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++
// 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; } |
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 |
Python3
# Python3 program to count # cross line in array def countCrossLine(arr, n): count_crossline = 0; i, key, j = 0, 0, 0; for i in range(1, n): 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 += 1; arr[j + 1] = key; return count_crossline; # Driver Code if __name__ == '__main__': arr = [4, 3, 1, 2]; n = len(arr); print(countCrossLine(arr, n)); # This code is contributed by PrinciRaj1992 |
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
<?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. ?> |
Javascript
<script> // Javascript program to count cross line in array // function return count of cross line in an array function countCrossLine( arr, n) { let count_crossline = 0; let 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 let arr = [ 4, 3, 1, 2 ]; let n = arr.length; document.write(countCrossLine(arr, n) + "</br"); </script> |
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
Below is the implementation of above idea.
C++
// 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; } |
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 |
Python3
# Python program to count cross line in array count_crossline=0 # Merges two subarrays of arr[]. # First subarray is arr[l..m] # Second subarray is arr[m+1..r] def merge(arr,l,m,r): global count_crossline n1=m-l+1 n2=r-m # create temp arrays L=[0]*n1 R=[0]*n2 # Copy data to temp arrays L[] and R[] for i in range(n1): L[i]=arr[l+i] for j in range(n2): 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 and j<n2): if(L[i]<=R[j]): arr[k]=L[i] i+=1 else: arr[k]=R[j] # ==================================== # ======= MAIN PORTION OF CODE ====== # =================================== # add all line which is cross by current element count_crossline+=(n1-i) j+=1 k+=1 # Copy the remaining elements of L[], if there # are any while(i<n1): arr[k]=L[i] i+=1 k+=1 # Copy the remaining elements of R[], if there # are any while(j<n2): arr[k]=R[j] j+=1 k+=1# l is for left index and r is right index of the # sub-array of arr to be sorted def mergeSort(arr,l,r): if(l<r): # Same as (l+r)/2, but avoids overflow for # large l and h 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 def countCrossLine(arr,n): mergeSort(arr,0,n-1) return count_crossline # driver program to test above function arr=[ 12, 11, 13, 5, 6, 7 ] n=len(arr) print(countCrossLine(arr,n)) # This code is contributed by Pushpesh Raj. |
C#
// C# program to count cross line in array using System; 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; Console.WriteLine(countCrossLine(arr, n)); } } // This code is contributed by PrinciRaj1992 |
Javascript
<script> // Javascript program to count cross line in array let count_crossline = 0; // Merges two subarrays of arr[]. // First subarray is arr[l..m] // Second subarray is arr[m+1..r] function merge(arr, l, m, r) { let i, j, k; let n1 = m - l + 1; let n2 = r - m; /* create temp arrays */ let L = []; let R = []; /* 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 */ function mergeSort(arr, l, r) { if (l < r) { // Same as (l+r)/2, but avoids overflow for // large l and h let m = l + Math.floor((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 function countCrossLine(arr, n) { mergeSort(arr, 0, n - 1); document.write(count_crossline); } // Driver code let arr = [12, 11, 13, 5, 6, 7]; let n = arr.length; countCrossLine(arr, n); // This code is contributed by code_hunt. </script> |
Output:
10
Time complexity: O(nlogn)
Auxiliary Space: O(n)
Reference: https://www.careercup.com/question?id=5669565693427712
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