarily work (it would if I found specifically the 2 in index 5 but I can't guarantee that unfortunately).
Question:
Now I'm stumped, can anyone come up with a solution/idea that works?
Here is a live javascript implementation of an algorithm that:
finds the 4 smallest elements (excluding first/last element from search)
finds the pairs of these 4 elements that are not adjacent in original array
finds from these pairs the one with the minimal sum
function findMinNonAdjacentPair(a) {
var mins = [];
// quick exits:
if (a.length < 5) return {error: "no solution, too few elements."};
if (a.some(isNaN)) return {error: "non-numeric values given."};
// collect 4 smallest values by their indexes
for (var i = 1; i < a.length - 1; i++) { // O(n)
if (mins.length < 4 || a[i] < a[mins[3]]) {
// need to keep record of this element in sorted list of 4 elements
for (var j = Math.min(mins.length - 1, 2); j >= 0; j--) { // O(1)
if (a[i] >= a[mins[j]]) break;
mins[j+1] = mins[j];
}
mins[j+1] = i;
}
}
// mins now has the indexes to the 4 smallest values
// Find the smallest sum
var result = {
sum: a[mins[mins.length-1]]*2+1 // large enough
}
for (var j = 0; j < mins.length-1; j++) { // O(1)
for (var k = j + 1; k < mins.length; k++) {
if (Math.abs(mins[j] - mins[k]) > 1) { // not adjacent
if (result.sum > a[mins[j]]+a[mins[k]]) {
result.sum = a[mins[j]]+a[mins[k]];
result.index1 = mins[j];
result.index2 = mins[k];
};
if (k < j + 3) return result; // cannot be improved
break; // exit inner loop: it cannot bring improvement
}
}
}
return result;
}
// Get I/O elements
var input = document.getElementById('in');
var output = document.getElementById('out');
var select = document.getElementById('pre');
function process() {
// translate input to array of numbers
var a = input.value.split(',').map(Number);
// call main function and display returned value
output.textContent = JSON.stringify(findMinNonAdjacentPair(a), null, 4);
}
// respond to selection from list
select.onchange = function() {
input.value = select.value;
process();
}
// respond to change in input box
input.oninput = process;
// and produce result upon load:
process();
Type comma-separated list of values (or select one):</br>
<input id="in" value="2, 2, 1, 2, 4, 2, 6"> <=
<select id="pre">
<option value="5, 2, 4, 6, 3, 7">5, 2, 4, 6, 3, 7</option>
<option value="1, 2, 3, 3, 2, 1">1, 2, 3, 3, 2, 1</option>
<option value="4, 2, 1, 2, 4">4, 2, 1, 2, 4</option>
<option value="2, 2, 1, 2, 4, 2, 6" selected>2, 2, 1, 2, 4, 2, 6</option>
</select>
</br>
Output:</br>
<pre id="out"></pre>
The algorithm has a few loops with following big-O complexities:
find 4 smallest values: O(n), as the inner loop runs at most 3 times, which is O(1)
find the smallest sum of non-adjacent pairs has a double loop: in total the body will run at most 4 times = O(1). NB: The number of possible pairs is 6, but the execution is guaranteed to break out of the loops sooner.
So the algorithm runs in O(n).
Find the smallest number beside the first and the last.
Find the second smallest that is not a neighbour of the first one and not the first or last one in the array. Then build the sum.
If the first element is the second or the penultimate element you already have the solution.
Otherwise calculate the sum of both neighbours of the first number. check if its smaller then the first sum
if not: take the first sum
otherwise take the second one
This will always work because if the first sum is not the answer that means the first number cannot be part of the solution. And that on the other hand means, the solution can just be the second sum.
This problem can be solved with about 10 lines of Java code.
You can start with an obvious but inefficient (O(N^2)) solution:
public class Main {
int solve(int[] array) {
int answer = Integer.MAX_VALUE;
for (int i = 3; i < array.length - 1; i++) {
for (int j = 1; j < i - 1; j++) {
if (array[i] + array[j] < answer) {
answer = array[i] + array[j];
}
}
}
return answer;
}
}
But then you can notice that you actually do not need the internal for loop because you can just preserve the minimum and update it with every new element if necessary, which is faster than finding the minimum anew every time. Therefore the final O(N) solution looks like this:
public class Main {
int solve(int[] array) {
int answer = Integer.MAX_VALUE;
int min = array[1];
for (int i = 3; i < array.length - 1; i++) {
min = Math.min(min, array[i - 2]);
if (array[i] + min < answer) {
answer = array[i] + min;
}
}
return answer;
}
}
Find the four smallest and consider all possibilities among those four. The smallest is nonadjacent to at least one of the second, third, or fourth smallest; the only other possibility that could be better is the second and third smallest (assuming that they are nonadjacent).
I think this does not need any deep reasoning, and can be solved in a single pass, keeping the optimal solution of the array elements processed so far:
public static int[] minimumSumOfNonAcjacentElements(int[] a) {
// the result for the sequence a[1:i]
int minSum = Integer.MAX_VALUE;
int minSumElement1 = Integer.MAX_VALUE;
int minSumElement2 = Integer.MAX_VALUE;
// the minimum element eligible for joining with a[i], i.e. from a[1 : i-2]
int minElement = a[1];
int prevElement = a[2]; // a[i - 1]
for (int i = 3; i + 1 < a.length; i++) {
int sum = minElement + a[i];
if (sum < minSum) {
minSum = sum;
minSumElement1 = minElement;
minSumElement2 = a[i];
}
if (prevElement < minElement) {
minElement = prevElement;
}
prevElement = a[i];
}
return new int[] {minSumElement1, minSumElement2};
}
Here's some test code, with the corner cases from OP's question:
private static void test(int minSumIndex1, int minSumIndex2, int... input) {
int[] result = minimumSumOfNonAcjacentElements(input);
if (result[0] == minSumIndex1 && result[1] == minSumIndex2) {
// ok
} else {
throw new AssertionError("Expected: " + minSumIndex1 + ", " + minSumIndex2 + ". Actual=" + Arrays.toString(result));
}
}
public static void main(String[] args) throws Exception {
test(2, 2, 4, 2, 1, 2, 4);
test(1, 2, 2, 2, 1, 2, 4, 2, 6);
test(1, 2, 0, 2, 1, 2, 4, 2, 0);
System.out.println("All tests passed.");
}
Use dynamic programming.
Remove or disregard the first and last elements of your array. Since they cannot participate in the solution, they are not important. Once you've done this, you can also ignore the "must not be the first or last element" constraint since we've already accounted for it.
Find the solution for the first three elements of (what's left of) the array (and without considering the "no first/last element" rule). There is only one solution in this case (array[0] + array[2]), so this is a trivial step.
Memoize the minimal element which is not the last element (i.e. min(array[0], array[1])).
Find the solution for the first four elements. We don't have to redo the whole problem; instead we just have to ask whether introducing the fourth element allows us to produce a smaller solution. We can do this by adding the fourth element to the minimal element we memoized in the previous step, and comparing the sum to the solution we found in the second step.
Update the memoized minimal element so that it is the minimum of the first three elements.
Continue widening and updating in this fashion until we have considered the entire array.
The whole algorithm is O(n), since both widening and updating are constant-time operations. The algorithm can be proved correct by simple induction. O(n) is also a lower bound since we have to consider every element of the array, so this algorithm is optimal.
Algorithm:
Find the minimum, avoiding the end indices. (1 O(n) pass)
Find the minimum, avoiding the end indices and the index of (1) and adjacent indices. (1 O(n) pass)
Find the minimum, avoiding the end indices and the index of (1) (1 O(n) pass)
Find the minimum, avoiding the end indices and the index of (3) and adjacent indices. (1 O(n) pass)
Return the minimum of the sums (1) + (2), (3) + (4), if they exist.
Passes 3 and 4 are meant to pass the case [4, 2, 1, 2, 4] = 4 by finding both 2s.
public static int minSumNonAdjNonEnd(int[] array)
{
// 1. Find minimum
int minIdx1 = -1;
int minValue1 = Integer.MAX_VALUE;
for (int i = 1; i < array.length - 1; i++)
{
if (array[i] < minValue1)
{
minIdx1 = i;
minValue1 = array[i];
}
}
// 2. Find minimum not among (1) or adjacents.
int minIdx2 = -1;
int minValue2 = Integer.MAX_VALUE;
for (int i = 1; i < array.length - 1; i++)
{
if ((i < minIdx1 - 1 || i > minIdx1 + 1) && (array[i] < minValue2))
{
minIdx2 = i;
minValue2 = array[i];
}
}
boolean sum1Exists = (minIdx1 > -1 && minIdx2 > -1);
int sum1 = minValue1 + minValue2;
// 3. Find minimum not among (1).
int minIdx3 = -1;
int minValue3 = Integer.MAX_VALUE;
for (int i = 1; i < array.length - 1; i++)
{
if ((i != minIdx1) && (array[i] < minValue3))
{
minIdx3 = i;
minValue3 = array[i];
}
}
// 4. Find minimum not among(3) or adjacents.
int minIdx4 = -1;
int minValue4 = Integer.MAX_VALUE;
for (int i = 1; i < array.length - 1; i++)
{
if ((i < minIdx3 - 1 || i > minIdx3 + 1) && (array[i] < minValue4))
{
minIdx4 = i;
minValue4 = array[i];
}
}
boolean sum2Exists = (minIdx3 > -1 && minIdx4 > -1);
int sum2 = minValue3 + minValue4;
if (sum1Exists)
{
if (sum2Exists)
return Math.min(sum1, sum2);
else
return sum1;
}
else
{
if (sum2Exists)
return sum2;
else
throw new IllegalArgumentException("impossible");
}
}
This performs 4 linear searches, for a complexity of O(n).
Test cases:
System.out.println(minSumNonAdjNonEnd(new int[] {5, 2, 4, 6, 3, 7}));
System.out.println(minSumNonAdjNonEnd(new int[] {1, 2, 3, 3, 2, 1}));
System.out.println(minSumNonAdjNonEnd(new int[] {4, 2, 1, 2, 4}));
System.out.println(minSumNonAdjNonEnd(new int[] {2, 2, 1, 2, 4, 2, 6}));
System.out.println(minSumNonAdjNonEnd(new int[] {2, 2, 3, 2}));
5
4
4
3
Exception in thread "main" java.lang.IllegalArgumentException: impossible
I don't know if my solution is correct because I just tested it with the data in the OP, and I don't even know if this is better or worse than the other ideas but I wanted to try it.
static void printMinimalSum(int[] A) {
// Looking for mins so we init this with max value
int[] mins = new int[]{Integer.MAX_VALUE, Integer.MAX_VALUE, Integer.MAX_VALUE};
// Indices, used just to print the solution
int[] indices = new int[]{-1, -1, -1};
// If the array has length 5 then there's only one solution with the 2nd and 4th elements
if (A.length == 5) {
mins[0] = A[1];
indices[0] = 1;
mins[1] = A[3];
indices[1] = 3;
} else {
// Loop on the array without considering the first and the last element
for (int i = 1; i < A.length - 1; i++) {
// We consider each element which is smaller than its neighbours
if ((i == 1 && A[i] < A[i + 1]) // 1: first element, compare it with the second one
|| (i == A.length - 2 && A[i] < A[i - 1]) // 2: last element, compare it with the previous one
|| (A[i] < A[i + 1] && A[i] < A[i - 1])) { // 3: mid element, compare it with both neighbors
// If the element is "legal" then we see if it's smaller than the 3 already saved
if (A[i] < mins[0]) {
mins[0] = A[i];
indices[0] = i;
} else if (A[i] < mins[1]) {
mins[1] = A[i];
indices[1] = i;
} else if (A[i] < mins[2]) {
mins[2] = A[i];
indices[2] = i;
}
}
}
}
// Compute the 3 sums between those 3 elements
int[] sums = new int[]{Math.abs(mins[0]+mins[1]), Math.abs(mins[0]+mins[2]), Math.abs(mins[1]+mins[2])};
// Find the smaller sum and print it
if (sums[0] < sums[1] || sums[0] < sums[2]){
System.out.println("Sum = " + sums[0] + " (elements = {" + mins[0] + "," + mins[1] + "}, indices = {" + indices[0] + "," + indices[1] + "}");
} else if (sums[1] < sums[0] || sums[1] < sums[2]){
System.out.println("Sum = " + sums[1] + " (elements = {" + mins[0] + "," + mins[2] + "}, indices = {" + indices[0] + "," + indices[2] + "}");
} else {
System.out.println("Sum = " + sums[2] + " (elements = {" + mins[1] + "," + mins[2] + "}, indices = {" + indices[1] + "," + indices[2] + "}");
}
}
public static void main(String[] args) {
printMinimalSum(new int[]{5, 2, 4, 6, 3, 7});
printMinimalSum(new int[]{1, 2, 3, 3, 2, 1});
printMinimalSum(new int[]{4, 2, 1, 2, 4});
printMinimalSum(new int[]{2, 2, 1, 2, 4, 2, 6});
}
Output is:
Sum = 5 (elements = {2,3}, indices = {1,4}
Sum = 4 (elements = {2,2}, indices = {1,4}
Sum = 4 (elements = {2,2}, indices = {1,3}
Sum = 3 (elements = {1,2}, indices = {2,5}
which seems fine.
edit: you're right, I completely ignored the adjacency constraint.
luckily I've thought of a solution.
The algorithm goes like this:
You run once over the array to find the smallest (O(n))
You run a second time to find the second smallest (O(n))
If second smallest is not adjacent to smallest we're done(O(1) - just an index check)
Otherwise run a third time to find third smallest (still O(n))
If not adjacent to smallest return smallest and third smallest
otherwise return second and third smallest
Elaborating on the above answer, you'd need a modified insertion-sort to track the smallest four values and the corresponding indexes (an array of 4 elements for each).
Once found the solution would be the first pair whose difference in indexes would be more than 1 and whose sum is the least.
The solution being one of (0,1) or (0,2) or (0,3) or (1,2) or (1,3) or (2,3) where the values indicate the indexes of the array that in turn tracks the position of the actual elements in the array.
Also you'd need to handle the special case for array-length 5 (arr\[1]+arr[3]) and an error for those arrays less than 5.
I think this should work:
Find the minimum 3 element and their indices. Since all of them can't be adjacent choose 2 among them.
If all of them are adjacent and the minimum number is in the middle of them, iterate through all elements, find the forth minimum element, choose minimum of min1+min4, min2+min3, whichever is smaller.
You can do this in one iteration too.
I have used dynamic programming to solve it.
Idea is to first create the array which tracks the minimum found till now as below:
Input array = [1, 3, 0, 5, 6]
Minimum array = [1, 1, 0, 0, 0]
Now using the minimum array and the input array we can use below:
DP[i] = min(DP[i-1], min(first_data, second_data))
where DP[i] means the minimum found till now which is sum of two previous alternate elements.
first_data = sum of current element in input array + sum of current-2 element in minimum array
second_data = sum of current-1 element in input array + sum of current-3 element in minimum array
import random
def get_min(numbers):
#disregard the first and last element
numbers = numbers[1:len(numbers)-1]
#for remembering the past results
DP = [0]*len(numbers)
#for keeping track of minimum till now found
table = [0]*len(numbers)
high_number = 1 << 30
min_number = numbers[0]
table[0] = min_number
for i in range(0, len(numbers)):
DP[i] = high_number
for i in range(1, len(numbers)):
if numbers[i] < min_number:
min_number = numbers[i]
table[i] = numbers[i]
else:
table[i] = min_number
for i in range(0, len(numbers)):
min_first, min_second = high_number, high_number
if i >= 2:
min_first = numbers[i] + table[i-2]
if i >= 3:
min_second = numbers[i-1] + table[i-3]
if i >= 1:
DP[i] = min(min(DP[i-1], min_first), min_second)
return DP[len(numbers)-1]
input = random.sample(range(100), 10)
print(input)
print(get_min(input))
How about that: you find k smallest numbers (or more precisely their indices) (k big enough, let say 10). It is sure, that the wanted pair is between them. Now you just check the possible 50 pairs and select the best which satisfies the constraints.
You don't need 10, less would do - but more than 3 :)
Edit: finding k smallest numbers is O(n), because you just keep the best 10 for example in a heap (add new element, delete maximum O(k*logk)=O(1) operations).
Then there will be a pair which satisfy the constraints (not next to each other). It is also clear that, if you build the sum with an element not from those k, it would be bigger than the best pair chosen from those k elements.
Checking at most k*k pairs is also O(1), thus the whole running time is O(n).
Here is the python implementation in O(N) time complexity
import math
def minSum(array):
_min = array[1]
result = math.inf
for i in range(3, len(array) - 1):
_min = min(_min, array[i-2])
if (_min + array[i]) < result:
result = _min + array[i]
return result
As we only need to track the minimum sum of two no adjacent values, we could do it by iterating over the array excluding the first and last element and keeping the track of min values and minimum sum. current min value will the two index before current value. For example if we are checking the current index i then minValue is the minimum value from index 1 to i-2.
Code:
int minSum(int[] A){
int minSum=Integer.MAX_VALUE;
int min= Integer.MAX_VALUE;
for(int i=3; i<A.length-1; i++){
min= Math.min(A[i-2], min);
minSum = Math.min(min+A[i], minSum);
}
return minSum;
}
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 */ }
}
...