# Minimum length Subarray starting from each index with maximum OR

• Difficulty Level : Medium
• Last Updated : 23 Aug, 2022

Given an array of size N. Consider all the subarrays starting at each index from [0, N – 1]. Determine the length of the smallest subarray starting from each index whose bitwise OR  is maximum.

Examples:

Input: N = 5, A = {1, 2, 3, 4, 5}
Output: {4, 3, 2, 2, 1}
Explanation:

• For i = 1, and size of subarray = 1, score of this subarray is 1.
For size = 2, the value is 1 | 2 = 3.
For size = 3, the value is 1 | 2 | 3 = 3.
For size = 4, the value is 1 | 2 | 3 | 4 = 7 and
for size = 5, the value of this subarray is 1 | 2 | 3 | 4 | 5 = 7.
You can see that the maximum value is 7, and the smallest size
of the subarray starting at 1 with this value is of size 4.
So the maximum answer starting from 1st index is equal to 4.
• For i = 2 and size = 1, the value is 2.
For size = 2, the value is 2 | 3 = 3.
For size = 3, the value is 2 | 3 | 4 = 7.
For size = 4, the value is 2 | 3 | 4 | 5 = 7.
You can see that the maximum score is 7, and the smallest size
of the subarray starting at 2, with this value is of size 3.
So the maximum answer starting from 2nd index is equal to 3.
• For i = 3 and size = 1, the value is 3.
For size = 2, the value is 3 | 4 = 7.
For size = 3, the value is 3 | 4 | 5 = 7.
So for i = 3 the size is 2
• For i = 4 and size = 1, the score is 4 and for size = 2, the score is 4 | 1 = 5.
So the size is equal to 2.
• For i = 5 only one subarray is there of size 1.

Input: N = 7, A = {2, 4, 3, 1, 5, 4, 6}
Output: {3, 2, 3, 4, 3, 2, 1}

Naive Approach: To solve the problem follow the below idea:

For each index find all the subarrays starting from that index and find the smallest one with maximum bitwise OR.

Follow the given steps to solve the problem using the above approach:

• Traverse the array using two nested for loops, to find every possible subarray
• Calculate the OR value of every subarray and update the maximum answer found so far for the current starting index.
• Push the minimum length subarray size, with maximum value into the answer array.
• Return the answer array.

Below is the implementation for the above approach:

## C++

 `// C++ program for the above approach`   `#include ` `using` `namespace` `std;`   `// This function is for finding the minimum` `// length maximum OR subarray, for each index` `vector<``int``> solve(``int` `arr[], ``int` `N)` `{` `    ``vector<``int``> len;`   `    ``for` `(``int` `i = 0; i < N; i++) {` `        ``int` `mxor = 0, mnlen = 0, OR = 0;` `        ``for` `(``int` `j = i; j < N; j++) {` `            ``OR = OR | arr[j];`   `            ``// Updating maximum value found` `            ``// so far` `            ``if` `(mxor < OR) {` `                ``mxor = OR;` `                ``mnlen = j - i + 1;` `            ``}` `        ``}` `        ``len.push_back(mnlen);` `    ``}`   `    ``return` `len;` `}`   `// Driver's code` `int` `main()` `{` `    ``int` `arr[] = { 1, 2, 3, 4, 5 };` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);`   `    ``// Function call` `    ``vector<``int``> mnlen = solve(arr, N);`   `    ``for` `(``int` `i = 0; i < N; i++)` `        ``cout << mnlen[i] << ``" "``;` `    ``return` `0;` `}`

## Java

 `/*package whatever //do not write package name here */`   `import` `java.io.*;` `import` `java.util.*;`   `class` `GFG {`   `  ``// This function is for finding the minimum length` `  ``// maximum OR subarray, for each index` `  ``static` `List solve(``int``[] arr, ``int` `N)` `  ``{` `    ``List len = ``new` `ArrayList<>();`   `    ``for` `(``int` `i = ``0``; i < N; i++) {` `      ``int` `mxor = ``0``, mnlen = ``0``, OR = ``0``;` `      ``for` `(``int` `j = i; j < N; j++) {` `        ``OR = OR | arr[j];`   `        ``// Updating maximum value found so far` `        ``if` `(mxor < OR) {` `          ``mxor = OR;` `          ``mnlen = j - i + ``1``;` `        ``}` `      ``}` `      ``len.add(mnlen);` `    ``}` `    ``return` `len;` `  ``}`   `  ``public` `static` `void` `main(String[] args)` `  ``{` `    ``int``[] arr = { ``1``, ``2``, ``3``, ``4``, ``5` `};` `    ``int` `N = arr.length;`   `    ``// Function call` `    ``List mnlen = solve(arr, N);`   `    ``for` `(``int` `i = ``0``; i < N; i++) {` `      ``System.out.print(mnlen.get(i) + ``" "``);` `    ``}` `  ``}` `}`   `// This code is contributed by lokesh (lokeshmvs21).`

## Python

 `# Python program for the above approach`   `# This function is for finding the minimum` `# length maximum OR subarray, for each index` `def` `solve(arr, N):` `    ``len` `=`  `list``()`   `    ``for` `i ``in` `range``(``0` `,N):` `        ``mxor ``=` `0` `        ``mnlen ``=` `0` `        ``OR ``=` `0` `        ``for` `j ``in` `range``(i, N): ` `            ``OR ``=` `OR | arr[j];`   `            ``# Updating maximum value found` `            ``# so far` `            ``if` `(mxor < OR):` `                ``mxor ``=` `OR` `                ``mnlen ``=` `j ``-` `i ``+` `1` `        ``len``.append(mnlen)`   `    ``return` `len`   `# Driver's code` `if` `__name__ ``=``=` `"__main__"``:` `  `  `    ``arr ``=` `[ ``1``, ``2``, ``3``, ``4``, ``5` `]` `    ``N ``=` `len``(arr)`   `    ``# Function call` `    ``mnlen ``=` `solve(arr, N)`   `    ``for` `i ``in` `range``(``0` `,N):` `        ``print` `mnlen[i],` `  `  `  ``# This code is contributed by hrithikgarg03188.`

## C#

 `// C# program for the above approach`   `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG {`   `  ``// This function is for finding the minimum length` `  ``// maximum OR subarray, for each index` `  ``static` `List<``int``> solve(``int``[] arr, ``int` `N)` `  ``{` `    ``List<``int``> len = ``new` `List<``int``>();`   `    ``for` `(``int` `i = 0; i < N; i++) {` `      ``int` `mxor = 0, mnlen = 0, OR = 0;` `      ``for` `(``int` `j = i; j < N; j++) {` `        ``OR = OR | arr[j];`   `        ``// Updating maximum value found so far` `        ``if` `(mxor < OR) {` `          ``mxor = OR;` `          ``mnlen = j - i + 1;` `        ``}` `      ``}` `      ``len.Add(mnlen);` `    ``}` `    ``return` `len;` `  ``}`   `  ``public` `static` `void` `Main()` `  ``{` `    ``int``[] arr = { 1, 2, 3, 4, 5 };` `    ``int` `N = arr.Length;`   `    ``// Function call` `    ``List<``int``> mnlen = solve(arr, N);`   `    ``for` `(``int` `i = 0; i < N; i++) {` `      ``Console.Write(mnlen[i] + ``" "``);` `    ``}` `  ``}` `}`   `// This code is contributed by Taranpreet`

## Javascript

 ``

Output

`4 3 2 2 1 `

Time Complexity: O(N2 )
Auxiliary Space: O(N)

Efficient approach: To solve the problem follow the below idea:

By observing the functionality of OR, bits can only be turned on using 1. So, start from the end and keep track of the minimum index that will keep the particular bit as a set and then we take a max of all the indexes that contain the set bits.

Follow the given steps to solve the problem using the above approach:

• Traverse the given array from end and update the minimum index of every set bit in the current element
• Take the maximum index of all the set bits so far, as the answer for the current index.
• Return the answer array

Below is the implementation for the above approach:

## C++

 `// C++ program for the above approach`   `#include ` `using` `namespace` `std;`   `// This function is for finding the` `// minimum length maximum OR subarray` `vector<``int``> solve(``int` `arr[], ``int` `N)` `{` `    ``vector<``int``> ans;` `    ``vector<``int``> nearest(32, -1);` `    ``for` `(``int` `i = N - 1; i >= 0; i--) {` `        ``for` `(``int` `j = 0; j < 32; j++) {`   `            ``// Nearest index where jth` `            ``// bit is set` `            ``if` `(arr[i] & (1 << j))` `                ``nearest[j] = i;` `        ``}`   `        ``int` `last_set_bit_index = i;`   `        ``// Finding Maximum index of all set bits` `        ``for` `(``int` `j = 0; j < 32; j++)` `            ``last_set_bit_index` `                ``= max(last_set_bit_index, nearest[j]);`   `        ``ans.push_back(last_set_bit_index - i + 1);` `    ``}`   `    ``// Reversing the answer vector` `    ``reverse(ans.begin(), ans.end());` `    ``return` `ans;` `}`   `// Driver code` `int` `main()` `{`   `    ``int` `arr[] = { 1, 2, 3, 4, 5 };` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);`   `    ``// Function call` `    ``vector<``int``> mnlen = solve(arr, N);` `    ``for` `(``int` `i = 0; i < N; i++)` `        ``cout << mnlen[i] << ``" "``;` `    ``return` `0;` `}`

## C#

 `// C# implementation of above approach` `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG {`   `  ``// This function is for finding the` `// minimum length maximum OR subarray` `static` `List<``int``> solve(``int``[] arr, ``int` `N)` `{` `     ``List<``int``> ans = ``new` `List<``int``>();` `    ``int``[] nearest = ``new` `int``;` `    ``for``(``int` `i =0; i<32; i++)` `    ``{` `        ``nearest[i] = -1;` `    ``}` `    `  `    ``for` `(``int` `i = N - 1; i >= 0; i--) {` `        ``for` `(``int` `j = 0; j < 32; j++) {` ` `  `            ``// Nearest index where jth` `            ``// bit is set` `            ``if` `((arr[i]  & (1 << j)) != 0)` `                ``nearest[j] = i;` `        ``}` ` `  `        ``int` `last_set_bit_index = i;` ` `  `        ``// Finding Maximum index of all set bits` `        ``for` `(``int` `j = 0; j < 32; j++)` `            ``last_set_bit_index` `                ``= Math.Max(last_set_bit_index, nearest[j]);` ` `  `        ``ans.Add(last_set_bit_index - i + 1);` `    ``}` ` `  `    ``// Reversing the answer vector` `    ``ans.Reverse();` `    ``return` `ans;` `}` `  ``// Driver Code` `  ``public` `static` `void` `Main()` `  ``{` `     ``int``[] arr = { 1, 2, 3, 4, 5 };` `    ``int` `N = arr.Length;` ` `  `    ``// Function call` `    ``List<``int``> mnlen = solve(arr, N);` `    ``for` `(``int` `i = 0; i < N; i++)` `        ``Console.Write(mnlen[i] + ``" "``);` `  ``}` `}`   `// This code is contributed by code_hunt.`

## Javascript

 ``

Output

`4 3 2 2 1 `

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

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