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# Remove one element to get minimum OR value

• Last Updated : 30 Jan, 2023

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:
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

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, the prefix and the suffix OR arrays can be calculated 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`

## Javascript

 ``

Output:

`3`

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

Space Optimized Approach:

Count frequency of set bit of each element in 32 size array and traverse the array from most significant bit and check if any bit has frequency one, if we found such element than remove that particular element. The left element will participate in our actual result.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach``#include ``using` `namespace` `std;` `int` `minOR(``int` `arr[], ``int` `n)``{``    ``// Helper array where helper[i] will store the number of``    ``// elements in the array having the i'th bit set``    ``int` `helper;``    ``memset``(helper, 0, ``sizeof``(helper));``    ``for` `(``int` `i = 0; i < n; i++) {``        ``for` `(``int` `j = 0; j < 32; j++) {``            ``// checking if the i'th bit is set``            ``if` `((arr[i] & (1 << j))) {``                ``helper[j]++;``            ``}``        ``}``    ``}` `    ``// Traverse the helper array from most significant bit``    ``// to find out the first index i where helper[i] = 1.``    ``// this means that there is only one element. in the``    ``// array having i'th bit as set. Let's say this element``    ``// as current. Since we are traversing the array from``    ``// most significant bit removing current element will``    ``// minimize the bitwise OR of the entire array``    ``int` `first_index = -1;` `    ``for` `(``int` `i = 31; i >= 0; i--) {``        ``if` `(helper[i] == 1) {``            ``first_index = i;``            ``break``;``        ``}``    ``}` `    ``// If there is no such element for which helper[i] = 1,``    ``// it means that either helper[i] = 0 or helper[i] >= 2``    ``// for i'th index. if helper[i] >=2 this means that we``    ``// need to atleast remove two elements from the array to``    ``// obtain minimum possible bitwise OR but in the``    ``// question we need to remove exactly one element.` `    ``int` `sum = 0;``    ``if` `(first_index == -1) {``        ``for` `(``int` `i = 31; i >= 0; i--) {``            ``sum = sum | arr[i];``        ``}``    ``}``    ``else` `{` `        ``// Calculating the bitwise OR of all the bits that``        ``// are set excluding "first_index``        ``for` `(``int` `i = 0; i < n; i++) {``            ``if` `((arr[i] & (1 << first_index))) {``                ``continue``;``            ``}``            ``else` `{``                ``sum = sum | arr[i];``            ``}``        ``}``    ``}``    ``return` `sum;``}` `// Driver code.``int` `main()``{` `    ``int` `a[] = { 4, 9, 8 };``    ``int` `n = ``sizeof``(a) / ``sizeof``(a);` `    ``int` `ans = minOR(a, n);``    ``cout << ans << endl;``}`

## Java

 `// Java implementation of the above approach``import` `java.util.*;``public` `class` `GFG {` `  ``static` `int` `minOR(``int``[] arr, ``int` `n)``  ``{``    ``// Helper array where helper[i] will store the``    ``// number of elements in the array having the i'th``    ``// bit set``    ``int``[] helper = ``new` `int``[``32``];``    ``for` `(``int` `i = ``0``; i < ``32``; i++) {``      ``helper[i] = ``0``;``    ``}` `    ``for` `(``int` `i = ``0``; i < n; i++) {``      ``for` `(``int` `j = ``0``; j < ``32``; j++) {` `        ``// checking if the i'th bit is set``        ``if` `((arr[i] & (``1` `<< j)) != ``0``) {``          ``helper[j]++;``        ``}``      ``}``    ``}` `    ``// Traverse the helper array from most significant``    ``// bit to find out the first index i where helper[i]``    ``// = 1. this means that there is only one element.``    ``// in the array having i'th bit as set. Let's say``    ``// this element as current. Since we are traversing``    ``// the array from most significant bit removing``    ``// current element will minimize the bitwise OR of``    ``// the entire array``    ``int` `first_index = -``1``;` `    ``for` `(``int` `i = ``31``; i >= ``0``; i--) {``      ``if` `(helper[i] == ``1``) {``        ``first_index = i;``        ``break``;``      ``}``    ``}` `    ``// If there is no such element for which helper[i] =``    ``// 1, it means that either helper[i] = 0 or``    ``// helper[i] >= 2 for i'th index. if helper[i] >=2``    ``// this means that we need to atleast remove two``    ``// elements from the array to obtain minimum``    ``// possible bitwise OR but in the question we need``    ``// to remove exactly one element.` `    ``int` `sum = ``0``;``    ``if` `(first_index == -``1``) {``      ``for` `(``int` `i = ``31``; i >= ``0``; i--) {``        ``sum = sum | arr[i];``      ``}``    ``}``    ``else` `{` `      ``// Calculating the bitwise OR of all the bits``      ``// that are set excluding "first_index``      ``for` `(``int` `i = ``0``; i < n; i++) {``        ``if` `((arr[i] & (``1` `<< first_index)) != ``0``) {``          ``continue``;``        ``}``        ``else` `{``          ``sum = sum | arr[i];``        ``}``      ``}``    ``}``    ``return` `sum;``  ``}` `  ``// Driver code.``  ``public` `static` `void` `main(String args[])``  ``{` `    ``int``[] a = { ``4``, ``9``, ``8` `};``    ``int` `n = a.length;` `    ``int` `ans = minOR(a, n);``    ``System.out.println(ans);``  ``}``}` `// This code is contributed by Samim Hossain Mondal.`

## C#

 `// C# implementation of the above approach``using` `System;``class` `GFG {` `  ``static` `int` `minOR(``int``[] arr, ``int` `n)``  ``{``    ``// Helper array where helper[i] will store the``    ``// number of elements in the array having the i'th``    ``// bit set``    ``int``[] helper = ``new` `int``;``    ``for` `(``int` `i = 0; i < 32; i++) {``      ``helper[i] = 0;``    ``}` `    ``for` `(``int` `i = 0; i < n; i++) {``      ``for` `(``int` `j = 0; j < 32; j++)``      ``{``        ` `        ``// checking if the i'th bit is set``        ``if` `((arr[i] & (1 << j)) != 0) {``          ``helper[j]++;``        ``}``      ``}``    ``}` `    ``// Traverse the helper array from most significant``    ``// bit to find out the first index i where helper[i]``    ``// = 1. this means that there is only one element.``    ``// in the array having i'th bit as set. Let's say``    ``// this element as current. Since we are traversing``    ``// the array from most significant bit removing``    ``// current element will minimize the bitwise OR of``    ``// the entire array``    ``int` `first_index = -1;` `    ``for` `(``int` `i = 31; i >= 0; i--) {``      ``if` `(helper[i] == 1) {``        ``first_index = i;``        ``break``;``      ``}``    ``}` `    ``// If there is no such element for which helper[i] =``    ``// 1, it means that either helper[i] = 0 or``    ``// helper[i] >= 2 for i'th index. if helper[i] >=2``    ``// this means that we need to atleast remove two``    ``// elements from the array to obtain minimum``    ``// possible bitwise OR but in the question we need``    ``// to remove exactly one element.` `    ``int` `sum = 0;``    ``if` `(first_index == -1) {``      ``for` `(``int` `i = 31; i >= 0; i--) {``        ``sum = sum | arr[i];``      ``}``    ``}``    ``else` `{` `      ``// Calculating the bitwise OR of all the bits``      ``// that are set excluding "first_index``      ``for` `(``int` `i = 0; i < n; i++) {``        ``if` `((arr[i] & (1 << first_index)) != 0) {``          ``continue``;``        ``}``        ``else` `{``          ``sum = sum | arr[i];``        ``}``      ``}``    ``}``    ``return` `sum;``  ``}` `  ``// Driver code.``  ``public` `static` `void` `Main()``  ``{` `    ``int``[] a = { 4, 9, 8 };``    ``int` `n = a.Length;` `    ``int` `ans = minOR(a, n);``    ``Console.WriteLine(ans);``  ``}``}` `// This code is contributed by Samim Hossain Mondal.`

## Javascript

 `// Javascript implementation of the above approach` `function` `minOR(arr, n)``{``    ``// Helper array where helper[i] will store the number of``    ``// elements in the array having the i'th bit set``    ``let helper=``new` `Array(32).fill(0);``    ``for` `(let i = 0; i < n; i++) {``        ``for` `(let j = 0; j < 32; j++) {``            ``// checking if the i'th bit is set``            ``if` `((arr[i] & (1 << j))) {``                ``helper[j]++;``            ``}``        ``}``    ``}` `    ``// Traverse the helper array from most significant bit``    ``// to find out the first index i where helper[i] = 1.``    ``// this means that there is only one element. in the``    ``// array having i'th bit as set. Let's say this element``    ``// as current. Since we are traversing the array from``    ``// most significant bit removing current element will``    ``// minimize the bitwise OR of the entire array``    ``let first_index = -1;` `    ``for` `(let i = 31; i >= 0; i--) {``        ``if` `(helper[i] == 1) {``            ``first_index = i;``            ``break``;``        ``}``    ``}` `    ``// If there is no such element for which helper[i] = 1,``    ``// it means that either helper[i] = 0 or helper[i] >= 2``    ``// for i'th index. if helper[i] >=2 this means that we``    ``// need to atleast remove two elements from the array to``    ``// obtain minimum possible bitwise OR but in the``    ``// question we need to remove exactly one element.` `    ``let sum = 0;``    ``if` `(first_index == -1) {``        ``for` `(let i = 31; i >= 0; i--) {``            ``sum = sum | arr[i];``        ``}``    ``}``    ``else` `{` `        ``// Calculating the bitwise OR of all the bits that``        ``// are set excluding "first_index``        ``for` `(let i = 0; i < n; i++) {``            ``if` `((arr[i] & (1 << first_index))) {``                ``continue``;``            ``}``            ``else` `{``                ``sum = sum | arr[i];``            ``}``        ``}``    ``}``    ``return` `sum;``}` `// Driver code.` `    ``let a = [ 4, 9, 8 ];``    ``let n = a.length;` `    ``let ans = minOR(a, n);``    ``console.log(ans);`

## Python3

 `# python implementation of the above approach` `def` `minOR(arr, n):``    ``# Helper array where helper[i] will store the number of``    ``# elements in the array having the i'th bit set``    ``helper``=``[``0``]``*``32``;``    ``for` `i ``in` `range``(``0``,n):``        ``for` `j ``in` `range``(``0``,``32``):``            ``# checking if the i'th bit is set``            ``if` `((arr[i] & (``1` `<< j))):``                ``helper[j]``+``=``1``;` `    ``# Traverse the helper array from most significant bit``    ``# to find out the first index i where helper[i] = 1.``    ``# this means that there is only one element. in the``    ``# array having i'th bit as set. Let's say this element``    ``# as current. Since we are traversing the array from``    ``# most significant bit removing current element will``    ``# minimize the bitwise OR of the entire array``    ``first_index ``=` `-``1``;``    ` `    ``i``=``31``;``    ``while``(i>``=``0``):``        ``if` `(helper[i] ``=``=` `1``):``            ``first_index ``=` `i;``            ``break``;``        ``i``=``i``-``1``;``        ` `    ``# If there is no such element for which helper[i] = 1,``    ``# it means that either helper[i] = 0 or helper[i] >= 2``    ``# for i'th index. if helper[i] >=2 this means that we``    ``# need to atleast remove two elements from the array to``    ``# obtain minimum possible bitwise OR but in the``    ``# question we need to remove exactly one element.` `    ``sum` `=` `0``;``    ``if` `(first_index ``=``=` `-``1``):``        ``i``=``31``;``        ``while``(i>``=``0``):``            ``sum` `=` `sum` `| arr[i];``            ``i``-``=``1``        ` `    ``else``:` `        ``# Calculating the bitwise OR of all the bits that``        ``# are set excluding "first_index``        ``for` `i ``in` `range``(``0``,n):``            ``if` `((arr[i] & (``1` `<< first_index))) :``                ``continue``;``            ` `            ``else``:``                ``sum` `=` `sum` `| arr[i];``            ` `    ``return` `sum``;` `# Driver code.` `a ``=` `[ ``4``, ``9``, ``8` `];``n ``=` `len``(a);` `ans ``=` `minOR(a, n);``print``(ans);`

Output

`9`

Time Complexity: O(N)
Auxiliary Space: O(1)

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