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# Replace every element of the array by its previous element

Given an array arr, the task is to replace each element of the array with the element that appears before it and replace the first element with -1

Examples:

Input: arr[] = {5, 1, 3, 2, 4}
Output: -1 5 1 3 2

Input: arr[] = {6, 8, 32, 12, 14, 10, 25 }
Output: -1 6 8 32 12 14 10

Approach: Traverse the array from n – 1 to 1 and update arr[i] = arr[i-1]. In the end set a = -1 and print the contents of the updated array.

Below is the implementation of the above approach:

## C++

 `// C++ program to replace every element of the array``// with the element that appears before it``#include ``using` `namespace` `std;` `// Function to print the array after replacing every element``// of the array with the element that appears before it``void` `updateArray(``int` `arr[], ``int` `n)``{``    ``// Update array``    ``for` `(``int` `i = n - 1; i > 0; i--)``        ``arr[i] = arr[i - 1];` `    ``// Change the first element to -1``    ``arr = -1;` `    ``// Print the updated array``    ``for` `(``int` `i = 0; i < n; i++)``        ``cout << arr[i] << ``" "``;``}` `// Driver program``int` `main()``{``    ``int` `arr[] = { 5, 1, 3, 2, 4 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);``    ``updateArray(arr, N);``    ``return` `0;``}`

## C

 `// C program to replace every element of the array``// with the element that appears before it``#include ` `// Function to print the array after replacing every element``// of the array with the element that appears before it``void` `updateArray(``int` `arr[], ``int` `n)``{``    ``// Update array``    ``for` `(``int` `i = n - 1; i > 0; i--)``        ``arr[i] = arr[i - 1];` `    ``// Change the first element to -1``    ``arr = -1;` `    ``// Print the updated array``    ``for` `(``int` `i = 0; i < n; i++)``      ``printf``(``"%d "``,arr[i]);``}` `// Driver program``int` `main()``{``    ``int` `arr[] = { 5, 1, 3, 2, 4 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);``    ``updateArray(arr, N);``    ``return` `0;``}` `// This code is contributed by kothavvsaakash.`

## Java

 `// Java program to replace every element``// of the array with the element that``// appears before it``class` `GFG``{` `// Function to print the array after``// replacing every element of the``// array with the element that appears``// before it``static` `void` `updateArray(``int` `arr[], ``int` `n)``{``    ``// Update array``    ``for` `(``int` `i = n - ``1``; i > ``0``; i--)``        ``arr[i] = arr[i - ``1``];` `    ``// Change the first element to -1``    ``arr[``0``] = -``1``;` `    ``// Print the updated array``    ``for` `(``int` `i = ``0``; i < n; i++)``        ``System.out.print(arr[i] + ``" "``);``}` `// Driver Code``public` `static` `void` `main(String []args)``{``    ``int` `arr[] = { ``5``, ``1``, ``3``, ``2``, ``4` `} ;``    ``int` `N = arr.length ;``    ``updateArray(arr, N);``}``}` `// This code is contributed by Ryuga`

## Python3

 `# Python 3 program to replace every element``# of the array with the element that appears``# before it` `# Function to print the array after replacing``# every element of the array with the element``# that appears before it``def` `updateArray(arr, n):``    ` `    ``# Update array``    ``i ``=` `n ``-` `1``    ``while``(i > ``0``):``        ``arr[i] ``=` `arr[i ``-` `1``]``        ``i ``-``=` `1` `    ``# Change the first element to -1``    ``arr[``0``] ``=` `-``1` `    ``# Print the updated array``    ``for` `i ``in` `range``(``0``, n, ``1``):``        ``print``(arr[i], end ``=` `" "``)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``arr ``=` `[``5``, ``1``, ``3``, ``2``, ``4``]``    ``N ``=` `len``(arr)``    ``updateArray(arr, N)` `# This code is contributed by``# Surendra_Gangwar`

## C#

 `// C# program to replace every element``// of the array with the element that``// appears before it``using` `System;` `class` `GFG``{` `// Function to print the array after``// replacing every element of the``// array with the element that appears``// before it``static` `void` `updateArray(``int``[] arr, ``int` `n)``{``    ``// Update array``    ``for` `(``int` `i = n - 1; i > 0; i--)``        ``arr[i] = arr[i - 1];` `    ``// Change the first element to -1``    ``arr = -1;` `    ``// Print the updated array``    ``for` `(``int` `i = 0; i < n; i++)``        ``Console.Write(arr[i] + ``" "``);``}` `// Driver Code``public` `static` `void` `Main()``{``    ``int``[] arr = { 5, 1, 3, 2, 4 } ;``    ``int` `N = arr.Length ;``    ``updateArray(arr, N);``}``}` `// This code is contributed``// by Akanksha Rai`

## PHP

 ` 0; ``\$i``--)``        ``\$arr``[``\$i``] = ``\$arr``[``\$i` `- 1];` `    ``// Change the first element to -1``    ``\$arr`` = -1;` `    ``// Print the updated array``    ``for` `(``\$i` `= 0; ``\$i` `< ``\$n``; ``\$i``++)``        ``echo` `\$arr``[``\$i``] ,``" "``;``}` `// Driver Code``\$arr` `= ``array``(5, 1, 3, 2, 4 );``\$N` `= sizeof(``\$arr``);``updateArray(``\$arr``, ``\$N``);` `// This code is contributed``// by Sach_Code``?>`

## Javascript

 ``

Output

`-1 5 1 3 2 `

Complexity Analysis:

• Time Complexity: O(n)
•  Auxiliary Space: O(1)

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