Dado un arreglo y un entero k, cuente todas las inversiones en todos los subarreglos de tamaño k.
Ejemplo:
Input : a[] = {7, 3, 2, 4, 1},
k = 3;
Output : 6
Explanation: subarrays of size 3 are -
{7, 3, 2}
{3, 2, 4}
{2, 4, 1}
and there inversion count are 3, 1, 2
respectively. So, total number of
inversions are 6.
Se recomienda encarecidamente consultar el recuento de inversión en una array utilizando BIT
El proceso de inversión de conteo es el mismo que en el artículo vinculado anteriormente. Primero, calculamos la inversión en los primeros k (k se le da el tamaño del subarreglo) elementos en el arreglo usando BIT. Ahora, después de esto, para cada subarreglo siguiente restamos la inversión del primer elemento del subarreglo anterior y sumamos las inversiones del siguiente elemento no incluido en el subarreglo anterior. Este proceso se repite hasta que se procesa el último subarreglo. En el ejemplo anterior, este algoritmo funciona así:
a[] = {7, 3, 2, 4, 1},
k = 3;
Inversion are counted for first subarray
A = {7, 3, 2} Let this be equal to invcount_A.
For counting the inversion of subarray B we
subtract the inversion of first element of A,
from invcount_A and add inversions of 4 (last
element of B) in the subarray B.
So, invcount_B = invcount_A - 2 + 0
= 3 - 2 + 0
= 1
Same process is repeated for next subarray
and sum of all inversion count is the answer.
Implementación:
C++
// C++ program to count inversions in all sub arrays
// of size k using Binary Indexed Tree
#include <bits/stdc++.h>
using namespace std;
// Returns 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;
}
}
// Returns inversion count of all subarray
// of size k in arr[0..n-1]
int getInvCount(int arr[], int k, int n)
{
int invcount = 0; // Initialize result
// 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;
// Get inversion count of first subarray
for (int i = k - 1; i >= 0; i--) {
// Get count of elements smaller than arr[i]
invcount += getSum(BIT, arr[i] - 1);
// Add current element to BIT
updateBIT(BIT, n, arr[i], 1);
}
// now calculate the inversion of all other subarrays
int ans = invcount;
int i = 0, j = k, icnt = 0, jcnt = 0;
while (j <= n - 1) {
// calculate value of inversion count of first element
// in previous subarray
icnt = getSum(BIT, arr[i] - 1);
updateBIT(BIT, n, arr[i], -1);
// calculating inversion count of last element in the
// current subarray
jcnt = getSum(BIT, n) - getSum(BIT, arr[j]);
updateBIT(BIT, n, arr[j], 1);
// calculating inversion count of current subarray
invcount = invcount - icnt + jcnt;
// adding current inversion to the answer
ans = ans + invcount;
i++, j++;
}
return ans;
}
int main()
{
int arr[] = { 7, 3, 2, 4, 1 };
int k = 3;
int n = sizeof(arr) / sizeof(int);
cout << "Number of inversions in all "
"subarrays of size " << k <<" are : "
<< getInvCount(arr, k, n);
return 0;
}
Java
// Java program to count inversions in all sub arrays
// of size k using Binary Indexed Tree
import java.util.*;
class GFG
{
// Returns sum of arr[0..index]. This function assumes
// that the array is preprocessed and partial sums of
// array elements are stored in BITree[].
static 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.
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 }
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++)
{
// lower_bound() Returns pointer to
// the first element greater than
// or equal to arr[i]
arr[i] = lower_bound(temp, 0, n,
arr[i]) + 1;
}
}
static int lower_bound(int[] a, int low,
int high, int element)
{
while(low < high)
{
int middle = low + (high - low) / 2;
if(element > a[middle])
low = middle + 1;
else
high = middle;
}
return low;
}
// Returns inversion count of all subarray
// of size k in arr[0..n-1]
static int getInvCount(int arr[], int k, int n)
{
int invcount = 0; // Initialize result
// 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;
// Get inversion count of first subarray
for (int i = k - 1; i >= 0; i--)
{
// Get count of elements smaller than arr[i]
invcount += getSum(BIT, arr[i] - 1);
// Add current element to BIT
updateBIT(BIT, n, arr[i], 1);
}
// now calculate the inversion of
// all other subarrays
int ans = invcount;
int i = 0, j = k, icnt = 0, jcnt = 0;
while (j <= n - 1)
{
// calculate value of inversion count of
// first element in previous subarray
icnt = getSum(BIT, arr[i] - 1);
updateBIT(BIT, n, arr[i], -1);
// calculating inversion count of
// last element in the current subarray
jcnt = getSum(BIT, n) - getSum(BIT, arr[j]);
updateBIT(BIT, n, arr[j], 1);
// calculating inversion count
// of current subarray
invcount = invcount - icnt + jcnt;
// adding current inversion to the answer
ans = ans + invcount;
i++; j++;
}
return ans;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 7, 3, 2, 4, 1 };
int k = 3;
int n = arr.length;
System.out.println("Number of inversions in all " +
"subarrays of size " + k +
" are : " + getInvCount(arr, k, n));
}
}
// This code is contributed by PrinciRaj1992
Python3
# Python3 program to count inversions in all sub arrays
# of size k using Binary Indexed Tree
from bisect import bisect_left as lower_bound
# Returns 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):
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.
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 to3, 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 = sorted(temp)
# 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
# Returns inversion count of all subarray
# of size k in arr[0..n-1]
def getInvCount(arr, k, n):
invcount = 0 # Initialize result
# 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 to3, 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)
for i in range(1, n + 1):
BIT[i] = 0
# Get inversion count of first subarray
for i in range(k - 1, -1, -1):
# Get count of elements smaller than arr[i]
invcount += getSum(BIT, arr[i] - 1)
# Add current element to BIT
updateBIT(BIT, n, arr[i], 1)
# now calculate the inversion of all other subarrays
ans = invcount
i = 0
j = k
icnt = 0
jcnt = 0
while (j <= n - 1):
# calculate value of inversion count of first element
# in previous subarray
icnt = getSum(BIT, arr[i] - 1)
updateBIT(BIT, n, arr[i], -1)
# calculating inversion count of last element in the
# current subarray
jcnt = getSum(BIT, n) - getSum(BIT, arr[j])
updateBIT(BIT, n, arr[j], 1)
# calculating inversion count of current subarray
invcount = invcount - icnt + jcnt
# adding current inversion to the answer
ans = ans + invcount
i += 1
j += 1
return ans
# Driver code
if __name__ == '__main__':
arr= [7, 3, 2, 4, 1]
k = 3
n = len(arr)
print("Number of inversions in all subarrays of size "
,k," are : ",getInvCount(arr, k, n))
# This code is contributed by mohit kumar 29
C#
// C# program to count inversions in all sub arrays
// of size k using Binary Indexed Tree
using System;
class GFG
{
// Returns sum of arr[0..index]. This function assumes
// that the array is preprocessed and partial sums of
// array elements are stored in BITree[].
static 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.
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 }
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++)
{
// lower_bound() Returns pointer to
// the first element greater than
// or equal to arr[i]
arr[i] = lower_bound(temp, 0, n,
arr[i]) + 1;
}
}
static int lower_bound(int[] a, int low,
int high, int element)
{
while(low < high)
{
int middle = low + (high - low) / 2;
if(element > a[middle])
low = middle + 1;
else
high = middle;
}
return low;
}
// Returns inversion count of all subarray
// of size k in arr[0..n-1]
static int getInvCount(int []arr, int k, int n)
{
int invcount = 0; // Initialize result
// 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;
// Get inversion count of first subarray
for (int i = k - 1; i >= 0; i--)
{
// Get count of elements smaller than arr[i]
invcount += getSum(BIT, arr[i] - 1);
// Add current element to BIT
updateBIT(BIT, n, arr[i], 1);
}
// now calculate the inversion of
// all other subarrays
int ans = invcount;
int x = 0, j = k, icnt = 0, jcnt = 0;
while (j <= n - 1)
{
// calculate value of inversion count of
// first element in previous subarray
icnt = getSum(BIT, arr[x] - 1);
updateBIT(BIT, n, arr[x], -1);
// calculating inversion count of
// last element in the current subarray
jcnt = getSum(BIT, n) - getSum(BIT, arr[j]);
updateBIT(BIT, n, arr[j], 1);
// calculating inversion count
// of current subarray
invcount = invcount - icnt + jcnt;
// adding current inversion to the answer
ans = ans + invcount;
x++; j++;
}
return ans;
}
// Driver Code
static public void Main ()
{
int []arr = { 7, 3, 2, 4, 1 };
int k = 3;
int n = arr.Length;
Console.Write("Number of inversions in all " +
"subarrays of size " + k +
" are : " + getInvCount(arr, k, n));
}
}
// This code is contributed by ajit.
Javascript
<script>
// Javascript program to count inversions in all sub arrays
// of size k using Binary Indexed Tree
// Returns sum of arr[0..index]. This function assumes
// that the array is preprocessed and partial sums of
// array elements are stored in BITree[].
function getSum(BITree, index)
{
let 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.
function 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 }
function convert(arr, n)
{
// Create a copy of arrp[] in temp and sort
// the temp array in increasing order
let temp = new Array(n);
for (let i = 0; i < n; i++)
temp[i] = arr[i];
temp.sort(function(a, b){return a - b});
// Traverse all array elements
for (let 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, 0, n, arr[i]) + 1;
}
}
function lower_bound(a, low, high, element)
{
while(low < high)
{
let middle = low + parseInt((high - low) / 2, 10);
if(element > a[middle])
low = middle + 1;
else
high = middle;
}
return low;
}
// Returns inversion count of all subarray
// of size k in arr[0..n-1]
function getInvCount(arr, k, n)
{
let invcount = 0; // Initialize result
// 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)
let BIT = new Array(n + 1);
for (let i = 1; i <= n; i++)
BIT[i] = 0;
// Get inversion count of first subarray
for (let i = k - 1; i >= 0; i--)
{
// Get count of elements smaller than arr[i]
invcount += getSum(BIT, arr[i] - 1);
// Add current element to BIT
updateBIT(BIT, n, arr[i], 1);
}
// now calculate the inversion of
// all other subarrays
let ans = invcount;
let x = 0, j = k, icnt = 0, jcnt = 0;
while (j <= n - 1)
{
// calculate value of inversion count of
// first element in previous subarray
icnt = getSum(BIT, arr[x] - 1);
updateBIT(BIT, n, arr[x], -1);
// calculating inversion count of
// last element in the current subarray
jcnt = getSum(BIT, n) - getSum(BIT, arr[j]);
updateBIT(BIT, n, arr[j], 1);
// calculating inversion count
// of current subarray
invcount = invcount - icnt + jcnt;
// adding current inversion to the answer
ans = ans + invcount;
x++; j++;
}
return ans;
}
let arr = [ 7, 3, 2, 4, 1 ];
let k = 3;
let n = arr.length;
document.write("Number of inversions in all " +
"subarrays of size " + k +
" are : " + getInvCount(arr, k, n));
</script>
Producción
Number of inversions in all subarrays of size 3 are : 6
Complejidad de tiempo: el conteo de inversión del primer subarreglo toma O(k log(n)) tiempo y para todos los demás subarreglo toma O((nk)log(n)). Entonces, la complejidad del tiempo total es: O (n log n).
Espacio auxiliar:- O(n)
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Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA