给定大小为N的数组arr [] 。任务是为给定数组中的每个元素arr [i]在右侧找到较小的元素，在左侧找到较大的元素。
例子：
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
先决条件：使用BIT计算数组中的求反
方法：我们已经在本文中讨论了在右侧计算较小元素的实现。在这里，我们将使用二进制索引树为数组中的每个元素计算右侧的较小元素和左侧的较大元素。首先，按照上一篇文章中的建议，从右向左遍历数组，并在右侧找到较小的元素。然后重置BIT数组，并从左到右遍历该数组，并在左侧找到更大的元素。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// 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[]
int getSum(int BITree[], int index)
{
int sum = 0; // Initialize result
// 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.
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 }
void convert(int arr[], int n)
{
// Create a copy of arrp[] in temp and sort the temp array
// in increasing order
int temp[n];
for (int i = 0; i < n; i++)
temp[i] = arr[i];
sort(temp, temp + n);
// Traverse all array elements
for (int i = 0; i < n; i++) {
// lower_bound() Returns pointer to the first element
// greater than or equal to arr[i]
arr[i] = lower_bound(temp, temp + n, arr[i]) - temp + 1;
}
}
// Function to find smaller_right array
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[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[n], greater_left[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);
}
cout << "Smaller right: ";
// Print smaller_right array
for (int i = 0; i < n; i++)
cout << smaller_right[i] << " ";
cout << endl;
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);
}
cout << "Greater left: ";
// Print greater_left array
for (int i = 0; i < n; i++)
cout << greater_left[i] << " ";
}
// Driver code
int main()
{
int arr[] = { 12, 1, 2, 3, 0, 11, 4 };
int n = sizeof(arr) / sizeof(arr[0]);
// Function call
findElements(arr, n);
return 0;
} 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);
}
}
// This code is contributed by kunalsg18elecPython3
# Python3 implementation of the approach
from bisect import bisect_left as lower_bound
# 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[]
def getSum(BITree, index):
# Initialize result
s = 0
# Traverse ancestors of BITree[index]
while index > 0:
# Add current element of BITree to sum
s += BITree[index]
# Move index to parent node in getSum View
index -= index & (-index)
return s
# 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.
def updateBIT(BITree, n, index, 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 }
def convert(arr, n):
# Create a copy of arrp[] in temp and
# sort the temp array in increasing order
temp = [0] * n
for i in range(n):
temp[i] = arr[i]
temp.sort()
# Traverse all array elements
for i in range(n):
# lower_bound() Returns pointer to the first element
# greater than or equal to arr[i]
arr[i] = lower_bound(temp, arr[i]) + 1
# Function to find smaller_right array
def findElements(arr, 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)
BIT = [0] * (n + 1)
# To store smaller elements in right side
# and greater elements on left side
smaller_right = [0] * n
greater_left = [0] * n
# Traverse all elements from right.
for i in range(n - 1, -1, -1):
# 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)
print("Smaller right:", end = " ")
for i in range(n):
print(smaller_right[i], end = " ")
print()
# Print smaller_right array
for i in range(1, n + 1):
BIT[i] = 0
# Find all left side greater elements
for i in range(n):
# 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)
print("Greater left:", end = " ")
# Print greater_left array
for i in range(n):
print(greater_left[i], end = " ")
print()
# Driver Code
if __name__ == "__main__":
arr = [12, 1, 2, 3, 0, 11, 4]
n = len(arr)
# Function call
findElements(arr, n)
# This code is contributed by
# sanjeev2552C#
// C# implementation of the
// above approach
using System;
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];
Array.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] = Array.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);
}
Console.Write("Smaller right: ");
// Print smaller_right array
for(int i = 0; i < n; i++)
Console.Write(smaller_right[i] + " ");
Console.WriteLine();
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);
}
Console.Write("Greater left: ");
// Print greater_left array
for(int i = 0; i < n; i++)
Console.Write(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);
}
}
// This code is contributed by Rajput-Ji输出：
Smaller right: 6 1 1 1 0 1 0
Greater left: 0 1 1 1 4 1 2
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