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Increment odd positioned elements by 1 and decrement even positioned elements by...

 3 years ago
source link: https://www.geeksforgeeks.org/increment-odd-positioned-elements-by-1-and-decrement-even-positioned-elements-by-1-in-an-array/
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Increment odd positioned elements by 1 and decrement even positioned elements by 1 in an Array

  • Difficulty Level : Easy
  • Last Updated : 26 Apr, 2021

Given an array arr[], the task is increment all the odd positioned elements by 1 and decrement all the even positioned elements by 1.
Examples: 

Input: arr[] = {3, 6, 8} 
Output: 4 5 9
Input: arr[] = {9, 7, 3} 
Output: 10 6 4

Approach: Traverse the array element by element and if the current element’s position is odd then increment it by 1 else decrement it by 1. Prin the contents of the updated array in the end.
Below is the implementation of the above approach:  

  • Python3
  • Javascript
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Utility function to print
// the contents of an array
void printArr(int arr[], int n)
{
for (int i = 0; i < n; i++)
cout << arr[i] << " ";
}
// Function to increment all the odd
// positioned elements by 1 and decrement
// all the even positioned elements by 1
void updateArr(int arr[], int n)
{
for (int i = 0; i < n; i++)
// If current element is odd positioned
if ((i + 1) % 2 == 1)
arr[i]++;
// If even positioned
else
arr[i]--;
// Print the updated array
printArr(arr, n);
}
// Driver code
int main()
{
int arr[] = { 3, 6, 8 };
int n = sizeof(arr) / sizeof(arr[0]);
updateArr(arr, n);
return 0;
}
Output: 
4 5 9

Time Complexity: O(n)

Auxiliary Space: O(1)

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