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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 9Input: 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. Print the contents of the updated array in the end.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void printArr(int arr[], int n)
{
for (int i = 0; i < n; i++)
cout << arr[i] << " ";
}
void updateArr(int arr[], int n)
{
for (int i = 0; i < n; i++)
if ((i + 1) % 2 == 1)
arr[i]++;
else
arr[i]--;
printArr(arr, n);
}
int main()
{
int arr[] = { 3, 6, 8 };
int n = sizeof(arr) / sizeof(arr[0]);
updateArr(arr, n);
return 0;
}
Java
class GfG
{
static void printArr(int arr[], int n)
{
for (int i = 0; i < n; i++)
System.out.print(arr[i] + " ");
}
static void updateArr(int arr[], int n)
{
for (int i = 0; i < n; i++)
if ((i + 1) % 2 == 1)
arr[i]++;
else
arr[i]--;
printArr(arr, n);
}
public static void main(String[] args)
{
int arr[] = { 3, 6, 8 };
int n = arr.length;
updateArr(arr, n);
}
}
Python3
def printArr(arr, n):
for i in range(0, n):
print(arr[i], end = " ");
def updateArr(arr, n):
for i in range(0, n):
if ((i + 1) % 2 == 1):
arr[i] += 1;
else:
arr[i] -= 1;
printArr(arr, n);
if __name__ == '__main__':
arr = [3, 6, 8];
n = len(arr);
updateArr(arr, n);
C#
class GfG
{
static void printArr(int []arr, int n)
{
for (int i = 0; i < n; i++)
System.Console.Write(arr[i] + " ");
}
static void updateArr(int []arr, int n)
{
for (int i = 0; i < n; i++)
if ((i + 1) % 2 == 1)
arr[i]++;
else
arr[i]--;
printArr(arr, n);
}
static void Main()
{
int []arr = { 3, 6, 8 };
int n = arr.Length;
updateArr(arr, n);
}
}
PHP
<?php
function printArr($arr, $n)
{
for ($i = 0; $i < $n; $i++)
echo $arr[$i] . " ";
}
function updateArr($arr, $n)
{
for ($i = 0; $i < $n; $i++)
if (($i + 1) % 2 == 1)
$arr[$i]++;
else
$arr[$i]--;
printArr($arr, $n);
}
$arr = array( 3, 6, 8 );
$n = count($arr);
updateArr($arr, $n);
?>
Javascript
<script>
function printArr(arr, n)
{
var i;
for (i = 0; i < n; i++)
document.write(arr[i] + " ");
}
function updateArr(arr, n)
{
var i;
for (i = 0; i < n; i++)
if ((i + 1) % 2 == 1)
arr[i]++;
else
arr[i]--;
printArr(arr, n);
}
var arr = [3, 6, 8];
var n = arr.length;
updateArr(arr, n);
</script>
Time Complexity: O(n)
Auxiliary Space: O(1)