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

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 make 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 j^th 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 j^th bit is set, update pre[i][j] as pre[i][j]=pre[i-1][j] and increment sum by 2^j. Otherwise, update pre[i][j] = pre[i-1][j] + 2^j – sum where (2^j-sum) is the value that must be added to set the j^th bit of arr[i].
5. Now, for each query {l, r}, the minimum sum required to set j^th 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 <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;
}

Java

// Java code to implement the approach
import java.util.Arrays;

class Main {
  public static void main(String[] args)
  {

    // Given array
    long[] arr = { 1, 2, 4, 8 };

    // Given Queries
    long[][] Q = { new long[] { 1, 4 }, new long[] { 2, 3 }, new long[] { 1, 3 } };

    // 2d Prefix vector
    long[][] pre = new long[100001][];
    Arrays.setAll(pre, i -> new long[32]);

    // Preprocessing
    processing(arr, pre);

    // Function call for queries
    long[] ans = processQuery(Q, pre);

    // Print the answer vector
    System.out.println(String.join(", ", Arrays.toString(ans)));
  }

  private static void processing(long[] v, long[][] pre) {
    // Number of elements
    long N = v.length;

    // Traverse all elements
    for (long i = 0; i < N; i++) {
      // Binary representation
      String b = Long.toBinaryString(v[(int) i]);
      b = "0".repeat((int) (32 - b.length())) + b;
      b = new StringBuilder(b).reverse().toString();

      long sum = 0;

      // Take previous values
      if (i != 0) {
        pre[(int) i] = Arrays.copyOf(pre[(int) (i - 1)], pre[(int) (i - 1)].length);
      }

      // Processing each bit and store it in 2d vector
      for (int j = 0; j < 32; j++) {
        if (b.charAt(j) == '1') {
          sum += (long) Math.pow(2, j);
        } else {
          pre[(int) i][j] += (long) Math.pow(2, j) - sum;
        }
      }
    }
  }

  private static long[] processQuery(long[][] Q, long[][] pre) {
    // Stores the sum for each query
    long[] ans = new long[Q.length];

    for (long i = 0; i < Q.length; i++) {
      // Update wrt 0-based index
      Q[(int) i][0] -= 1;
      Q[(int) i][1] -= 1;

      // Initialize answer
      long min1 = Long.MAX_VALUE;

      // Find minimum sum for each bit
      if (Q[(int) i][0] == 0) {
        for (long j = 0; j < 32; j++) {
          min1 =(long) Math.min(pre[(int) Q[(int) i][1]][(int) j], min1);
        }
      } else {
        for (long j = 0; j < 32; j++) {
          min1 =(long) Math.min(
            pre[(int) Q[(int) i][1]][(int) j] - pre[(int) (Q[(int) i][0] - 1)][(int) j], min1);
        }
      }

      // Store the answer for each query
      ans[(int)i] = min1;
    }

    return ans;
  }
}

// This code is contributed by phasing17

Python3

from typing import List, Tuple

def processing(v: List[int], pre: List[List[int]]) -> None:
    # Number of elements
    N = len(v)

    # Traverse all elements
    for i in range(N):
        # Binary representation
        b = bin(v[i])[2:]
        b = "0"*(32-len(b)) + b
        b = b[::-1]

        sum = 0

        # Take previous values
        if i != 0:
            pre[i] = pre[i-1][:]

        # Processing each bit and store it in 2d vector
        for j in range(32):
            if b[j] == "1":
                sum += 1 << j
            else:
                pre[i][j] += (1 << j) - sum


def process_query(Q: List[Tuple[int, int]], pre: List[List[int]]) -> List[int]:
    # Stores the sum for each query
    ans = []

    for i in range(len(Q)):
        # Update wrt 0-based index
        Q[i] = (Q[i][0]-1, Q[i][1]-1)

        # Initialize answer
        min1 = float("inf")

        # Find minimum sum for each bit
        if Q[i][0] == 0:
            for j in range(32):
                min1 = min(pre[Q[i][1]][j], min1)
        else:
            for j in range(32):
                min1 = min(pre[Q[i][1]][j] - pre[Q[i][0]-1][j], min1)

        # Store the answer for each query
        ans.append(min1)

    return ans

# Given array
arr = [1, 2, 4, 8]

# Given Queries
Q = [(1, 4), (2, 3), (1, 3)]

# 2d Prefix vector
pre = [[0]*32 for _ in range(100001)]

# Preprocessing
processing(arr, pre)

# Function call for queries
ans = process_query(Q, pre)

# Print the answer vector
print(ans)

# This code is contributed by phasing17.

C#

// C# code to implement the approach
using System;
using System.Linq;

class Program {
  static void Main(string[] args)
  {
    // Given array
    long[] arr = { 1, 2, 4, 8 };

    // Given Queries
    long[][] Q
      = { new long[] { 1, 4 }, new long[] { 2, 3 },
         new long[] { 1, 3 } };

    // 2d Prefix vector
    long[][] pre = new long[100001][]
      .Select(x => new long[32])
      .ToArray();

    // Preprocessing
    Processing(arr, pre);

    // Function call for queries
    long[] ans = ProcessQuery(Q, pre);

    // Prlong the answer vector
    Console.WriteLine(string.Join(", ", ans));
  }

  private static void Processing(long[] v, long[][] pre)
  {
    // Number of elements
    long N = v.Length;

    // Traverse all elements
    for (long i = 0; i < N; i++) {
      // Binary representation
      string b = Convert.ToString(v[i], 2);
      b = "0".PadLeft(32 - b.Length, '0') + b;
      b = new string(b.Reverse().ToArray());

      long sum = 0;

      // Take previous values
      if (i != 0) {
        pre[i] = pre[i - 1].ToArray();
      }

      // Processing each bit and store it in 2d vector
      for (int j = 0; j < 32; j++) {
        if (b[j] == '1') {
          sum += (long)Math.Pow(2, j);
        }
        else {
          pre[i][j] += (long)Math.Pow(2, j) - sum;
        }
      }
    }
  }

  private static long[] ProcessQuery(long[][] Q,
                                     long[][] pre)
  {
    // Stores the sum for each query
    long[] ans = new long[Q.Length];

    for (long i = 0; i < Q.Length; i++) {
      // Update wrt 0-based index
      Q[i][0] -= 1;
      Q[i][1] -= 1;

      // Initialize answer
      long min1 = long.MaxValue;

      // Find minimum sum for each bit
      if (Q[i][0] == 0) {
        for (long j = 0; j < 32; j++) {
          min1 = Math.Min(pre[Q[i][1]][j], min1);
        }
      }
      else {
        for (long j = 0; j < 32; j++) {
          min1 = Math.Min(
            pre[Q[i][1]][j]
            - pre[Q[i][0] - 1][j],
            min1);
        }
      }

      // Store the answer for each query
      ans[i] = min1;
    }

    return ans;
  }
}

// This code is contributed by phasing17

Javascript

// JS program to implement the above approach

const processing = (v, pre) => {
  // Number of elements
  const N = v.length;

  // Traverse all elements
  for (let i = 0; i < N; i++) {
    // Binary representation
    let b = v[i].toString(2);
    b = "0".repeat(32 - b.length) + b;
    b = b.split("").reverse().join("");

    let sum = 0;

    // Take previous values
    if (i !== 0) {
      pre[i] = pre[i - 1].slice();
    }

    // Processing each bit and store it in 2d vector
    for (let j = 0; j < 32; j++) {
      if (b[j] === "1") {
        sum += 2 ** j;
      } else {
        pre[i][j] += 2 ** j - sum;
      }
    }
  }
};

const processQuery = (Q, pre) => {
  // Stores the sum for each query
  const ans = [];

  for (let i = 0; i < Q.length; i++) {
    // Update wrt 0-based index
    Q[i] = [Q[i][0] - 1, Q[i][1] - 1];

    // Initialize answer
    let min1 = Number.POSITIVE_INFINITY;

    // Find minimum sum for each bit
    if (Q[i][0] === 0) {
      for (let j = 0; j < 32; j++) {
        min1 = Math.min(pre[Q[i][1]][j], min1);
      }
    } else {
      for (let j = 0; j < 32; j++) {
        min1 = Math.min(pre[Q[i][1]][j] - pre[Q[i][0] - 1][j], min1);
      }
    }

    // Store the answer for each query
    ans.push(min1);
  }

  return ans;
};

// Given array
const arr = [1, 2, 4, 8];

// Given Queries
const Q = [[1, 4], [2, 3], [1, 3]];

// 2d Prefix vector
const pre = new Array(100001).fill(0).map(() => new Array(32).fill(0));

// Preprocessing
processing(arr, pre);

// Function call for queries
const ans = processQuery(Q, pre);

// Print the answer vector
console.log(ans);

// This code is implemented by Phasing17

Output:

3 2 2

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