Minimum operations required to modify the array such that parity of adjacent elements is different
Given an array A[], the task is to find the minimum number of operations required to convert the array into B[] such that for every index in B (except the last) parity(b[i]) != parity(b[i + 1]) where parity(x) = x % 3.
Below is the operation to be performed:
Any element from the set {1, 2} can be added to any element of the array.
Examples:
Input: A[] = {2, 1, 3, 0}
Output: 1
1 can be added to 0 in a single operation and the array becomes {2, 1, 3, 1}.
Input: A[] = {2, 2, 2, 2}
Output: 2
Approach: An optimal approach is to update the element which is in between two other elements so that it can be updated such that its parity can be different from the element previous to it and the element next to it in a single operation (as the number of operations needs to be minimized).
So, start traversing the array from A[1] to A[n – 2] and for every element A[i] such that parity(A[i]) == parity(A[i – 1]) or parity(A[i]) == parity(A[i + 1]), update A[i] such that its parity is different from both the numbers surrounding it and keep a track of the count of operations performed. Print the count in the end.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int parity( int a)
{
return a % 3;
}
int solve( int array[], int size)
{
int operations = 0;
for ( int i = 0; i < size - 1; i++) {
if (parity(array[i]) == parity(array[i + 1])) {
operations++;
if (i + 2 < size) {
int pari1 = parity(array[i]);
int pari2 = parity(array[i + 2]);
if (pari1 == pari2) {
if (pari1 == 0)
array[i + 1] = 1;
else if (pari1 == 1)
array[i + 1] = 0;
else
array[i + 1] = 1;
}
else {
if ((pari1 == 0 && pari2 == 1)
|| (pari1 == 1 && pari2 == 0))
array[i + 1] = 2;
if ((pari1 == 1 && pari2 == 2)
|| (pari1 == 2 && pari2 == 1))
array[i + 1] = 0;
if ((pari1 == 2 && pari2 == 0)
|| (pari1 == 0 && pari2 == 2))
array[i + 1] = 1;
}
}
}
}
return operations;
}
int main()
{
int array[] = { 2, 1, 3, 0 };
int size = sizeof (array) / sizeof (array[0]);
cout << solve(array, size) << endl;
return 0;
}
|
Java
class GFG
{
static int parity( int a)
{
return a % 3 ;
}
static int solve( int []array, int size)
{
int operations = 0 ;
for ( int i = 0 ; i < size - 1 ; i++)
{
if (parity(array[i]) == parity(array[i + 1 ]))
{
operations++;
if (i + 2 < size)
{
int pari1 = parity(array[i]);
int pari2 = parity(array[i + 2 ]);
if (pari1 == pari2)
{
if (pari1 == 0 )
array[i + 1 ] = 1 ;
else if (pari1 == 1 )
array[i + 1 ] = 0 ;
else
array[i + 1 ] = 1 ;
}
else
{
if ((pari1 == 0 && pari2 == 1 ) ||
(pari1 == 1 && pari2 == 0 ))
array[i + 1 ] = 2 ;
if ((pari1 == 1 && pari2 == 2 )
|| (pari1 == 2 && pari2 == 1 ))
array[i + 1 ] = 0 ;
if ((pari1 == 2 && pari2 == 0 )
|| (pari1 == 0 && pari2 == 2 ))
array[i + 1 ] = 1 ;
}
}
}
}
return operations;
}
public static void main (String arg[])
{
int []array = { 2 , 1 , 3 , 0 };
int size = array.length;
System.out.println(solve(array, size));
}
}
|
Python3
def parity(a):
return a % 3
def solve(array, size):
operations = 0
for i in range ( 0 , size - 1 ):
if parity(array[i]) = = parity(array[i + 1 ]):
operations + = 1
if i + 2 < size:
pari1 = parity(array[i])
pari2 = parity(array[i + 2 ])
if pari1 = = pari2:
if pari1 = = 0 :
array[i + 1 ] = 1
elif pari1 = = 1 :
array[i + 1 ] = 0
else :
array[i + 1 ] = 1
else :
if ((pari1 = = 0 and pari2 = = 1 ) or
(pari1 = = 1 and pari2 = = 0 )):
array[i + 1 ] = 2
if ((pari1 = = 1 and pari2 = = 2 ) or
(pari1 = = 2 and pari2 = = 1 )):
array[i + 1 ] = 0
if ((pari1 = = 2 and pari2 = = 0 ) and
(pari1 = = 0 and pari2 = = 2 )):
array[i + 1 ] = 1
return operations
if __name__ = = "__main__" :
array = [ 2 , 1 , 3 , 0 ]
size = len (array)
print (solve(array, size))
|
C#
using System;
class GFG
{
static int parity( int a)
{
return a % 3;
}
static int solve( int []array, int size)
{
int operations = 0;
for ( int i = 0; i < size - 1; i++)
{
if (parity(array[i]) == parity(array[i + 1]))
{
operations++;
if (i + 2 < size)
{
int pari1 = parity(array[i]);
int pari2 = parity(array[i + 2]);
if (pari1 == pari2)
{
if (pari1 == 0)
array[i + 1] = 1;
else if (pari1 == 1)
array[i + 1] = 0;
else
array[i + 1] = 1;
}
else
{
if ((pari1 == 0 && pari2 == 1) ||
(pari1 == 1 && pari2 == 0))
array[i + 1] = 2;
if ((pari1 == 1 && pari2 == 2)
|| (pari1 == 2 && pari2 == 1))
array[i + 1] = 0;
if ((pari1 == 2 && pari2 == 0)
|| (pari1 == 0 && pari2 == 2))
array[i + 1] = 1;
}
}
}
}
return operations;
}
public static void Main()
{
int []array = { 2, 1, 3, 0 };
int size = array.Length;
Console.WriteLine(solve(array, size));
}
}
|
PHP
<?php
function parity( $a )
{
return $a % 3;
}
function solve( $array , $size )
{
$operations = 0;
for ( $i = 0; $i < $size - 1; $i ++)
{
if (parity( $array [ $i ]) == parity( $array [ $i + 1]))
{
$operations ++;
if ( $i + 2 < $size )
{
$pari1 = parity( $array [ $i ]);
$pari2 = parity( $array [ $i + 2]);
if ( $pari1 == $pari2 )
{
if ( $pari1 == 0)
$array [ $i + 1] = 1;
else if ( $pari1 == 1)
$array [ $i + 1] = 0;
else
$array [ $i + 1] = 1;
}
else
{
if (( $pari1 == 0 && $pari2 == 1) ||
( $pari1 == 1 && $pari2 == 0))
$array [ $i + 1] = 2;
if (( $pari1 == 1 && $pari2 == 2) ||
( $pari1 == 2 && $pari2 == 1))
$array [ $i + 1] = 0;
if (( $pari1 == 2 && $pari2 == 0) ||
( $pari1 == 0 && $pari2 == 2))
$array [ $i + 1] = 1;
}
}
}
}
return $operations ;
}
$array = array (2, 1, 3, 0 );
$size = count ( $array );
echo solve( $array , $size );
?>
|
Javascript
<script>
function parity(a)
{
return a % 3;
}
function solve(array, size)
{
var operations = 0;
for ( var i = 0; i < size - 1; i++) {
if (parity(array[i]) == parity(array[i + 1])) {
operations++;
if (i + 2 < size) {
var pari1 = parity(array[i]);
var pari2 = parity(array[i + 2]);
if (pari1 == pari2) {
if (pari1 == 0)
array[i + 1] = 1;
else if (pari1 == 1)
array[i + 1] = 0;
else
array[i + 1] = 1;
}
else {
if ((pari1 == 0 && pari2 == 1)
|| (pari1 == 1 && pari2 == 0))
array[i + 1] = 2;
if ((pari1 == 1 && pari2 == 2)
|| (pari1 == 2 && pari2 == 1))
array[i + 1] = 0;
if ((pari1 == 2 && pari2 == 0)
|| (pari1 == 0 && pari2 == 2))
array[i + 1] = 1;
}
}
}
}
return operations;
}
var array = [2, 1, 3, 0];
var size = array.length;
document.write( solve(array, size));
</script>
|
Time Complexity: O(N), since there runs a loop from 0 to (n – 1).
Auxiliary Space: O(1), since no extra space has been taken.
Last Updated :
22 Jun, 2022
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