Interview Question

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of 0 vote

- Each number in the array can only be used once
- X is <= length of the array

This can be solved with a costly backtracking algorithm.

public static ArrayList<int[]> getSumPairs(int[] arr, int x){
	if(arr == null){
		throw new NullPointerException();
	if(x < 2){
		throw new IllegalArgumentException();
	Worker worker = new Worker(arr, x);
	return worker.getResults();

static class Worker{
	int[] arr;
	int[] vals;
	int limit;
	int index;
	int sum;
	ArrayList<int[]> results;

	Worker(int[] arr, int x){
		this.arr = arr;
		this.limit = x;
		this.vals = new int[x];
		this.results = new ArrayList<int[]>();

	void execute(){

	void executeRecur(int arrPos){
		if(this.index > 2 && this.sum % this.limit == 0){
			int[] pair = new int[this.index];
			for(int i = 0; i < this.index; i++){
				pair[i] = this.vals[i];
		if(this.index == this.limit){
		if(arrPos == this.arr.length){
		int localPosition = this.index;
		for(int i = arrPos; i < this.arr.length; i++){
			this.vals[localPosition] = this.arr[i];

	ArrayList<int[]> getResults(){
		return this.results;

- zortlord December 31, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
of 0 vote

First of all, the example output in the question is wrong. Here's the correct output
[4 6 8] [7 6 8] [9 4 8]
[3 4 8] [4 8] [9 7 8]
[3 7 8] [7 8] [3 9 6]
[9 6] [3 6] [3 9]
No of groups: 12

OK, let's analyze the problem. The silly part of the problem is that we have to find and report ALL
the groups. That yields to non-polynomial time complexity algorithm in worst case.
If we had to find any group with a given criteria or prove that no such group exists or solve optimization problem
(e.g. find the group of a maximum length divisible by X) then we could provide a nice polynomial (or as some people say pseudo-polynomial) time
algorithm based on dynamic programming with the complexity O(X*N).
For example, if we want to maximise the length of the group, we could use the following recursive relationship for DP:
dp[i][s] = MAX( dp[i-1][r], dp[i-1][(x+s-arr[i-1])%x] + 1)
where dp[i][s] holds the maximum length of subset formed from the first i elements of the array with a sum s modulo x.
Note, that in my solution all sums are computed using modulo X. so s is a remainder of a division of a sum to X.
dp[arr.size()][0] will contain the length of a largest subset with a sum divisible by x.

Ok, let's solve the original problem. We want to print all groups efficiently and try to keep running time (pseudo)polynomial
if there is a few groups to report. The algorithm below is a hybridization of a dynamic programming with a
search with pruning and backtracking. dp[i][s] relationship mentioned above is used to avoid recursing into the costly branches
which don't lead to printable results.
Space complexity O(X*N), time complexity O(X*(N+k)) where k is the # of groups which will be found and printed.
If no groups exist with the given criteria or O(k) == O(N) then algorithm terminates on O(X*N) time.
If O(k) = O(N^c) for some constant c the algorithm terminates in polynomial time O(X*N^c)


class FindGroupsWithSumDivisibleByX
	int x;
	const vector<int>& arr;
	int count;

	// dp[i][j] holds maximum length of a subsequence of a first i elements of x
	// which sum to j modulo X 
	vector<vector<int>> dp;
	FindGroupsWithSumDivisibleByX(const vector<int>& arr, int x): arr(arr), x(x)
		dp.resize(arr.size() + 1, vector<int>(x, INT_MIN));

	void PrintAllGroups()
		count = 0;
		vector<int> output(arr.size());
		Backtrack(output, 0, arr.size(), 0);
		cout << "No of groups: " << count << endl;

	int Backtrack(vector<int>& output, int pos, int i, int sumModuloX)
		if (i == 0) {
			PrintGroup(output, pos, sumModuloX);
			return 0;
		int& cache = dp[i][sumModuloX];
		if (cache != INT_MIN) {
			if (pos >= x || cache == 0 || (pos == 0 && cache == 1)) {
				// print accumulated group (if matches the criteria)
				PrintGroup(output, pos, sumModuloX);
				// pruning the current recursive search branch
				return cache;
		output[pos] = arr[i - 1];
		// case a: adding arr[i-1] to the group
		int a = Backtrack(output, pos + 1, i - 1, (x + sumModuloX - (arr[i - 1] % x)) % x) + 1;
		// case b: not adding arr[i-1] to the group
		int b = Backtrack(output, pos, i - 1, sumModuloX);
		cache = max(a, b);
		return cache;

	void PrintGroup(vector<int>& output, int pos, int sumModuloX)
		if (sumModuloX == 0 && pos > 1 && pos <= x) {
			for (int j = pos - 1; j >= 0; j--)
				cout << output[j] << " ";
			cout << endl;

Some notes:
1) negative numbers: the code supports negative numbers, however according to C++ standard the sign of the % result for negative param is
implementation defined.
As written above, the code will work on mainstream compilers but if you care about portability
you have to replace arr[i]%x to abs(arr[i%x])*sgn(arr[i]).
2) space complexity can be optimized further: technically the program uses only 2 bits of the values stored in dp array,
so the storage can be reduced. This can be even reduced to a single bit-matrix without changing the asymptotic complexity
but at the cost of larger constants
3) I also took an assumption that numbers in array are unique. If this assumption isn't correct we need to also take care to avoid reporting duplicated groups unless this is permitted

- 0xF4 January 01, 2015 | Flag Reply

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