Skip to content
Related Articles

Related Articles

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.

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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program for the above approach
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the min sum required
// to set the jth bit of all integer
void processing(vector<int> v,
                vector<vector<long long int> >& 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<vector<int> > Q,
              vector<vector<long long int> >& 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<vector<int> > Q
        = { { 1, 4 }, { 2, 3 }, { 1, 3 } };
  
    // 2d Prefix vector
    vector<vector<long long int> > pre(
        100001, vector<long long int>(32, 0));
  
    // Preprocessing
    processing(arr, pre);
  
    // Function call for queries
    process_query(Q, pre);
  
    return 0;
}

chevron_right


Output:

3 2 2

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

Auxiliary Space: O(N*32)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up
Recommended Articles
Page :