# Remove one element to get minimum OR value

Given an array arr[] of N elements, the task is to remove one element from the array such that the OR value of the array is minimized. Print the minimized value.

Examples:

Input: arr[] = {1, 2, 3}
Output: 3
All possible ways of deleting one element and their
corresponding OR values will be:
a) Remove 1 -> (2 | 3) = 3
b) Remove 2 -> (1 | 3) = 3
c) Remove 3 -> (1 | 2) = 3
Thus, the answer will be 3.

Input: arr[] = {2, 2, 2}
Output: 2

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive approach: One way will be to remove each element one by one and then finding the OR of the remaining elements. The time complexity of this approach will be O(N2).

Efficient approach: To solve the problem efficiently, the value of (OR(arr[0…i-1]) | OR(arr[i+1…N-1])) has to be determined for any element arr[i]. To do so, he prefix and the suffix OR arrays can be calcualted say pre[] and suf[] where pre[i] stores OR(arr[0…i]) and suff[i] stores OR(arr[i…N-1]). Then the OR value of the array after deleting the ith element can be calculated as (pre[i-1] | suff[i+1]) and the answer will be minimum of all the possible OR values.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the minimized OR ` `// after removing an element from the array ` `int` `minOR(``int``* arr, ``int` `n) ` `{ ` `    ``// Base case ` `    ``if` `(n == 1) ` `        ``return` `0; ` ` `  `    ``// Prefix and suffix OR array ` `    ``int` `pre[n], suf[n]; ` `    ``pre = arr, suf[n - 1] = arr[n - 1]; ` ` `  `    ``// Computing prefix/suffix OR arrays ` `    ``for` `(``int` `i = 1; i < n; i++) ` `        ``pre[i] = (pre[i - 1] | arr[i]); ` `    ``for` `(``int` `i = n - 2; i >= 0; i--) ` `        ``suf[i] = (suf[i + 1] | arr[i]); ` ` `  `    ``// To store the final answer ` `    ``int` `ans = min(pre[n - 2], suf); ` ` `  `    ``// Finding the final answer ` `    ``for` `(``int` `i = 1; i < n - 1; i++) ` `        ``ans = min(ans, (pre[i - 1] | suf[i + 1])); ` ` `  `    ``// Returning the final answer ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 2, 3 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``); ` ` `  `    ``cout << minOR(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG ` `{ ` ` `  `// Function to return the minimized OR ` `// after removing an element from the array ` `static` `int` `minOR(``int` `[]arr, ``int` `n) ` `{ ` `    ``// Base case ` `    ``if` `(n == ``1``) ` `        ``return` `0``; ` ` `  `    ``// Prefix and suffix OR array ` `    ``int` `[]pre = ``new` `int``[n]; ` `    ``int` `[]suf = ``new` `int``[n]; ` `    ``pre[``0``] = arr[``0``]; ` `    ``suf[n - ``1``] = arr[n - ``1``]; ` ` `  `    ``// Computing prefix/suffix OR arrays ` `    ``for` `(``int` `i = ``1``; i < n; i++) ` `        ``pre[i] = (pre[i - ``1``] | arr[i]); ` `    ``for` `(``int` `i = n - ``2``; i >= ``0``; i--) ` `        ``suf[i] = (suf[i + ``1``] | arr[i]); ` ` `  `    ``// To store the final answer ` `    ``int` `ans = Math.min(pre[n - ``2``], suf[``1``]); ` ` `  `    ``// Finding the final answer ` `    ``for` `(``int` `i = ``1``; i < n - ``1``; i++) ` `        ``ans = Math.min(ans, (pre[i - ``1``] |  ` `                             ``suf[i + ``1``])); ` ` `  `    ``// Returning the final answer ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr[] = { ``1``, ``2``, ``3` `}; ` `    ``int` `n = arr.length; ` ` `  `    ``System.out.print(minOR(arr, n)); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return the minimized OR ` `# after removing an element from the array ` `def` `minOR(arr, n): ` `     `  `    ``# Base case ` `    ``if` `(n ``=``=` `1``): ` `        ``return` `0` ` `  `    ``# Prefix and suffix OR array ` `    ``pre ``=` `[``0``] ``*` `n ` `    ``suf ``=` `[``0``] ``*` `n ` `    ``pre[``0``] ``=` `arr[``0``] ` `    ``suf[n ``-` `1``] ``=` `arr[n ``-` `1``] ` ` `  `    ``# Computing prefix/suffix OR arrays ` `    ``for` `i ``in` `range``(``1``, n): ` `        ``pre[i] ``=` `(pre[i ``-` `1``] | arr[i]) ` `    ``for` `i ``in` `range``(n ``-` `2``, ``-``1``, ``-``1``): ` `        ``suf[i] ``=` `(suf[i ``+` `1``] | arr[i]) ` ` `  `    ``# To store the final answer ` `    ``ans ``=` `min``(pre[n ``-` `2``], suf[``1``]) ` ` `  `    ``# Finding the final answer ` `    ``for` `i ``in` `range``(``1``, n ``-` `1``): ` `        ``ans ``=` `min``(ans, (pre[i ``-` `1``] | suf[i ``+` `1``])) ` ` `  `    ``# Returning the final answer ` `    ``return` `ans ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``arr ``=` `[``1``, ``2``, ``3``] ` `    ``n ``=` `len``(arr) ` ` `  `    ``print``(minOR(arr, n)) ` ` `  `# This code is contributed by Mohit Kumar `

## C#

 `// C# implementation of the approach  ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `    ``// Function to return the minimized OR  ` `    ``// after removing an element from the array  ` `    ``static` `int` `minOR(``int` `[]arr, ``int` `n)  ` `    ``{  ` `        ``// Base case  ` `        ``if` `(n == 1)  ` `            ``return` `0;  ` `     `  `        ``// Prefix and suffix OR array  ` `        ``int` `[]pre = ``new` `int``[n]; ` `        ``int` `[]suf = ``new` `int``[n];  ` `         `  `        ``pre = arr; ` `        ``suf[n - 1] = arr[n - 1];  ` `     `  `        ``// Computing prefix/suffix OR arrays  ` `        ``for` `(``int` `i = 1; i < n; i++)  ` `            ``pre[i] = (pre[i - 1] | arr[i]);  ` `             `  `        ``for` `(``int` `i = n - 2; i >= 0; i--)  ` `            ``suf[i] = (suf[i + 1] | arr[i]);  ` `     `  `        ``// To store the final answer  ` `        ``int` `ans = Math.Min(pre[n - 2], suf);  ` `     `  `        ``// Finding the final answer  ` `        ``for` `(``int` `i = 1; i < n - 1; i++)  ` `            ``ans = Math.Min(ans, (pre[i - 1] |  ` `                                 ``suf[i + 1]));  ` `     `  `        ``// Returning the final answer  ` `        ``return` `ans;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``static` `public` `void` `Main () ` `    ``{  ` `        ``int` `[]arr = { 1, 2, 3 };  ` `        ``int` `n = arr.Length;  ` `     `  `        ``Console.WriteLine(minOR(arr, n));  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

Output:

```3
```

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