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# Queries to minimize sum added to given ranges in an array to make their Bitwise AND non-zero

• Last Updated : 05 Nov, 2020

Given an array arr[] consisting of N integers an array Q[][] consisting of queries of the form {l, r}. For each query {l, r}, the task is to determine the minimum sum of all values that must be added to each array element in that range such that the Bitwise AND of all numbers in that range exceeds 0.
Note: Different values can be added to different integers in the given range.

Examples:

Input: arr[] = {1, 2, 4, 8}, Q[][] = {{1, 4}, {2, 3}, {1, 3}}
Output: 3 2 2
Explanation: Binary representation of array elements are as follows:
1 – 0001
2 – 0010
4 – 0100
8 – 1000
For first query {1, 4}, add 1 to all numbers present in the range except the first number. Therefore, Bitwise AND = (1 & 3 & 5 & 9) = 1 and minimum sum of elements added = 3.
For second query {2, 3}, add 2 to 3rd element. Therefore, Bitwise AND = (2 & 6) = 2 and minimum sum of elements added = 2.
For third query {1, 3}, add 1 to 2nd and 3rd elements. Therefore, Bitwise AND = (1 & 3 & 5) = 1 and minimum sum of elements added = 2.

Input: arr[] = {4, 6, 5, 3}, Q[][] = {{1, 4}}
Output: 1
Explanation: Optimal way to mmake the Bitwise AND non-zero is to add 1 to the last element. Therefore, bitwise AND = (4 & 6 & 5 & 4) = 4 and minimum sum of elements added = 1.

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

Approach: The idea is to first observe that the Bitwise AND of all the integers in the range [l, r] can only be non-zero if a bit at a particular index is set for each integer in that range.
Follow the steps below to solve the problem:

1. Initialize a 2D vector pre where pre[i][j] stores the minimum sum of integers to be added to all integers from index 0 to index i such that, for each of them, their jth bit is set.
2. For each element at index i, check for each of its bit from j = 0 to 31.
3. Initialize a variable sum with 0
4. If jth bit is set, update pre[i][j] as pre[i][j]=pre[i-1][j] and increment sum by 2j. Otherwise, update pre[i][j] = pre[i-1][j] + 2j – sum where (2j-sum) is the value that must be added to set the jth bit of arr[i].
5. Now, for each query {l, r}, the minimum sum required to set jth bit of all elements in the given range is pre[r][j] – pre[l-1][j].
6. For each query {l, r}, find the answer for each bit from j = 0 to 31 and print the minimum amongst them.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach`` ` `#include ``using` `namespace` `std;`` ` `// Function to find the min sum required``// to set the jth bit of all integer``void` `processing(vector<``int``> v,``                ``vector >& pre)``{`` ` `    ``// Number of elements``    ``int` `N = v.size();`` ` `    ``// Traverse all elements``    ``for` `(``int` `i = 0; i < N; i++) {`` ` `        ``// Binary representation``        ``bitset<32> b(v[i]);`` ` `        ``long` `long` `int` `sum = 0;`` ` `        ``// Take previous values``        ``if` `(i != 0) {`` ` `            ``pre[i] = pre[i - 1];``        ``}`` ` `        ``// Processing each bit and``        ``// store it in 2d vector``        ``for` `(``int` `j = 0; j < 32; j++) {`` ` `            ``if` `(b[j] == 1) {`` ` `                ``sum += 1ll << j;``            ``}``            ``else` `{`` ` `                ``pre[i][j]``                    ``+= (1ll << j) - sum;``            ``}``        ``}``    ``}``}`` ` `// Function to print the minimum``// sum for each query``long` `long` `int``process_query(vector > Q,``              ``vector >& pre)``{`` ` `    ``// Stores the sum for each query``    ``vector<``int``> ans;`` ` `    ``for` `(``int` `i = 0; i < Q.size(); i++) {`` ` `        ``// Update wrt 0-based index``        ``--Q[i][0], --Q[i][1];`` ` `        ``// Initizlize answer``        ``long` `long` `int` `min1 = INT_MAX;`` ` `        ``// Find minimum sum for each bit``        ``if` `(Q[i][0] == 0) {`` ` `            ``for` `(``int` `j = 0; j < 32; j++) {`` ` `                ``min1 = min(pre[Q[i][1]][j], min1);``            ``}``        ``}``        ``else` `{`` ` `            ``for` `(``int` `j = 0; j < 32; j++) {`` ` `                ``min1 = min(pre[Q[i][1]][j]``                               ``- pre[Q[i][0] - 1][j],``                           ``min1);``            ``}``        ``}`` ` `        ``// Store the answer for``        ``// each query``        ``ans.push_back(min1);``    ``}`` ` `    ``// Print the answer vector``    ``for` `(``int` `i = 0; i < ans.size(); i++) {``        ``cout << ans[i] << ``" "``;``    ``}``}`` ` `// Driver Code``int` `main()``{`` ` `    ``// Given array``    ``vector<``int``> arr = { 1, 2, 4, 8 };`` ` `    ``// Given Queries``    ``vector > Q``        ``= { { 1, 4 }, { 2, 3 }, { 1, 3 } };`` ` `    ``// 2d Prefix vector``    ``vector > pre(``        ``100001, vector<``long` `long` `int``>(32, 0));`` ` `    ``// Preprocessing``    ``processing(arr, pre);`` ` `    ``// Function call for queries``    ``process_query(Q, pre);`` ` `    ``return` `0;``}`
Output:
```3 2 2
```

Time Complexity: O(N*32 + sizeof(Q)*32)

Auxiliary Space: O(N*32)

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