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Finding the sum of common elements between n number of arrays in java

I have a program that sums the common elements of two arrays. For that I used two for loops and if I have three then I could use three for loops. But how to sum the common elements of n number of arrays where n is coming during run time.
I don't know how to change the number of loops during run time or is there any other relevant concept for this ?
Here is the code I've tried for summing twoarrays:
import java.util.Scanner;
public class Sample {
public static void main(String... args)
{
Scanner sc=new Scanner(System.in);
int arr1[]={1,2,3,4,5},arr2[]={4,5,6,7,8},sum=0;
for (int i=0;i<arr1.length;i++)
{
for (int j=0;j<arr2.length;j++)
{
if (arr1[i]==arr2[j])
{
sum+=(arr1[i]);
}
}
}
}
}

There can be different implementation for that. You can use the following approach. Here is the pseudo code
use a 2D array to store the array. if the number of array is n and size is m then the array will be input[n][m]
Use a ArrayList commonItems to store the common items of. Initiate it with the elements of input[0]
Now iterate through the array for i = 1 to n-1. compare with every input[i], store only the common items of commonItems and input[i] at each step. You can do it by converting the input[i] into a list and by using retainAll method.
At the end of the iteration the commonItem list will contains the common numbers only. Now sum the value of this list.

There is actually a more general method, that also answers the question "how to change the number of loops during run time?".
The general question
We are looking for a way to implement something equivalent to this:
for (i1 = 0; i1 < k1; i1++) {
for (i2 = 0; i2 < k2; i2++) {
for (i3 = 0; i3 < k3; i3++) {
...
for (in = 0; in < kn; in++) {
f(x1[i1], x2[i2], ... xn[in]);
}
...
}
}
}
where, n is given at runtime and f is a function taking a list of n parameters, processing the current n-tuple.
A general solution
There is a general solution, based on the concept of recursion.
This is one implementation that produces the desired behavior:
void process(int idx, int n, int[][] x, int[] k, Object[] ntuple) {
if (idx == n) {
// we have a complete n-tuple,
// with an element from each of the n arrays
f(ntuple);
return;
}
// this is the idx'th "for" statement
for (int i = 0; i < k[idx]; i++) {
ntuple[idx] = x[idx][i];
// with this recursive call we make sure that
// we also generate the rest of the for's
process(idx + 1, n, x, k, ntuple);
}
}
The function assumes that the n arrays are stored in a matrix x, and the first call should look like this:
process(0, n, x, k, new Object[n]);
Practical considerations
The solution above has a high complexity (it is O(k1⋅k2⋅..⋅kn)), but sometimes it is possible to avoid going until the deepest loop.
Indeed, in the specific problem mentioned in this post (which requires summing common elements across all arrays), we can skip generating some tuples e.g. if already x2[i2] ≠ x1[i1].
In the recursive solution, those situations can easily be pruned. The specific code for this problem would probably look like this:
void process(int idx, int n, int[][] x, int[] k, int value) {
if (idx == n) {
// all elements from the current tuple are equal to "value".
// add this to the global "sum" variable
sum += value;
return;
}
for (int i = 0; i < k[idx]; i++) {
if (idx == 0) {
// this is the outer "for", set the new value
value = x[0][i];
} else {
// check if the current element from the idx'th for
// has the same value as all previous elements
if (x[idx][i] == value) {
process(idx + 1, n, x, k, value);
}
}
}
}

Assuming that the index of the element is not important: a[1] = 2 and a[5] = 2, you only need two nested loops.
First you need to put n-1 arrays in a list of sets. Then loop over nth array and check if each element exists in all of the sets in the list. If it does exist then add to total.

Java - possible permutations of an array of numbers which would result in an identical binary search tree

Given an array of ints which would generate a certain BST, how many variations of that array would result in an identical BST? I have found a few solutions in C++ and python, but nothing in Java. I think I understand the concept of how to develop the correct code.
I'm doing this for a certain Google foobar challenge. When I throw any possible arrays that I could think of I get the correct answer, but when I try to verify my code with Google I get an ArithmeticException. I cannot find where this would possibly occur in my code.
I need to return the answer in a String and the parameter can be an array with a maximum of 50 integers.
This is the code I currently have:
public static String answer(int[] seq) {
if (seq.length <= 1) {
return Integer.toString(1);
}
ArrayList<Integer> rightList = new ArrayList<>();
ArrayList<Integer> leftList = new ArrayList<>();
int root = seq[0];
for (int i : seq) {
if (i > root) {
leftList.add(i);
} else if (i < root) {
rightList.add(i);
}
}
int[] rightArray = new int[rightList.size()];
int[] leftArray = new int[leftList.size()];
int i = 0;
for (int j : rightList) {
rightArray[i++] = j;
}
int k = 0;
for (int l : leftList) {
leftArray[k++] = l;
}
int recurseLeft = Integer.parseInt(answer(leftArray));
int recurseRight = Integer.parseInt(answer(rightArray));
return Integer.toString(recurseLeft * recurseRight
* interleave(leftList.size(), rightList.size()));
}
private static int interleave(int a, int b) {
return (factorial(a + b)) / ((factorial(a) * factorial(b)));
}
private static int factorial(int n) {
return (n <= 1 ? 1 : n * factorial(n - 1));
}
Can someone help find either a bug or a possible array of integers that would cause an ArithmeticException?

Can someone help find either a bug or a possible array of integers
that would cause an ArithmeticException?
The ArithmeticException is likely thrown because you divide a number by 0. Adding the stacktrace would have helped to identify where it occurs, but you're performing a division in the interleave method.
factorial(a) * factorial(b) is an integer multiplication. If the result is too large to fit for the max value an integer can have, it will overflow.
For instance 34! mod 241 = 0. So it suffices that you have a degenerated BST where all the elements are superior than the root (which is the first element of your array here) with 35 elements to get an exception.
Hence the following array:
int[] arr = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,3‌​0,31,32,33,34,35};
will throw it.

ArithmeticException is thrown when you try to divide by zero. The only place in your code to use division is interleave function

how to improve this code?

I have developed a code for expressing the number in terms of the power of the 2 and I am attaching the same code below.
But the problem is that the expressed output should of minimum length.
I am getting output as 3^2+1^2+1^2+1^2 which is not minimum length.
I need to output in this format:
package com.algo;
import java.util.Scanner;
public class GetInputFromUser {
public static void main(String[] args) {
// TODO Auto-generated method stub
int n;
Scanner in = new Scanner(System.in);
System.out.println("Enter an integer");
n = in.nextInt();
System.out.println("The result is:");
algofunction(n);
}
public static int algofunction(int n1)
{
int r1 = 0;
int r2 = 0;
int r3 = 0;
//System.out.println("n1: "+n1);
r1 = (int) Math.sqrt(n1);
r2 = (int) Math.pow(r1, 2);
// System.out.println("r1: "+r1);
//System.out.println("r2: "+r2);
System.out.print(r1+"^2");
r3 = n1-r2;
//System.out.println("r3: "+r3);
if (r3 == 0)
return 1;
if(r3 == 1)
{
System.out.print("+1^2");
return 1;
}
else {
System.out.print("+");
algofunction(r3);
return 1;
}
}
}

Dynamic programming is all about defining the problem in such a way that if you knew the answer to a smaller version of the original, you could use that to answer the main problem more quickly/directly. It's like applied mathematical induction.
In your particular problem, we can define MinLen(n) as the minimum length representation of n. Next, say, since we want to solve MinLen(12), suppose we already knew the answer to MinLen(1), MinLen(2), MinLen(3), ..., MinLen(11). How could we use the answer to those smaller problems to figure out MinLen(12)? This is the other half of dynamic programming - figuring out how to use the smaller problems to solve the bigger one. It doesn't help you if you come up with some smaller problem, but have no way of combining them back together.
For this problem, we can make the simple statement, "For 12, it's minimum length representation DEFINITELY has either 1^2, 2^2, or 3^2 in it." And in general, the minimum length representation of n will have some square less than or equal to n as a part of it. There is probably a better statement you can make, which would improve the runtime, but I'll say that it is good enough for now.
This statement means that MinLen(12) = 1^2 + MinLen(11), OR 2^2 + MinLen(8), OR 3^2 + MinLen(3). You check all of them and select the best one, and now you save that as MinLen(12). Now, if you want to solve MinLen(13), you can do that too.
Advice when solo:
The way I would test this kind of program myself is to plug in 1, 2, 3, 4, 5, etc, and see the first time it goes wrong. Additionally, any assumptions I happen to have thought were a good idea, I question: "Is it really true that the largest square number less than n will be in the representation of MinLen(n)?"
Your code:
r1 = (int) Math.sqrt(n1);
r2 = (int) Math.pow(r1, 2);
embodies that assumption (a greedy assumption), but it is wrong, as you've clearly seen with the answer for MinLen(12).
Instead you want something more like this:
public ArrayList<Integer> minLen(int n)
{
// base case of recursion
if (n == 0)
return new ArrayList<Integer>();
ArrayList<Integer> best = null;
int bestInt = -1;
for (int i = 1; i*i <= n; ++i)
{
// Check what happens if we use i^2 as part of our representation
ArrayList<Integer> guess = minLen(n - i*i);
// If we haven't selected a 'best' yet (best == null)
// or if our new guess is better than the current choice (guess.size() < best.size())
// update our choice of best
if (best == null || guess.size() < best.size())
{
best = guess;
bestInt = i;
}
}
best.add(bestInt);
return best;
}
Then, once you have your list, you can sort it (no guarantees that it came in sorted order), and print it out the way you want.
Lastly, you may notice that for larger values of n (1000 may be too large) that you plug in to the above recursion, it will start going very slow. This is because we are constantly recalculating all the small subproblems - for example, we figure out MinLen(3) when we call MinLen(4), because 4 - 1^2 = 3. But we figure it out twice for MinLen(7) -> 3 = 7 - 2^2, but 3 also is 7 - 1^2 - 1^2 - 1^2 - 1^2. And it gets much worse the larger you go.
The solution to this, which lets you solve up to n = 1,000,000 or more, very quickly, is to use a technique called Memoization. This means that once we figure out MinLen(3), we save it somewhere, let's say a global location to make it easy. Then, whenever we would try to recalculate it, we check the global cache first to see if we already did it. If so, then we just use that, instead of redoing all the work.
import java.util.*;
class SquareRepresentation
{
private static HashMap<Integer, ArrayList<Integer>> cachedSolutions;
public static void main(String[] args)
{
cachedSolutions = new HashMap<Integer, ArrayList<Integer>>();
for (int j = 100000; j < 100001; ++j)
{
ArrayList<Integer> answer = minLen(j);
Collections.sort(answer);
Collections.reverse(answer);
for (int i = 0; i < answer.size(); ++i)
{
if (i != 0)
System.out.printf("+");
System.out.printf("%d^2", answer.get(i));
}
System.out.println();
}
}
public static ArrayList<Integer> minLen(int n)
{
// base case of recursion
if (n == 0)
return new ArrayList<Integer>();
// new base case: problem already solved once before
if (cachedSolutions.containsKey(n))
{
// It is a bit tricky though, because we need to be careful!
// See how below that we are modifying the 'guess' array we get in?
// That means we would modify our previous solutions! No good!
// So here we need to return a copy
ArrayList<Integer> ans = cachedSolutions.get(n);
ArrayList<Integer> copy = new ArrayList<Integer>();
for (int i: ans) copy.add(i);
return copy;
}
ArrayList<Integer> best = null;
int bestInt = -1;
// THIS IS WRONG, can you figure out why it doesn't work?:
// for (int i = 1; i*i <= n; ++i)
for (int i = (int)Math.sqrt(n); i >= 1; --i)
{
// Check what happens if we use i^2 as part of our representation
ArrayList<Integer> guess = minLen(n - i*i);
// If we haven't selected a 'best' yet (best == null)
// or if our new guess is better than the current choice (guess.size() < best.size())
// update our choice of best
if (best == null || guess.size() < best.size())
{
best = guess;
bestInt = i;
}
}
best.add(bestInt);
// check... not needed unless you coded wrong
int sum = 0;
for (int i = 0; i < best.size(); ++i)
{
sum += best.get(i) * best.get(i);
}
if (sum != n)
{
throw new RuntimeException(String.format("n = %d, sum=%d, arr=%s\n", n, sum, best));
}
// New step: Save the solution to the global cache
cachedSolutions.put(n, best);
// Same deal as before... if you don't return a copy, you end up modifying your previous solutions
//
ArrayList<Integer> copy = new ArrayList<Integer>();
for (int i: best) copy.add(i);
return copy;
}
}
It took my program around ~5s to run for n = 100,000. Clearly there is more to be done if we want it to be faster, and to solve for larger n. The main issue now is that in storing the entire list of results of previous answers, we use up a lot of memory. And all of that copying! There is more you can do, like storing only an integer and a pointer to the subproblem, but I'll let you do that.
And by the by, 1000 = 30^2 + 10^2.

Comparing Two Arrays & Get the Percent that Match - Java

Background: Very new at Java, have little understanding. Would prefer a "point in the right direction" with explanation, if possible, than a copy/paste answer without explanation. If I want to stop being a novice, I need to learn! :)
Anyway, my goal is, as simply as possible, to be given 2 arrays numberList and winningNumbers, compare them, and return the percentage that numberList matches winningNumbers. Both array lengths will always be 10.
I have no idea where to start. I have been googling and going at this for 2 hours. My idea is to write a for loop that compares each individually integer in a string to one in the other, but I am not sure how to do that, or if there is a simpler method. I have little knowledge of arrays, and the more I google the more confused I become.
So far the only thing I have is
public double getPercentThatMatch(int[] winningNumbers) {}
numberList is preset.

one way you could approach it is to:
1) convert both lists to sets.
2) subtract one from the other. ie if 4 are the same, the resulting set will have the 6 values not the same
3) 10 - (size of resulting set) * 100 = %

Here's a runnable example of how you would compare the two arrays of ints to get a percent match.
public class LotteryTicket {
int[] numberList;
LotteryTicket(int... numbers) {
numberList = numbers;
}
public int getPercentThatMatch(int[] winningNumbers) {
Arrays.sort(numberList);
Arrays.sort(winningNumbers);
int i = 0, n = 0, match = 0;
while (i < numberList.length && n < winningNumbers.length) {
if (numberList[i] < winningNumbers[n]) {
i++;
} else if (numberList[i] > winningNumbers[n]) {
n++;
} else {
match++;
i++;
n++;
}
}
return match * 100 / winningNumbers.length;
}
public static void main(String[] args)
{
int[] winningNumbers = { 12, 10, 4, 3, 2, 5, 6, 7, 9, 1 };
LotteryTicket ticket = new LotteryTicket(5, 2, 6, 7, 8, 4, 3, 1, 9, 0);
int percentMatching = ticket.getPercentThatMatch(winningNumbers);
System.out.println(percentMatching + "%");
}
}
Output:
80%

Since you wanted to be pointed in the right direction, rather than havving proper code, and assuming you want to use arrays to solve the problem, try to put something like this in your method:
(loop through arrayA){
(loop through arrayB){
if (current arrayA number is equal to current arrayB number){
then increase match counter by one, since this exists.
also break out of current arrayB loop. (Check next arrayA now.)
}
}
}
When done: return 100*matchCount/totalCount, as a double
So for every index in one array, you check against every other index of the other array. Increase a counter each time there's a match, and you'll be able to get a ratio of matches. If you use an integer as a counter, remember that division with integers acts funky, so you'd need to throw to a double:
double aDoubleNumber = (double) intNumber / anotherIntNumber

The problem would be easier if we consider them set. Let you have two set -
Set<Integer> s1 = //a HashSet of Integer;
Set<Integer> s2 = //a HashSet of Integer;
Now make a copy of s1 for example s11 and do the following thing -
s1.retainAll(s2);
Now s1 contains only element of both sets - that is the intersection.
After that you can easily calculate the percentage
Edit: You can convert the array to a set easily by using the following code snippet (I am assuming you have array of int) -
Set<Integer> s1 = new HashSet<Integer>(Arrays.asList(somePrimiteiveIntArray));
I think this trick will works for other primitive type also.
Hope this will help.
Thanks a lot.

I am going to attempt to beat a dead horse and explain the easiest (conceptual) way to approach this problem I will include some code but leave a lot up to interpretation.
You have two arrays so I would change the overall method to something like this:
public double getPercentage(int[] arrayA, int[] arrayB) {
double percentage=0;
for(/*go through the first array*/) {
for(/*go through second array*/) {
if(arrayA[i]==arrayB[j]) { /*note the different indices*/
percentage++; /*count how many times you have matching values*/
/* NOTE: This only works if you don't have repeating values in arrayA*/
}
}
}
return (percentage/arrayA.length)*100; /*return the amount of times over the length times 100*/
}
You are going to move through the first array with the first loop and the second array with the second loop. So you go through every value in arrayB for each value in arrayA to check.

In my approach I tried storing the winning numbers in a Hashset (one pass iteration, O(n) )
And when iterating on the numberList, I would check for presence of number in Hashset and if so, I will increment the counter. (one pass iteration, so O(n) )
The percentage is thus calculated by dividing the counter with size of array.
See if the sample code makes sense:
import java.util.HashSet;
public class Arraycomparison {
public static void main(String ... args){
int[] arr0 = {1,4,2,7,6,3,5,0,3,9,3,5,7};
int[] arr1 = {5,2,4,1,3,7,8,3,2,6,4,4,1};
HashSet set = new HashSet();
for(int j = 0; j < arr1.length; j++){
set.add(arr1[j]);
}
double counter = 0;
for(int i = 0; i < arr0.length; i++){
if(set.contains(arr0[i])){
counter++;
}
}
System.out.println("Match percentage between arrays : " + counter/arr0.length*100);
}
}

You should use List over array, because that's a convenient way, but with array:
public class Winner {
public static void main(String... args) {
double result = getPercentThatMatch(new int[]{1,2,3,4,5}, new int[]{2,3,4,5,6});
System.out.println("Result="+result+"%");
}
public static double getPercentThatMatch(int[] winningNumbers,
int[] numberList) { // it is confusing to call an array as List
int match = 0;
for (int win : winningNumbers) {
for (int my : numberList ){
if (win == my){
System.out.println(win + " == " + my);
match++;
}
}
}
int max = winningNumbers.length; // assume that same length
System.out.println("max:"+max);
System.out.println("match:"+match);
double devide = match / max; // it won't be good, because the result will be intm so Java will trunc it!
System.out.println("int value:"+devide);
devide = (double) match / max; // you need to cast to float or double
System.out.println("float value:"+devide);
double percent = devide * 100;
return percent;
}
}
Hope this helps. ;)

//For unique elements
getpercentage(arr1, arr2){
res = arr1.filter(element=>arr2.includes(element))
return res.lenght/arr2.lenght * 100;
}
//For duplicate elements
getpercentage(arr1, arr2){
const setA = Set(arr1);
const setB = Set(arr2);
Let res = [ ];
for(let i of setB){
if(setA.has(i)){
res.push(i);
}
}
return res.lenght/setA.size* 100;

find a solution to subset sum using dynamic programming

What I want to do
I want to find a subset of an array that sums to a target T. I also want to use to a dynamic programming approach (and a bottom-up solution at that) to do this.
What I currently have
Currently I only found a way to see if amongst all subsets of size N, whether or not there is at least one subset that has the desired sum. See code below.
public boolean solve(int[] numbers, int target) {
//Safeguard against invalid parameters
if ((target < 0) || (sum(numbers) < target)){
return false;
}
boolean [][] table = new boolean [target + 1] [numbers.length + 1] ;
for (int i = 0; i <= numbers.length; ++i) {
table[0][i] = true;
}
/* Base cases have been covered.
* Now look set subsets [1..n][target] to be true or false.
* n represents the number of elements from the start that have a subset
* that sums to target
*/
for (int i = 1; i <= target; ++i){
for (int j = 1; j <= numbers.length; ++j){
/* Mark index j as one of the numbers in the array
* which is part of the solution with the given subtarget */
table [i][j] = table[i][j-1];
if (i >= numbers[j-1])
table[i][j] = table[i][j] || table[i - numbers[j-1]] [j-1];
}
}
return table[target][numbers.length];
}
Where I am stuck
Right now, I know if there is a solution, but I can't think of a way to actually output a solution.
I am not looking for anyone to provide me specific code, but pseudocode is welcome as are hints to how a solution may be saved.

The algorithm you provided can stay the same, you don't need to store anything else besides the DP-table table[][]. You just need an additional post-processing phase in which you step "backwards" through table[][] to get the solution set.
Just to recall:
You've computed the table table[i][j], which stores for every value 0<=i<=t(:=target) and every 0<=j<=n(:=numbers.length) whether there is a subset of numbers in numbers[0..j-1] that sum to i.
Consider the subset S corresponding to table[i][j] (, which is true). Note that:
The subset S contains the number numbers[j] only if table[ i-numbers[j] ][j-1] is true.
(Proof: recursively take the solution subset S' for table[ i-numbers[j] ][j-1], and add numbers[j])
On the other hand, this subset S does not contain the number numbers[j] only if table[ i-numbers[j] ][j-1] is false.
(Proof: assume S contains numbers[j], trow numbers[j] out of S, this implies table[ i-numbers[j] ][j-1], contradiction)
So to get the subset, simply use the above property to check whether numbers[n-1] is in the subset summing to t.
If so, recursively compute whether numbers[n-2] is in the subset summing to t-numbers[n-1],
else recursively compute whether numbers[n-2], is in the subset summing to t

Here are the two Java solutions for the subset sum problem.
First using Recursive Approach.
Second using Dynamic Programming Approach.
/*
Question: Given a set of non-negative integers, and a value sum, determine if there is a subset of the given set
with sum equal to given sum.
Examples: set[] = {3, 34, 4, 12, 5, 2}, sum = 9
Output: True //There is a subset (4, 5) with sum 9.
Let isSubSetSum(int set[], int n, int sum) be the function to find whether there is a subset of set[] with
sum equal to sum. n is the number of elements in set[].
*/
package SubsetSumProblem;
import java.util.Scanner;
public class UsingResursiveAndDPApproach {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
try{
System.out.println("Enter the number of elements in the array");
int n =in.nextInt();
System.out.println("Enter the elements of the array");
int[] a=new int[n];
for(int i=0;i<n;i++)
a[i]=in.nextInt();
System.out.println("Enter the sum, which you need to find");
int sum = in.nextInt();
System.out.println("Using recursion, the result is: "+usingRecursion(a,a.length,sum));
System.out.println("Using Dynamic Programming, the result is: "+usingDP(a,sum));
}
finally{
in.close();
}
}
private static boolean usingRecursion(int[] a,int length, int sum) {
// 1. Base Cases
if(sum==0)
return true;
if(length==0 && sum!=0)
return false;
// 2. To avoid unnecessary steps, we will optimize the recursion method by avoiding
// recursive calls to areas where we are definite that we can SAFELY ignore the case since
// the SOLUTION does not exist there.
// If last element is greater than sum, then ignore it
if(a[a.length-1]>sum)
return usingRecursion(a,length-1,sum);
// 3. This is the recursion step where we will call the method again and again
/* else, check if sum can be obtained by any of the following
(a) including the last element
(b) excluding the last element */
return (usingRecursion(a, length-1, sum-a[length-1])|| usingRecursion(a, length-1, sum));
}
/*
Analysis:
Time Complexity = O(2^n)
Space Complexity = // Don't know
*/
private static boolean usingDP(int[] a, int sum) {
// using boolean matrix for DP
boolean dp[][] = new boolean[a.length+1][sum+1]; // +1 in row and column
// if the length of the array is variable (and sum is 0) then fill TRUE, since the SUM=0
for(int row=0;row<dp.length;row++){
dp[row][0] = true; // NOTE: dp[length=VARIABLE][sum=0], thus we satisfy the condition where length is VARIABLE
// and the SUM=0
}
// if the SUM is variable and length is 0 then FALSE, since (sum=variable && length=0)
for(int column=1;column<dp[0].length;column++){
dp[0][column] = false; // NOTE: dp[length=0][sum=VARIABLE], thus we satisfy the condition where
// (length=0 && sum=variable)
}
for(int i=1;i<dp.length;i++){
for(int j=1;j<dp[0].length;j++){
/* Check if sum can be obtained by any of the following
(a) including the last element
(b) excluding the last element */
// VERY VERY IMP: This is same as "excluding the last element" which is represented in DP
dp[i][j] = dp[i-1][j]; // the current position[i][j] would be same as previous position.
// the previous position means that SUM is ACHIEVED OR NOT-ACHIEVED
// int the previous position then it will ofcourse be ACHIEVED or NOT-ACHIEVED
// in the current position.
// VERY VERY IMP: This is same as "including the last element" which is represented in DP
// if the column[ sum is represented in column of the matrix i.e this sum exist] > = sum-a[last_index]
// then decrease the sum
if(j>=a[i-1]) // i.e sum >= array[last index element]. If it is true then include this last element by
// deducting it from the total sum
dp[i][j] = dp[i][j] || dp[i-1][j-a[i-1]]; // VERY VERY IMP NOTE: Here dp[i][j] on R.H.S represent
// dp[i-1][j] which we have assigned in the previous step
}
}
return dp[a.length][sum];
}
/*
Analysis:
Time Complexity = O(a.length*sum)
Space Complexity = O(a.length*sum)
*/
}

Here is my solution is an iterative dp, but with only one dimension: Hope it can help you.
#include <iostream>
#include <cstring>
using namespace std;
const int maxN=1000;
int memo[maxN];
int pi[maxN];
int main(){
int a[]={7,8,5,1,4};
memset(memo,-1,sizeof memo);
memset(pi,-1,sizeof pi);
int n;
cin>>n;
memo[0]=0;
pi[0]=0;
for(int i=0;i<(int)sizeof(a)/4;i++){
for(int num=n;num>=0;num--){
if(num-a[i]>=0 and memo[num-a[i]]!=-1 and (memo[num]==-1 or memo[num]>1+memo[num-a[i]])){
memo[num]=1+memo[num-a[i]];
pi[num]=num-a[i];
}
}
}
int N=n;
while(N!=0){
cout<<N-pi[N]<<" ";
N=pi[N];
}
cout<<endl;
cout<<memo[n]<<endl;
return 0;
}