Given an array arr[] of N positive integers with an equal number of even and odd elements. The task is to use in-place swapping to interchange positions of even and odd elements in the array.
Examples:
Input: arr[] = {1, 3, 2, 4}
Output: 2 4 1 3
Explanation:
Before rearranging the given array, indices 0 and 1 had odd elements and indices 2 and 3 had even elements.
After rearrangement, array becomes {2, 4, 1, 3} where indices 0 and 1 have even elements and indices 2 and 3 have odd elements.Input: arr[] = {2, 2, 1, 3}
Output: 1 3 2 2
Explanation:
Before rearranging the given array, indices 0 and 1 had even elements and indices 2 and 3 had odd elements.
After rearrangement, array becomes {1, 3, 2, 2} where indices 0 and 1 have odd elements and indices 2 and 3 have even elements.
Naive Approach: The simplest approach is to iterate over array elements using two loops, the outer loop picks each element of the array and the inner loop is to find the opposite parity element for the picked element and swap them. After swapping elements, mark the picked element as negative so as not to pick it again. Finally, make all elements positive and print the array.
Below is the implementation of the above approach:
// C++ program for the above approach #include<bits/stdc++.h> using namespace std;
// Function to replace each even // element by odd and vice-versa // in a given array void replace( int arr[], int n)
{ // Traverse array
for ( int i = 0; i < n; i++)
{
for ( int j = i + 1; j < n; j++)
{
// If current element is even
// then swap it with odd
if (arr[i] >= 0 && arr[j] >= 0 &&
arr[i] % 2 == 0 &&
arr[j] % 2 != 0)
{
// Perform Swap
int tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
// Change the sign
arr[j] = -arr[j];
break ;
}
// If current element is odd
// then swap it with even
else if (arr[i] >= 0 && arr[j] >= 0 &&
arr[i] % 2 != 0 &&
arr[j] % 2 == 0)
{
// Perform Swap
int tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
// Change the sign
arr[j] = -arr[j];
break ;
}
}
}
// Marked element positive
for ( int i = 0; i < n; i++)
arr[i] = abs (arr[i]);
// Print final array
for ( int i = 0; i < n; i++)
cout << arr[i] << " " ;
} // Driver Code int main()
{ // Given array arr[]
int arr[] = { 1, 3, 2, 4 };
int n = sizeof (arr) / sizeof (arr[0]);
// Function Call
replace(arr,n);
} // This code is contributed by SURENDRA_GANGWAR |
// Java program for the above approach import java.util.*;
import java.lang.*;
class GFG {
// Function to replace each even
// element by odd and vice-versa
// in a given array
static void replace( int [] arr)
{
// Stores length of array
int n = arr.length;
// Traverse array
for ( int i = 0 ; i < n; i++) {
for ( int j = i + 1 ; j < n; j++) {
// If current element is even
// then swap it with odd
if (arr[i] >= 0
&& arr[j] >= 0
&& arr[i] % 2 == 0
&& arr[j] % 2 != 0 ) {
// Perform Swap
int tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
// Change the sign
arr[j] = -arr[j];
break ;
}
// If current element is odd
// then swap it with even
else if (arr[i] >= 0
&& arr[j] >= 0
&& arr[i] % 2 != 0
&& arr[j] % 2 == 0 ) {
// Perform Swap
int tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
// Change the sign
arr[j] = -arr[j];
break ;
}
}
}
// Marked element positive
for ( int i = 0 ; i < n; i++)
arr[i] = Math.abs(arr[i]);
// Print final array
for ( int i = 0 ; i < n; i++)
System.out.print(arr[i] + " " );
}
// Driver Code
public static void main(String[] args)
{
// Given array arr[]
int [] arr = { 1 , 3 , 2 , 4 };
// Function Call
replace(arr);
}
} |
# Python3 program for the # above approach # Function to replace each even # element by odd and vice-versa # in a given array def replace(arr, n):
# Traverse array
for i in range (n):
for j in range (i + 1 , n):
# If current element is
# even then swap it with odd
if (arr[i] > = 0 and
arr[j] > = 0 and
arr[i] % 2 = = 0 and
arr[j] % 2 ! = 0 ):
# Perform Swap
tmp = arr[i]
arr[i] = arr[j]
arr[j] = tmp
# Change the sign
arr[j] = - arr[j]
break
# If current element is odd
# then swap it with even
elif (arr[i] > = 0 and
arr[j] > = 0 and
arr[i] % 2 ! = 0 and
arr[j] % 2 = = 0 ):
# Perform Swap
tmp = arr[i]
arr[i] = arr[j]
arr[j] = tmp
# Change the sign
arr[j] = - arr[j]
break
# Marked element positive
for i in range (n):
arr[i] = abs (arr[i])
# Print final array
for i in range (n):
print (arr[i], end = " " )
# Driver Code if __name__ = = "__main__" :
# Given array arr[]
arr = [ 1 , 3 , 2 , 4 ]
n = len (arr)
# Function Call
replace(arr, n)
# This code is contributed by Chitranayal |
// C# program for the above approach using System;
class GFG{
// Function to replace each even // element by odd and vice-versa // in a given array static void replace( int [] arr)
{ // Stores length of array
int n = arr.Length;
// Traverse array
for ( int i = 0; i < n; i++)
{
for ( int j = i + 1; j < n; j++)
{
// If current element is even
// then swap it with odd
if (arr[i] >= 0 &&
arr[j] >= 0 &&
arr[i] % 2 == 0 &&
arr[j] % 2 != 0)
{
// Perform Swap
int tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
// Change the sign
arr[j] = -arr[j];
break ;
}
// If current element is odd
// then swap it with even
else if (arr[i] >= 0 &&
arr[j] >= 0 &&
arr[i] % 2 != 0 &&
arr[j] % 2 == 0)
{
// Perform Swap
int tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
// Change the sign
arr[j] = -arr[j];
break ;
}
}
}
// Marked element positive
for ( int i = 0; i < n; i++)
arr[i] = Math.Abs(arr[i]);
// Print readonly array
for ( int i = 0; i < n; i++)
Console.Write(arr[i] + " " );
} // Driver Code public static void Main(String[] args)
{ // Given array []arr
int [] arr = { 1, 3, 2, 4 };
// Function Call
replace(arr);
} } // This code is contributed by Rajput-Ji |
<script> // Javascript program to implement // the above approach // Function to replace each even
// element by odd and vice-versa
// in a given array
function replace(arr)
{
// Stores length of array
let n = arr.length;
// Traverse array
for (let i = 0; i < n; i++) {
for (let j = i + 1; j < n; j++) {
// If current element is even
// then swap it with odd
if (arr[i] >= 0
&& arr[j] >= 0
&& arr[i] % 2 == 0
&& arr[j] % 2 != 0) {
// Perform Swap
let tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
// Change the sign
arr[j] = -arr[j];
break ;
}
// If current element is odd
// then swap it with even
else if (arr[i] >= 0
&& arr[j] >= 0
&& arr[i] % 2 != 0
&& arr[j] % 2 == 0) {
// Perform Swap
let tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
// Change the sign
arr[j] = -arr[j];
break ;
}
}
}
// Marked element positive
for (let i = 0; i < n; i++)
arr[i] = Math.abs(arr[i]);
// Print final array
for (let i = 0; i < n; i++)
document.write(arr[i] + " " );
}
// Driver Code
// Given array arr[]
let arr = [ 1, 3, 2, 4 ];
// Function Call
replace(arr);
</script> |
2 4 1 3
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, the idea is to use Two Pointer Approach. Traverse the array by picking each element that is greater than 0 and search for the opposite parity element greater than 0 from the current index up to the end of the array. If found, swap the elements and multiply them with -1. Follow the below steps to solve the problem:
- Initialize the variables e and o with -1 that will store the currently found even and odd number respectively that is not yet taken.
- Traverse the given array over the range [0, N – 1] and do the following:
- Pick the element arr[i] if it is greater than 0.
- If arr[i] is even, increment o + 1 by 1 and find the next odd number that is not yet marked is found. Mark the current and the found number by multiplying them with -1 and swap them.
- If arr[i] is odd, increment e + 1 by 1 and find the next even number that is not yet marked is found. Mark the current and the found number by multiplying them with -1 and swap them.
- After traversing the whole array, multiply its elements with -1 to again make them positive and print that array.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
#define N 3 #define M 4 // Function to replace odd elements // with even elements and vice versa void swapEvenOdd( int arr[], int n)
{ int o = -1, e = -1;
// Traverse the given array
for ( int i = 0; i < n; i++)
{
// If arr[i] is visited
if (arr[i] < 0)
continue ;
int r = -1;
if (arr[i] % 2 == 0)
{
o++;
// Find the next odd element
while (arr[o] % 2 == 0 || arr[o] < 0)
o++;
r = o;
}
else
{
e++;
// Find next even element
while (arr[e] % 2 == 1 || arr[e] < 0)
e++;
r = e;
}
// Mark them visited
arr[i] *= -1;
arr[r] *= -1;
// Swap them
int tmp = arr[i];
arr[i] = arr[r];
arr[r] = tmp;
}
// Print the final array
for ( int i = 0; i < n; i++)
{
cout << (-1 * arr[i]) << " " ;
}
} // Driver Code int main()
{ // Given array arr[]
int arr[] = { 1, 3, 2, 4 };
// Length of the array
int n = sizeof (arr) / sizeof (arr[0]);
// Function Call
swapEvenOdd(arr, n);
} // This code is contributed by Rajput-Ji |
// Java program for the above approach import java.io.*;
class GFG {
// Function to replace odd elements
// with even elements and vice versa
static void swapEvenOdd( int arr[])
{
// Length of the array
int n = arr.length;
int o = - 1 , e = - 1 ;
// Traverse the given array
for ( int i = 0 ; i < n; i++) {
// If arr[i] is visited
if (arr[i] < 0 )
continue ;
int r = - 1 ;
if (arr[i] % 2 == 0 ) {
o++;
// Find the next odd element
while (arr[o] % 2 == 0
|| arr[o] < 0 )
o++;
r = o;
}
else {
e++;
// Find next even element
while (arr[e] % 2 == 1
|| arr[e] < 0 )
e++;
r = e;
}
// Mark them visited
arr[i] *= - 1 ;
arr[r] *= - 1 ;
// Swap them
int tmp = arr[i];
arr[i] = arr[r];
arr[r] = tmp;
}
// Print the final array
for ( int i = 0 ; i < n; i++) {
System.out.print(
(- 1 * arr[i]) + " " );
}
}
// Driver Code
public static void main(String[] args)
{
// Given array arr[]
int arr[] = { 1 , 3 , 2 , 4 };
// Function Call
swapEvenOdd(arr);
}
} |
# Python3 program for the above approach # Function to replace odd elements # with even elements and vice versa def swapEvenOdd(arr):
# Length of the array
n = len (arr)
o = - 1
e = - 1
# Traverse the given array
for i in range (n):
# If arr[i] is visited
if (arr[i] < 0 ):
continue
r = - 1
if (arr[i] % 2 = = 0 ):
o + = 1
# Find the next odd element
while (arr[o] % 2 = = 0 or
arr[o] < 0 ):
o + = 1
r = o
else :
e + = 1
# Find next even element
while (arr[e] % 2 = = 1 or
arr[e] < 0 ):
e + = 1
r = e
# Mark them visited
arr[i] * = - 1
arr[r] * = - 1
# Swap them
tmp = arr[i]
arr[i] = arr[r]
arr[r] = tmp
# Print the final array
for i in range (n):
print (( - 1 * arr[i]), end = " " )
# Driver Code if __name__ = = '__main__' :
# Given array arr
arr = [ 1 , 3 , 2 , 4 ]
# Function Call
swapEvenOdd(arr)
# This code is contributed by Amit Katiyar |
// C# program for the above approach using System;
class GFG{
// Function to replace odd elements // with even elements and vice versa static void swapEvenOdd( int []arr)
{ // Length of the array
int n = arr.Length;
int o = -1, e = -1;
// Traverse the given array
for ( int i = 0; i < n; i++)
{
// If arr[i] is visited
if (arr[i] < 0)
continue ;
int r = -1;
if (arr[i] % 2 == 0)
{
o++;
// Find the next odd element
while (arr[o] % 2 == 0 ||
arr[o] < 0)
o++;
r = o;
}
else
{
e++;
// Find next even element
while (arr[e] % 2 == 1 ||
arr[e] < 0)
e++;
r = e;
}
// Mark them visited
arr[i] *= -1;
arr[r] *= -1;
// Swap them
int tmp = arr[i];
arr[i] = arr[r];
arr[r] = tmp;
}
// Print the readonly array
for ( int i = 0; i < n; i++)
{
Console.Write((-1 * arr[i]) + " " );
}
} // Driver Code public static void Main(String[] args)
{ // Given array []arr
int []arr = { 1, 3, 2, 4 };
// Function Call
swapEvenOdd(arr);
} } // This code is contributed by Amit Katiyar |
<script> // JavaScript program for the above approach // Function to replace odd elements // with even elements and vice versa function swapEvenOdd(arr, n)
{ let o = -1, e = -1;
// Traverse the given array
for (let i = 0; i < n; i++)
{
// If arr[i] is visited
if (arr[i] < 0)
continue ;
let r = -1;
if (arr[i] % 2 == 0)
{
o++;
// Find the next odd element
while (arr[o] % 2 == 0 || arr[o] < 0)
o++;
r = o;
}
else
{
e++;
// Find next even element
while (arr[e] % 2 == 1 || arr[e] < 0)
e++;
r = e;
}
// Mark them visited
arr[i] *= -1;
arr[r] *= -1;
// Swap them
let tmp = arr[i];
arr[i] = arr[r];
arr[r] = tmp;
}
// Print the final array
for (let i = 0; i < n; i++)
{
document.write((-1 * arr[i]) + " " );
}
} // Driver Code // Given array arr[]
let arr = [ 1, 3, 2, 4 ];
// Length of the array
let n = arr.length;
// Function Call
swapEvenOdd(arr, n);
//This code is contributed by Mayank Tyagi </script> |
2 4 1 3
Time Complexity: O(N)
Auxiliary Space: O(N)
Efficient Approach 2 :
Another way to do this would be by using a stack. Follow the below steps:
- Take element at index from the array arr[] and push to a stack
- Iterate arr[] from index i = 1 to end and do the following:
- If arr[i] and item on top of the stack are not both even or not both odd, pop and swap
- Else push item to stack
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to replace odd elements // with even elements and vice versa void swapEvenOdd( int arr[], int N)
{ stack<pair< int , int > > stack;
// Push the first element to stack
stack.push({ 0, arr[0] });
// iterate the array and swap even and odd
for ( int i = 1; i < N; i++)
{
if (!stack.empty())
{
if (arr[i] % 2 != stack.top().second % 2)
{
// pop and swap
pair< int , int > pop = stack.top();
stack.pop();
int index = pop.first, val = pop.second;
arr[index] = arr[i];
arr[i] = val;
}
else
stack.push({ i, arr[i] });
}
else
stack.push({ i, arr[i] });
}
// print the arr[]
for ( int i = 0; i < N; i++)
cout << arr[i] << " " ;
} // Driven Program int main()
{ // Given array arr[]
int arr[] = { 1, 3, 2, 4 };
// Stores the length of array
int N = sizeof (arr) / sizeof (arr[0]);
// Function Call
swapEvenOdd(arr, N);
return 0;
} // This code is contributed by Kingash. |
// Java program for the above approach import java.io.*;
import java.lang.*;
import java.util.*;
class GFG {
// Function to replace odd elements
// with even elements and vice versa
static void swapEvenOdd( int arr[], int N)
{
Stack< int []> stack = new Stack<>();
// Push the first element to stack
stack.push( new int [] { 0 , arr[ 0 ] });
// iterate the array and swap even and odd
for ( int i = 1 ; i < N; i++)
{
if (!stack.isEmpty())
{
if (arr[i] % 2 != stack.peek()[ 1 ] % 2 )
{
// pop and swap
int pop[] = stack.pop();
int index = pop[ 0 ], val = pop[ 1 ];
arr[index] = arr[i];
arr[i] = val;
}
else
stack.push( new int [] { i, arr[i] });
}
else
stack.push( new int [] { i, arr[i] });
}
// print the arr[]
for ( int i = 0 ; i < N; i++)
System.out.print(arr[i] + " " );
}
// Driver code
public static void main(String[] args)
{
// Given array arr
int arr[] = { 1 , 3 , 2 , 4 };
// length of the arr
int N = arr.length;
// Function Call
swapEvenOdd(arr, N);
}
} // This code is contributed by Kingash. |
# Python3 program for the above approach # Function to replace odd elements # with even elements and vice versa def swapEvenOdd(arr):
stack = []
# Push the first element to stack
stack.append(( 0 , arr[ 0 ],))
# iterate the array and swap even and odd
for i in range ( 1 , len (arr)):
if stack:
if arr[i] % 2 ! = stack[ - 1 ][ 1 ] % 2 :
#pop and swap
index, val = stack.pop( - 1 )
arr[index] = arr[i]
arr[i] = val
else :
stack.append((i, arr[i],))
else :
stack.append((i, arr[i],))
return arr
# Driver Code if __name__ = = '__main__' :
# Given array arr
arr = [ 1 , 3 , 2 , 4 ]
# Function Call
print (swapEvenOdd(arr))
# This code is contributed by Arunabha Choudhury |
// C# program for the above approach using System;
using System.Collections.Generic;
public class GFG
{ // Function to replace odd elements
// with even elements and vice versa
static void swapEvenOdd( int []arr, int N) {
Stack< int []> stack = new Stack< int []>();
// Push the first element to stack
stack.Push( new int [] { 0, arr[0] });
// iterate the array and swap even and odd
for ( int i = 1; i < N; i++)
{
if (stack.Count != 0)
{
if (arr[i] % 2 != stack.Peek()[1] % 2)
{
// pop and swap
int []pop = stack.Pop();
int index = pop[0], val = pop[1];
arr[index] = arr[i];
arr[i] = val;
}
else
stack.Push( new int [] { i, arr[i] });
}
else
stack.Push( new int [] { i, arr[i] });
}
// print the []arr
for ( int i = 0; i < N; i++)
Console.Write(arr[i] + " " );
}
// Driver code
public static void Main(String[] args)
{
// Given array arr
int []arr = { 1, 3, 2, 4 };
// length of the arr
int N = arr.Length;
// Function Call
swapEvenOdd(arr, N);
}
} // This code is contributed by gauravrajput1 |
<script> // Javascript program for the above approach // Function to replace odd elements // with even elements and vice versa function swapEvenOdd(arr, N)
{ let stack = [];
// Push the first element to stack
stack.push([0, arr[0]]);
// Iterate the array and swap even and odd
for (let i = 1; i < N; i++)
{
if (stack.length != 0)
{
if (arr[i] % 2 !=
stack[stack.length - 1][1] % 2)
{
// pop and swap
let pop = stack.pop();
let index = pop[0], val = pop[1];
arr[index] = arr[i];
arr[i] = val;
}
else
stack.push([i, arr[i]]);
}
else
stack.push([i, arr[i]]);
}
// print the arr[]
for (let i = 0; i < N; i++)
document.write(arr[i] + " " );
} // Driver code // Given array arr let arr = [ 1, 3, 2, 4 ]; // Length of the arr let N = arr.length; // Function Call swapEvenOdd(arr, N); // This code is contributed by avanitrachhadiya2155 </script> |
[4, 2, 3, 1]
Time Complexity: O(N)
Auxiliary Space: O(N)