Given an array arr[] of size N . The task is to find smaller elements on the right side and greater elements on the left side for each element arr[i] in the given array.
Examples:
Input: arr = {12, 1, 2, 3, 0, 11, 4}
Output:
Smaller right: 6 1 1 1 0 1 0
Greater left: 0 1 1 1 4 1 2
Input: arr = {5, 4, 3, 2, 1}
Output:
Smaller right: 4 3 2 1 0
Greater left: 0 1 2 3 4
Input: arr = {1, 2, 3, 4, 5}
Output:
Smaller right: 0 0 0 0 0
Greater left: 0 0 0 0 0
Approach: We have already discussed the implementation to count smaller elements on the right side in this post. Here, we will use Binary Indexed Tree to count smaller elements on the right side and greater elements on the left side for each element in the array. First, traverse the array from right to left and find smaller elements on the right side as suggested in the previous post. Then reset the BIT array and traverse the array from left to right and find greater elements on the left side.
Below is the implementation of the above approach:
- JAVA
// Java implementation of the approach
import java.io.*;
import java.util.*;
class GFG{
// Function to return the sum of
// arr[0..index]. This function
// assumes that the array is
// preprocessed and partial sums
// of array elements are stored in BITree[]
public static int getSum( int BITree[], int index)
{
// Initialize result
int sum = 0 ;
// Traverse ancestors of BITree[index]
while (index > 0 )
{
// Add current element of BITree to sum
sum += BITree[index];
// Move index to parent node in getSum View
index -= index & (-index);
}
return sum;
}
// Updates a node in Binary Index Tree
// (BITree) at given index in BITree.
// The given value 'val' is added to BITree[i]
// and all of its ancestors in tree.
public static void updateBIT( int BITree[], int n,
int index, int val)
{
// Traverse all ancestors and add 'val'
while (index <= n)
{
// Add 'val' to current node of BI Tree
BITree[index] += val;
// Update index to that of parent
// in update View
index += index & (-index);
}
}
// Converts an array to an array with values
// from 1 to n and relative order of smaller
// and greater elements remains same.
// For example, {7, -90, 100, 1} is converted to
// {3, 1, 4, 2 }
public static void convert( int arr[], int n)
{
// Create a copy of arrp[] in temp
// and sort the temp array in
// increasing order
int [] temp = new int [n];
for ( int i = 0 ; i < n; i++)
temp[i] = arr[i];
Arrays.sort(temp);
// Traverse all array elements
for ( int i = 0 ; i < n; i++)
{
// Arrays.binarySearch() returns index
// to the first element greater than
// or equal to arr[i]
arr[i] = Arrays.binarySearch(temp, arr[i]) + 1 ;
}
}
// Function to find smaller_right array
public static void findElements( int arr[], int n)
{
// Convert arr[] to an array with values
// from 1 to n and relative order of smaller
// and greater elements remains same. For
// example, {7, -90, 100, 1} is converted to
// {3, 1, 4, 2 }
convert(arr, n);
// Create a BIT with size equal to
// maxElement+1 (Extra one is used
// so that elements can be directly be
// used as index)
int [] BIT = new int [n + 1 ];
for ( int i = 1 ; i <= n; i++)
BIT[i] = 0 ;
// To store smaller elements in right side
// and greater elements on left side
int [] smaller_right = new int [n];
int [] greater_left = new int [n];
// Traverse all elements from right.
for ( int i = n - 1 ; i >= 0 ; i--)
{
// Get count of elements smaller than arr[i]
smaller_right[i] = getSum(BIT, arr[i] - 1 );
// Add current element to BIT
updateBIT(BIT, n, arr[i], 1 );
}
System.out.print( "Smaller right: " );
// Print smaller_right array
for ( int i = 0 ; i < n; i++)
System.out.print(smaller_right[i] + " " );
System.out.println();
for ( int i = 1 ; i <= n; i++)
BIT[i] = 0 ;
// Find all left side greater elements
for ( int i = 0 ; i < n; i++)
{
// Get count of elements greater than arr[i]
greater_left[i] = i - getSum(BIT, arr[i]);
// Add current element to BIT
updateBIT(BIT, n, arr[i], 1 );
}
System.out.print( "Greater left: " );
// Print greater_left array
for ( int i = 0 ; i < n; i++)
System.out.print(greater_left[i] + " " );
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 12 , 1 , 2 , 3 , 0 , 11 , 4 };
int n = arr.length;
// Function call
findElements(arr, n);
}
}
Output:
Smaller right: 6 1 1 1 0 1 0
Greater left: 0 1 1 1 4 1 2