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.
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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.
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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++ implementation of the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the number of permutations // possible of the original array to satisfy // the given absolute differences int totalways( int * arr, int n)
{ // To store the count of each
// a[i] in a map
unordered_map< int , int > cnt;
for ( int i = 0; i < n; ++i) {
cnt[arr[i]]++;
}
// if n is odd
if (n % 2 == 1) {
int start = 0, endd = n - 1;
// check the count of each whether
// it satisfy the given criteria or not
for ( int i = start; i <= endd; i = i + 2) {
if (i == 0) {
// there is only 1 way
// for middle element.
if (cnt[i] != 1) {
return 0;
}
}
else {
// for others there are 2 ways.
if (cnt[i] != 2) {
return 0;
}
}
}
// now find total ways
int ways = 1;
start = 2, endd = n - 1;
for ( int i = start; i <= endd; i = i + 2) {
ways = ways * 2;
}
return ways;
}
// When n is even.
else if (n % 2 == 0) {
// there will be no middle element so
// for each a[i] there will be 2 ways
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;
}
} // Driver Code int main()
{ int N = 5;
int arr[N] = { 2, 4, 4, 0, 2 };
cout<<totalways(arr, N);
return 0;
} |
// Java implementation of the above approach import java.util.*;
class GFG
{ // Function to find the number of permutations // possible of the original array to satisfy // the given absolute differences static int totalways( int [] arr, int n)
{ // To store the count of each
// a[i] in a map
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 is odd
if (n % 2 == 1 )
{
int start = 0 , endd = n - 1 ;
// check the count of each whether
// it satisfy the given criteria or not
for ( int i = start; i <= endd; i = i + 2 )
{
if (i == 0 )
{
// there is only 1 way
// for middle element.
if (cnt.get(i) != 1 )
{
return 0 ;
}
}
else
{
// for others there are 2 ways.
if (cnt.get(i) != 2 )
{
return 0 ;
}
}
}
// now find total ways
int ways = 1 ;
start = 2 ; endd = n - 1 ;
for ( int i = start; i <= endd; i = i + 2 )
{
ways = ways * 2 ;
}
return ways;
}
// When n is even.
else if (n % 2 == 0 )
{
// there will be no middle element so
// for each a[i] there will be 2 ways
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;
} // Driver Code public static void main(String[] args)
{ int N = 5 ;
int []arr = { 2 , 4 , 4 , 0 , 2 };
System.out.println(totalways(arr, N));
} } // This code is contributed by Princi Singh |
# Python3 implementation of the above approach # Function to find the number of permutations # possible of the original array to satisfy # the given absolute differences def totalways(arr, n):
# To store the count of each
# a[i] in a map
cnt = dict ()
for i in range (n):
cnt[arr[i]] = cnt.get(arr[i], 0 ) + 1
# if n is odd
if (n % 2 = = 1 ):
start, endd = 0 , n - 1
# check the count of each whether
# it satisfy the given criteria or not
for i in range (start, endd + 1 , 2 ):
if (i = = 0 ):
# there is only 1 way
# for middle element.
if (cnt[i] ! = 1 ):
return 0
else :
# for others there are 2 ways.
if (cnt[i] ! = 2 ):
return 0
# now find total ways
ways = 1
start = 2
endd = n - 1
for i in range (start, endd + 1 , 2 ):
ways = ways * 2
return ways
# When n is even.
elif (n % 2 = = 0 ):
# there will be no middle element so
# for each a[i] there will be 2 ways
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
# Driver Code N = 5
arr = [ 2 , 4 , 4 , 0 , 2 ]
print (totalways(arr, N))
# This code is contributed by Mohit Kumar |
// C# implementation of the above approach using System;
using System.Collections.Generic;
class GFG
{ // Function to find the number of permutations // possible of the original array to satisfy // the given absolute differences static int totalways( int [] arr, int n)
{ // To store the count of each
// a[i] in a map
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 is odd
if (n % 2 == 1)
{
int start = 0, endd = n - 1;
// check the count of each whether
// it satisfy the given criteria or not
for ( int i = start; i <= endd; i = i + 2)
{
if (i == 0)
{
// there is only 1 way
// for middle element.
if (cnt[i] != 1)
{
return 0;
}
}
else
{
// for others there are 2 ways.
if (cnt[i] != 2)
{
return 0;
}
}
}
// now find total ways
int ways = 1;
start = 2; endd = n - 1;
for ( int i = start; i <= endd; i = i + 2)
{
ways = ways * 2;
}
return ways;
}
// When n is even.
else if (n % 2 == 0)
{
// there will be no middle element so
// for each a[i] there will be 2 ways
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;
} // Driver Code public static void Main(String[] args)
{ int N = 5;
int []arr = { 2, 4, 4, 0, 2 };
Console.WriteLine(totalways(arr, N));
} } // This code is contributed by 29AjayKumar |
<script> // JavaScript implementation of the above approach
// Function to find the number of permutations
// possible of the original array to satisfy
// the given absolute differences
function totalways(arr, n) {
// To store the count of each
// a[i] in a map
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 is odd
if (n % 2 === 1) {
var start = 0,
endd = n - 1;
// check the count of each whether
// it satisfy the given criteria or not
for ( var i = start; i <= endd; i = i + 2) {
if (i === 0) {
// there is only 1 way
// for middle element.
if (cnt[i] !== 1) {
return 0;
}
} else {
// for others there are 2 ways.
if (cnt[i] !== 2) {
return 0;
}
}
}
// now find total ways
var ways = 1;
start = 2;
endd = n - 1;
for ( var i = start; i <= endd; i = i + 2) {
ways = ways * 2;
}
return ways;
}
// When n is even.
else if (n % 2 === 0) {
// there will be no middle element so
// for each a[i] there will be 2 ways
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;
}
// Driver Code
var N = 5;
var arr = [2, 4, 4, 0, 2];
document.write(totalways(arr, N));
</script> |
4
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.