Interview Question for Software Engineers


Country: United States
Interview Type: In-Person




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package com.home.careercup;

public class InsertionSortForInversionCounting {

    public static void main(String[] args) {
        final int a [] = {3,3,3,3};
        System.out.println(inversionCount(a));
    }

    /*
    find number of positions any element moves forward during
    sorting using selection sort.
    This is not the most optimal.
    Other faster  ( O(n log (n)) methods are possible ( hint: divide and conquer)
     */
    static int inversionCount(int arr[]) {
        int n = arr.length;
        int forwardMoves = 0;
        for (int i = 1; i < n; ++i) {
            int key = arr[i];
            int j = i - 1;

            while (j >= 0 && arr[j] > key) {
                arr[j + 1] = arr[j];
                j = j - 1;
                forwardMoves++;
            }
            arr[j + 1] = key;
        }
        return forwardMoves;
    }
}

- Makarand September 12, 2017 | Flag Reply
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x ==> value at index
i ==> index
abs(i-x) ==> each index off by this much value when the numbers are sorted

(Sum of all abs(i-x) )/2 ==> number of intersections (divide by 2 bcos each intersection is counted twice)

- GP September 12, 2017 | Flag Reply
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#include "stdafx.h"
#include <algorithm>
#include <unordered_map>
#include <vector>
using namespace std;

int count_crossings_after_sort(vector<int> in) // no depes aloud !
{
	unordered_map<int, int> value_to_old_index;

	for (int i = 0; i < in.size(); i++)
	{
		value_to_old_index[in[i]] = i;
	}

	sort(in.begin(), in.end());

	int c = 0;

	for (int i = 0; i < in.size() - 1; i++)
	{
		for (int j = i + 1; j < in.size(); j++)
		{
			if (value_to_old_index[in[i]] > value_to_old_index[in[j]])
			{
				c++;
			}
		}
	}


	return c;
}



int _tmain(int argc, _TCHAR* argv[])
{
	vector<int> in = { 3, 5, 2, 1 };

	int r = count_crossings_after_sort(in);
	return 0;
}

- milota@mindgrip.com November 21, 2017 | Flag Reply


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