# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 // c++ program to count cross line in array #include 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; }

## Java

 // 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

## C#

 // 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

## PHP

 = 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. ?>

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++

 // c++ program to count cross line in array #include 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; }

## Java

 // 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

Output:

>10

Time complexity: O(nlogn)

Reference: https://www.careercup.com/question?id=5669565693427712
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Improved By : vt_m, Sam007, princiraj1992

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