Number of possible permutations when absolute difference between number of elements to the right and left are given
Given an array of N elements where each element i, the absolute difference between total elements to the right and left of it are given. Find the number of possible ordering of the actual array elements.
Examples:
Input : N = 5, arr[] = {2, 4, 4, 0, 2}
Output : 4
There are four possible orders, as follows:
2, 1, 4, 5, 3
2, 5, 4, 1, 3
3, 1, 4, 5, 2
3, 5, 4, 1, 2
Input : N = 7, arr[] = {6, 4, 0, 2, 4, 0, 2}
Output : 0
No any valid order is possible hence answer is 0.
Approach: Divide the problem into two parts. When N is odd and when N is even.
- Case 1: When N is odd.
Consider N = 7, there are 7 empty spaces and the absolute difference between the elements to the left and right must be like [6 4 2 0 2 4 6]. Observe that the element which is at the middle must have absolute difference 0, while other elements are from 2 to N-1 and each of their counts should be 2. If it doesn’t fulfill it then there is no valid order else for each element i from 2 to N-1 we have 2 ways to fill the spaces, hence total ways will be the product of all the ways.
- Case 2: When N is even.
Consider N = 6, There are 6 spaces and it will be like [5 3 1 1 3 5], where a[i] gives the absolute difference between the number of elements to the left and right. For each a[i] we have 2 ways, hence answer will be the product of all the ways.
Below is the implementation of the approach:
C++
#include <bits/stdc++.h>
using namespace std;
int totalways( int * arr, int n)
{
unordered_map< int , int > cnt;
for ( int i = 0; i < n; ++i) {
cnt[arr[i]]++;
}
if (n % 2 == 1) {
int start = 0, endd = n - 1;
for ( int i = start; i <= endd; i = i + 2) {
if (i == 0) {
if (cnt[i] != 1) {
return 0;
}
}
else {
if (cnt[i] != 2) {
return 0;
}
}
}
int ways = 1;
start = 2, endd = n - 1;
for ( int i = start; i <= endd; i = i + 2) {
ways = ways * 2;
}
return ways;
}
else if (n % 2 == 0) {
int start = 1, endd = n - 1;
for ( int i = 1; i <= endd; i = i + 2) {
if (cnt[i] != 2)
return 0;
}
int ways = 1;
for ( int i = start; i <= endd; i = i + 2) {
ways = ways * 2;
}
return ways;
}
}
int main()
{
int N = 5;
int arr[N] = { 2, 4, 4, 0, 2 };
cout<<totalways(arr, N);
return 0;
}
|
Java
import java.util.*;
class GFG
{
static int totalways( int [] arr, int n)
{
HashMap<Integer,
Integer>cnt = new HashMap<Integer,
Integer>();
for ( int i = 0 ; i < n; i++)
{
if (cnt.containsKey(arr[i]))
{
cnt.put(arr[i], cnt.get(arr[i])+ 1 );
}
else
{
cnt.put(arr[i], 1 );
}
}
if (n % 2 == 1 )
{
int start = 0 , endd = n - 1 ;
for ( int i = start; i <= endd; i = i + 2 )
{
if (i == 0 )
{
if (cnt.get(i) != 1 )
{
return 0 ;
}
}
else
{
if (cnt.get(i) != 2 )
{
return 0 ;
}
}
}
int ways = 1 ;
start = 2 ; endd = n - 1 ;
for ( int i = start; i <= endd; i = i + 2 )
{
ways = ways * 2 ;
}
return ways;
}
else if (n % 2 == 0 )
{
int start = 1 , endd = n - 1 ;
for ( int i = 1 ; i <= endd; i = i + 2 )
{
if (cnt.get(i) != 2 )
return 0 ;
}
int ways = 1 ;
for ( int i = start; i <= endd; i = i + 2 )
{
ways = ways * 2 ;
}
return ways;
}
return Integer.MIN_VALUE;
}
public static void main(String[] args)
{
int N = 5 ;
int []arr = { 2 , 4 , 4 , 0 , 2 };
System.out.println(totalways(arr, N));
}
}
|
Python3
def totalways(arr, n):
cnt = dict ()
for i in range (n):
cnt[arr[i]] = cnt.get(arr[i], 0 ) + 1
if (n % 2 = = 1 ):
start, endd = 0 , n - 1
for i in range (start, endd + 1 , 2 ):
if (i = = 0 ):
if (cnt[i] ! = 1 ):
return 0
else :
if (cnt[i] ! = 2 ):
return 0
ways = 1
start = 2
endd = n - 1
for i in range (start, endd + 1 , 2 ):
ways = ways * 2
return ways
elif (n % 2 = = 0 ):
start = 1
endd = n - 1
for i in range ( 1 , endd + 1 , 2 ):
if (cnt[i] ! = 2 ):
return 0
ways = 1
for i in range (start, endd + 1 , 2 ):
ways = ways * 2
return ways
N = 5
arr = [ 2 , 4 , 4 , 0 , 2 ]
print (totalways(arr, N))
|
C#
using System;
using System.Collections.Generic;
class GFG
{
static int totalways( int [] arr, int n)
{
Dictionary< int ,
int > cnt = new Dictionary< int ,
int >();
for ( int i = 0 ; i < n; i++)
{
if (cnt.ContainsKey(arr[i]))
{
cnt[arr[i]] = cnt[arr[i]] + 1;
}
else
{
cnt.Add(arr[i], 1);
}
}
if (n % 2 == 1)
{
int start = 0, endd = n - 1;
for ( int i = start; i <= endd; i = i + 2)
{
if (i == 0)
{
if (cnt[i] != 1)
{
return 0;
}
}
else
{
if (cnt[i] != 2)
{
return 0;
}
}
}
int ways = 1;
start = 2; endd = n - 1;
for ( int i = start; i <= endd; i = i + 2)
{
ways = ways * 2;
}
return ways;
}
else if (n % 2 == 0)
{
int start = 1, endd = n - 1;
for ( int i = 1; i <= endd; i = i + 2)
{
if (cnt[i] != 2)
return 0;
}
int ways = 1;
for ( int i = start; i <= endd; i = i + 2)
{
ways = ways * 2;
}
return ways;
}
return int .MinValue;
}
public static void Main(String[] args)
{
int N = 5;
int []arr = { 2, 4, 4, 0, 2 };
Console.WriteLine(totalways(arr, N));
}
}
|
Javascript
<script>
function totalways(arr, n) {
var cnt = {};
for ( var i = 0; i < n; i++) {
if (cnt.hasOwnProperty(arr[i])) {
cnt[arr[i]] = cnt[arr[i]] + 1;
} else {
cnt[arr[i]] = 1;
}
}
if (n % 2 === 1) {
var start = 0,
endd = n - 1;
for ( var i = start; i <= endd; i = i + 2) {
if (i === 0) {
if (cnt[i] !== 1) {
return 0;
}
} else {
if (cnt[i] !== 2) {
return 0;
}
}
}
var ways = 1;
start = 2;
endd = n - 1;
for ( var i = start; i <= endd; i = i + 2) {
ways = ways * 2;
}
return ways;
}
else if (n % 2 === 0) {
var start = 1,
endd = n - 1;
for ( var i = 1; i <= endd; i = i + 2) {
if (cnt[i] !== 2) return 0;
}
var ways = 1;
for ( var i = start; i <= endd; i = i + 2) {
ways = ways * 2;
}
return ways;
}
return -2147483648;
}
var N = 5;
var arr = [2, 4, 4, 0, 2];
document.write(totalways(arr, N));
</script>
|
Time Complexity: O(N), where N is the size of the given array.
Auxiliary Space: O(N) for using a hashmap to store the frequency of the given elements.
Last Updated :
05 Sep, 2022
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