Given an array arr[] consisting of N integers and integer K, the task is to find the smallest integer X with exactly K set bits such that the sum of Bitwise AND of X with every array element arr[i] is maximum.
Examples:
Input: arr[] = {3, 4, 5, 1}, K = 1
Output: 4
Explanation: Consider the value of X as 4. Then, the sum of Bitwise AND of X and array elements = 4 & 3 + 4 & 4 + 4 & 5 + 4 & 1 = 0 + 4 + 4 + 0 = 8, which is maximum.Input: arr[] = {1, 3, 4, 5, 2, 5}, K = 2
Output: 5
Approach: The given problem can be solved by using the Greedy Approach. Follow the steps below to solve the problem:
- Initialize a variable, say X as 0, to store the resultant value of X.
- Initialize an array, say count[] of size 30, to store at every Ith index, the number array elements having the ith bit set.
- Traverse the given array and if the ith bit is set, then update count[i] by 1.
- Initialize a vector of pairs, say ans, to store the contribution of each bit and values, i.e. if the Ith bit is set, then store the value {i, count[i]2i} in Ans.
- Sort the vector of pairs in descending order of the second elements.
- Traverse the vector Ans over the range [0, K – 1], and update the value of X as the Bitwise OR of X and 2(Ans[i].first).
- After completing the above steps, print the value of X as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Comparator function to sort the// vector of pairsbool comp(pair<int, int>& a, pair<int, int>& b){ // If the second is not the same // then sort in decreasing order if (a.second != b.second) return a.second > b.second; // Otherwise return a.first < b.first;}// Function to find the value of X// such that Bitwise AND of all array// elements with X is maximumint maximizeSum(int arr[], int n, int k){ // Stores the count of set bit at // each position vector<int> cnt(30, 0); // Stores the resultant value of X int X = 0; // Calculate the count of set bits // at each position for (int i = 0; i < n; i++) { for (int j = 0; j < 30; j++) { // If the jth bit is set if (arr[i] & (1 << j)) cnt[j]++; } } // Stores the contribution // of each set bit vector<pair<int, int> > v; // Store all bit and amount of // contribution for (int i = 0; i < 30; i++) { // Find the total contribution int gain = cnt[i] * (1 << i); v.push_back({ i, gain }); } // Sort V[] in decreasing // order of second parameter sort(v.begin(), v.end(), comp); // Choose exactly K set bits for (int i = 0; i < k; i++) { X |= (1 << v[i].first); } // Print the answer cout << X;}// Driver Codeint main(){ int arr[] = { 3, 4, 5, 1 }; int K = 1; int N = sizeof(arr) / sizeof(arr[0]); maximizeSum(arr, N, K); return 0;} |
Java
// Java program for the above approachimport java.io.*;import java.lang.*;import java.util.*;class GFG{// Function to find the value of X// such that Bitwise AND of all array// elements with X is maximumstatic void maximizeSum(int arr[], int n, int k){ // Stores the count of set bit at // each position int cnt[] = new int[30]; // Stores the resultant value of X int X = 0; // Calculate the count of set bits // at each position for(int i = 0; i < n; i++) { for(int j = 0; j < 30; j++) { // If the jth bit is set if ((arr[i] & (1 << j)) != 0) cnt[j]++; } } // Stores the contribution // of each set bit ArrayList<int[]> v = new ArrayList<>(); // Store all bit and amount of // contribution for(int i = 0; i < 30; i++) { // Find the total contribution int gain = cnt[i] * (1 << i); v.add(new int[] { i, gain }); } // Sort V[] in decreasing // order of second parameter Collections.sort(v, (a, b) -> { // If the second is not the same // then sort in decreasing order if (a[1] != b[1]) return b[1] - a[1]; // Otherwise return a[0] - b[0]; }); // Choose exactly K set bits for(int i = 0; i < k; i++) { X |= (1 << v.get(i)[0]); } // Print the answer System.out.println(X);}// Driver Codepublic static void main(String[] args){ int arr[] = { 3, 4, 5, 1 }; int K = 1; int N = arr.length; maximizeSum(arr, N, K);}}// This code is contributed by Kingash |
Python3
# Python3 program for the above approach# Function to find the value of X# such that Bitwise AND of all array# elements with X is maximumdef maximize_sum(arr, n, k): # Stores the count of set bit at # each position cnt = [0] * 30 # Stores the resultant value of X X = 0 # Calculate the count of set bits # at each position for i in range(n): for j in range(30): # If the jth bit is set if arr[i] & (1 << j): cnt[j] += 1 # Stores the contribution # of each set bit v = [] # Store all bit and amount of # contribution for i in range(30): # Find the total contribution gain = cnt[i] * (1 << i) v.append([i, gain]) # Sort V[] in decreasing # order of second parameter v.sort(key=lambda x: (x[1], x[0]), reverse=True) # Choose exactly K set bits for i in range(k): X |= (1 << v[i][0]) # Print the answer print(X)# Driver Codearr = [3, 4, 5, 1]K = 1N = len(arr)maximize_sum(arr, N, K)# This code is contributed by phasing17 |
C#
using System;using System.Linq;using System.Collections.Generic;class GFG{ // Function to find the value of X such that Bitwise AND of all array elements with X is maximum static void MaximizeSum(int[] arr, int n, int k) { // Stores the count of set bit at each position int[] cnt = new int[30]; // Stores the resultant value of X int X = 0; // Calculate the count of set bits at each position for (int i = 0; i < n; i++) { for (int j = 0; j < 30; j++) { // If the jth bit is set if ((arr[i] & (1 << j)) != 0) cnt[j]++; } } // Store all bit and amount of contribution List<int[]> v = new List<int[]>(); for (int i = 0; i < 30; i++) { // Find the total contribution int gain = cnt[i] * (1 << i); v.Add(new int[] { i, gain }); } // Sort V[] in decreasing order of second parameter v.Sort((a, b) => { // If the second is not the same then sort in decreasing order if (a[1] != b[1]) return b[1] - a[1]; // Otherwise return a[0] - b[0]; }); // Choose exactly K set bits for (int i = 0; i < k; i++) { X |= (1 << v[i][0]); } // Print the answer Console.WriteLine(X); } // Driver Code static public void Main(string[] args) { int[] arr = { 3, 4, 5, 1 }; int K = 1; int N = arr.Length; MaximizeSum(arr, N, K); }} |
Javascript
<script>// JavaScript program for the above approach// Function to find the value of X// such that Bitwise AND of all array// elements with X is maximumfunction maximizeSum(arr, n, k) { // Stores the count of set bit at // each position let cnt = new Array(30).fill(0); // Stores the resultant value of X let X = 0; // Calculate the count of set bits // at each position for (let i = 0; i < n; i++) { for (let j = 0; j < 30; j++) { // If the jth bit is set if (arr[i] & (1 << j)) cnt[j]++; } } // Stores the contribution // of each set bit let v = new Array(); // Store all bit and amount of // contribution for (let i = 0; i < 30; i++) { // Find the total contribution let gain = cnt[i] * (1 << i); v.push([i, gain]); } // Sort V[] in decreasing // order of second parameter v.sort((a, b) => { // If the second is not the same // then sort in decreasing order if (a[1] != b[1]) return b[1] - a[1]; // Otherwise return a[0] - b[0]; }); // Choose exactly K set bits for (let i = 0; i < k; i++) { X |= (1 << v[i][0]); } // Print the answer document.write(X);}// Driver Codelet arr = [3, 4, 5, 1];let K = 1;let N = arr.length;maximizeSum(arr, N, K);</script> |
Output:
4
Time Complexity: O(N)
Auxiliary Space: O(1)
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