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package com.interview.array;

/**

* Date 12/29/15

* @author Tushar Roy

*

* Given input array find number of inversions where i < j < k and input[i] > input[j] > input[k]

*

* http://www.geeksforgeeks.org/count-inversions-of-size-three-in-a-give-array/

*/

public class CountInversionOfSize3 {

/**

* Time complexity of this method is O(n^2)

* Space complexity is O(1)

*/

public int findInversions(int input[]) {

int inversion = 0;

for (int i = 1; i < input.length - 1 ; i++) {

int larger = 0;

for (int k = 0; k < i; k++) {

if (input[k] > input[i]) {

larger++;

}

}

int smaller = 0;

for (int k = i+1; k < input.length; k++) {

if (input[k] < input[i]) {

smaller++;

}

}

inversion += smaller*larger;

}

return inversion;

}

public static void main(String args[]) {

int input[] = {9, 6, 4, 5, 8};

CountInversionOfSize3 ci = new CountInversionOfSize3();

System.out.print(ci.findInversions(input));

}

}