Counting cross lines in an array

Given an unsorted array of distinct elements. Task is to count number of cross lines formed in an array elements after sorting the array elements.
Note: Draw a line between same array elements before sorting and after sorting the array elements.

Examples :

Input :  arr[] = { 3, 2, 1, 4, 5 }
Output : 3
      before sort: 3  2  1  4  5
                    \ | /   |  | 
                     \|/    |  |
                    / | \   |  |
      After sort : 1  2  3  4  5 
      line (1 to 1) cross line (2 to 2)
      line (1 to 1) cross line (3 to 3)
      line (2 to 2) cross line (3 to 3)
Note: the line between two 4s and the line 
between two 5s don't cross any other lines; 

Input : arr[] = { 5, 4, 3, 1 }
Output : 6



Simple solution of this problem is based on the insertion sort. we simply pick each array elements one-by-one and try to find it’s proper position in the sorted array.during finding it’s appropriate position of an element we have to cross all the element_line whose value is greater than current element.

Below is the implementation of above idea :

C++

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// c++ program to count cross line in array
#include <bits/stdc++.h>
using namespace std;
  
// function return count of cross line in an array
int countCrossLine(int arr[], int n)
{
    int count_crossline = 0;
    int i, key, j;
    for (i = 1; i < n; i++) {
        key = arr[i];
        j = i - 1;
  
        /* Move elements of arr[0..i-1], that are
          greater than key, to one position ahead
          of their current position */
        while (j >= 0 && arr[j] > key) {
            arr[j + 1] = arr[j];
            j = j - 1;
  
            // increment cross line by one
            count_crossline++;
        }
        arr[j + 1] = key;
    }
    return count_crossline;
}
  
// driver program to test above function
int main()
{
    int arr[] = { 4, 3, 1, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << countCrossLine(arr, n) << endl;
    return 0;
}

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Java

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// Java program to count 
// cross line in array
  
class GFG 
{
    static int countCrossLine(int arr[], 
                              int n)
    {
        int count_crossline = 0;
        int i, key, j;
        for (i = 1; i < n; i++)
        {
            key = arr[i];
            j = i - 1;
      
            // Move elements of arr[0..i-1], 
            // that are greater than key, 
            // to one position ahead of 
            // their current position 
            while (j >= 0 && arr[j] > key) 
            {
                arr[j + 1] = arr[j];
                j = j - 1;
      
                // increment cross
                // line by one
                count_crossline++;
            }
            arr[j + 1] = key;
        }
          
        return count_crossline;
    
      
    // Driver Code
    public static void main(String args[]) 
    {
        int arr[] = new int[]{ 4, 3, 1, 2 };
        int n = arr.length;
        System.out.print(countCrossLine(arr, n));
    }
}
  
// This code is contributed by Sam007

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C#

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// C# program to count cross line in array
using System;
  
class GFG {
  
    // function return count of cross line
    // in an array
    static int countCrossLine(int []arr, int n)
    {
        int count_crossline = 0;
        int i, key, j;
        for (i = 1; i < n; i++) {
            key = arr[i];
            j = i - 1;
      
            /* Move elements of arr[0..i-1], 
            that are greater than key, to one
            position ahead of their current
            position */
            while (j >= 0 && arr[j] > key) {
                arr[j + 1] = arr[j];
                j = j - 1;
      
                // increment cross line by one
                count_crossline++;
            }
            arr[j + 1] = key;
        }
          
        return count_crossline;
    
      
    // Driver code
    public static void Main()
    {
        int []arr = new int[]{ 4, 3, 1, 2 };
        int n = arr.Length;
        Console.Write(countCrossLine(arr, n));
    }
}
  
// This code is contributed by Sam007

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PHP

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<?php
// PHP program to count 
// cross line in array
  
// Function return count 
// of cross line in an array
function countCrossLine($arr, $n)
{
      
    $count_crossline = 0;
    $i; $key; $j;
    for ($i = 1; $i < $n; $i++)
    {
        $key = $arr[$i];
        $j = $i - 1;
  
        /* Move elements of arr[0..i-1], that are
           greater than key, to one position ahead
           of their current position */
        while ($j >= 0 and $arr[$j] > $key
        {
            $arr[$j + 1] = $arr[$j];
            $j = $j - 1;
  
            // increment cross line by one
            $count_crossline++;
        }
        $arr[$j + 1] = $key;
    }
    return $count_crossline;
}
  
    // Driver Code
    $arr = array( 4, 3, 1, 2 );
    $n = count($arr);
    echo countCrossLine($arr, $n);
      
// This code is contributed by anuj_67.
?>

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Output:

5

Time complexity: O(n2)
Auxiliary space: O(1)

Efficient solution based on the merge sortand inversions count.

   lets we have arr[] { 2, 4, 1, 3 }
                         \   \ /   /
                          \  / \ /
                          / \  / \
   After sort   arr[] { 1, 2, 3, 4 }
   and here all inversion are (2, 1), (4, 1), (4, 3)
   that mean line 1 : cross line 4, 2 
             line 2 : cross line 1 
             line 4 : cross line 3, 1
             line 3 : cross line 3
 so total unique cross_line are: 3  

During mer
Below is the implementation of above idea.

C++

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// c++ program to count cross line in array
#include <bits/stdc++.h>
using namespace std;
  
// Merges two subarrays of arr[].
// First subarray is arr[l..m]
// Second subarray is arr[m+1..r]
void merge(int arr[], int l, int m, int r, int* count_crossline)
{
    int i, j, k;
    int n1 = m - l + 1;
    int n2 = r - m;
  
    /* create temp arrays */
    int L[n1], R[n2];
  
    /* Copy data to temp arrays L[] and R[] */
    for (i = 0; i < n1; i++)
        L[i] = arr[l + i];
    for (j = 0; j < n2; j++)
        R[j] = arr[m + 1 + j];
  
    /* Merge the temp arrays back into arr[l..r]*/
    i = 0; // Initial index of first subarray
    j = 0; // Initial index of second subarray
    k = l; // Initial index of merged subarray
    while (i < n1 && j < n2) {
        if (L[i] <= R[j]) {
            arr[k] = L[i];
            i++;
        }
        else {
            arr[k] = R[j];
  
            //====================================//
            //======= MAIN PORTION OF CODE ======//
            //===================================//
            // add all line which is cross by current element
            *count_crossline += (n1 - i);
            j++;
        }
        k++;
    }
  
    /* Copy the remaining elements of L[], if there
    are any */
    while (i < n1) {
        arr[k] = L[i];
        i++;
        k++;
    }
  
    /* Copy the remaining elements of R[], if there
    are any */
    while (j < n2) {
        arr[k] = R[j];
        j++;
        k++;
    }
}
  
/* l is for left index and r is right index of the
sub-array of arr to be sorted */
void mergeSort(int arr[], int l, int r, int* count_crossline)
{
    if (l < r) {
  
        // Same as (l+r)/2, but avoids overflow for
        // large l and h
        int m = l + (r - l) / 2;
  
        // Sort first and second halves
        mergeSort(arr, l, m, count_crossline);
        mergeSort(arr, m + 1, r, count_crossline);
  
        merge(arr, l, m, r, count_crossline);
    }
}
  
// function return count of cross line in an array
int countCrossLine(int arr[], int n)
{
    int count_crossline = 0;
    mergeSort(arr, 0, n - 1, &count_crossline);
  
    return count_crossline;
}
  
// driver program to test above function
int main()
{
    int arr[] = { 12, 11, 13, 5, 6, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << countCrossLine(arr, n) << endl;
    return 0;
}

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Java

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// Java program to count cross line in array
import java.util.*;
  
class GFG 
{
    static int count_crossline;
      
    // Merges two subarrays of arr[].
    // First subarray is arr[l..m]
    // Second subarray is arr[m+1..r]
    static void merge(int arr[], int l, int m, int r)
    {
        int i, j, k;
        int n1 = m - l + 1;
        int n2 = r - m;
  
        /* create temp arrays */
        int[] L = new int[n1];
        int[] R = new int[n2];
  
        /* Copy data to temp arrays L[] and R[] */
        for (i = 0; i < n1; i++)
        {
            L[i] = arr[l + i];
        }
        for (j = 0; j < n2; j++) 
        {
            R[j] = arr[m + 1 + j];
        }
  
        /* Merge the temp arrays back into arr[l..r]*/
        i = 0; // Initial index of first subarray
        j = 0; // Initial index of second subarray
        k = l; // Initial index of merged subarray
        while (i < n1 && j < n2) 
        {
            if (L[i] <= R[j])
            {
                arr[k] = L[i];
                i++;
            
            else 
            {
                arr[k] = R[j];
  
                //====================================//
                //======= MAIN PORTION OF CODE ======//
                //===================================//
                // add all line which is cross by current element
                count_crossline += (n1 - i);
                j++;
            }
            k++;
        }
  
        /* Copy the remaining elements of L[],
        if there are any */
        while (i < n1)
        {
            arr[k] = L[i];
            i++;
            k++;
        }
  
        /* Copy the remaining elements of R[],
        if there are any */
        while (j < n2) 
        {
            arr[k] = R[j];
            j++;
            k++;
        }
    }
  
    /* l is for left index and r is right index of the
    sub-array of arr to be sorted */
    static void mergeSort(int arr[], int l, int r)
    {
        if (l < r)
        {
  
            // Same as (l+r)/2, but avoids overflow for
            // large l and h
            int m = l + (r - l) / 2;
  
            // Sort first and second halves
            mergeSort(arr, l, m);
            mergeSort(arr, m + 1, r);
  
            merge(arr, l, m, r);
        }
    }
  
    // function return count of cross line in an array
    static int countCrossLine(int arr[], int n) 
    {
        mergeSort(arr, 0, n - 1);
  
        return count_crossline;
    }
  
    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = {12, 11, 13, 5, 6, 7};
        int n = arr.length;
        System.out.println(countCrossLine(arr, n));
    }
}
  
// This code is contributed by PrinciRaj1992 

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Output:

>10

Time complexity: O(nlogn)

Reference: https://www.careercup.com/question?id=5669565693427712
This article is contributed by Nishant Singh . If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : vt_m, Sam007, princiraj1992