# Rearrange array such that even positioned are greater than odd in C++

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We are given an integer type array containing both positive and negative numbers, let's say, arr[] of any given size. The task is to rearrange an array in such a manner that all the elements at an even position or index should be greater than the elements at an odd position or index and print the result.

## Let us see various input output scenarios for this −

Input − int arr[] = {2, 1, 4, 3, 6, 5, 8, 7}

Output − Array before Arrangement: 2 1 4 3 6 5 8 7 Rearrangement of an array such that even positioned are greater than odd is: 1 2 3 4 5 6 7 8

Explanation − we are given an integer array of size 8 containing both positive and negative elements. Now, we will rearrange the array in such a manner that all the elements at an even position are greater than the elements at an odd position and the array formed after performing this operation is 1 2 3 4 5 6 7 8.

Input − int arr[] = {-3, 2, -4, -1}

Output − Array before Arrangement: -3 2 -4 -1 Rearrangement of an array such that even positioned are greater than odd is: -4 -3 -1 2

Explanation −we are given an integer array of size 8 containing both positive and negative elements. Now, we will rearrange the array in such a manner that all the elements at an even position are greater than the elements at an odd position and the array formed after performing this operation is -4 -3 -1 2.

## Approach used in the below program is as follows

• Input an array of integer type elements and calculate the size of an array.

• Sort an array using the sort method of C++ STL by passing array and size of an array to the sort function.

• Declare an integer variable and set it with the call to the function Rearrangement(arr, size)

• Inside the function Rearrangement(arr, size)

• Declare an integer type array let’s say, ptr[size] of same size of array arr[size]

• Declare temporary integer type variables i.e. first to 0 and last to size -1.

• Start loop FOR from i to 0 till i is less than the size of an array. Inside the loop, check IF (i + 1) % 2 equals to 0 then set ptr[i] to arr[last--].

• ELSE, set ptr[i] to arr[first++].

• Print the result.

## Example

#include <bits/stdc++.h>
using namespace std;
void Rearrangement(int* arr, int size){
int ptr[size];
int first = 0;
int last = size - 1;
for (int i = 0; i < size; i++){
if((i + 1) % 2 == 0){
ptr[i] = arr[last--];
}
else{
ptr[i] = arr[first++];
}
}
}
int main(){
//input an array
int arr[] = {2, 1, 4, 3, 6, 5, 8, 7};
int size = sizeof(arr) / sizeof(arr[0]);
//print the original Array
cout<<"Array before Arrangement: ";
for (int i = 0; i < size; i++){
cout << arr[i] << " ";
}
//sort an Array
sort(arr, arr + size);
//calling the function to rearrange the array
Rearrangement(arr, size);
//print the array after rearranging the values
cout<<"Rearrangement of an array such that even positioned are greater than odd is: ";
for(int i = 0; i < size; i++){
cout<< arr[i] << " ";
}
return 0;
}

## Output

If we run the above code it will generate the following Output

Array before Arrangement: 2 1 4 3 6 5 8 7
Rearrangement of an array such that even positioned are greater than odd is: 1 2 3 4 5 6 7 8
Updated on 02-Nov-2021 06:43:15