Table of Contents
Let us look at various input and output situations:
Approach used in the below program is as follows
Home Backend Development C++ Rearrange an array so that every odd-indexed element is greater than its previous element

Rearrange an array so that every odd-indexed element is greater than its previous element

Sep 01, 2023 am 09:45 AM
element rearrange odd index

Rearrange an array so that every odd-indexed element is greater than its previous element

We have an array of positive integer type, assumed to be arr[], with any size. The task is to rearrange the array so that all elements at odd index positions have greater values ​​than elements at even index positions, and print the result.

Let us look at various input and output situations:

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

Output − Array before sorting: 2 1 5 4 3 7 8 Rearrange the array so that every odd-indexed element is greater than its previous element: 1 4 2 5 3 8 7

Explanation - We are given an array of integers of size 7 . Now, if the element at even index is larger, we will swap the element at even index and the element at odd index

Arr[0] > arr[1] = call swap = {1, 2, 5, 4, 3, 7, 8}
Arr[2] > arr[3] = call swap = {1, 2, 4, 5, 3, 7, 8}
Arr[6] > arr[5] = call swap = {1, 2, 4, 5, 3, 8, 7}
Arr[2] > arr[1] = call swap = {1, 4, 2, 5, 3, 8, 7}
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input− int arr[] = {3, 2, 6, 9}

Output− Array before sorting: 3 2 6 9 Rearrangement of an array such that every odd indexed element is greater than it previous is: 2 3 6 9

Explanation − we are given an integer array of size 4. Now, we will swap the elements at even index with the elements at odd index if even indexed elements are greater i.e. Arr[0] > arr[1] = call swap = {2, 3, 6, 9}. No need to further call the swap method as all the elements at positions satisfies the conditions

Approach used in the below program is as follows

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

  • ##Print the array before arrangement and call the function Rearrangement(arr, size)

  • ##Inside the function Rearrangement(arr, size)
    • Create a variable of integer type let's say, ptr and set it with size-1.
    • ##Start loop FOR, from i to 0 till i less than ptr and i = i 1. Inside the loop, check if arr[i] is greater than arr[i 1] then call swap(arr[i], arr[i 1]).
    • Check IF size & 1 then start loop FOR from i to ptr till i greater than 0 and i = i - 2. Inside the loop, check IF arr[i] greater than arr[i - 1] then call swap(arr[i], arr[i-1])
    Print the array after the rearrangement of values ​​of an array.
  • Example
  • #include <iostream>
    using namespace std;
    void Rearrangement(int arr[], int size){
       int ptr = size - 1;
       for(int i = 0; i < ptr; i = i+2){
          if(arr[i] > arr[i+1]){
             swap(arr[i], arr[i+1]);
          }
       }
       if(size & 1){
          for(int i = ptr; i > 0; i = i-2){
             if(arr[i] > arr[i-1]){
                swap(arr[i], arr[i-1]);
             }
          }
       }
    }
    int main(){
       //input an array
       int arr[] = {2, 1, 5, 4, 3, 7, 8};
       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] << " ";
       }
       //calling the function to rearrange the array
       Rearrangement(arr, size);
       //print the array after rearranging the values
       cout<<"\nRearrangement of an array such that every odd indexed element is greater than it previous is: ";
       for(int i = 0; i < size; i++){
          cout<< arr[i] << " ";
       }
       return 0;
    }
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Output

If we run the above code, the following output will be generated

Array before Arrangement: 2 1 5 4 3 7 8
Rearrangement of an array such that every odd indexed element is greater than it previous is: 1 4 2 5 3 8 7
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