Given a square symmetric matrix mat[][] of size N, the task is to find the minimum sum possible of an array arr[] of size N, such that for i != j, the value of Bitwise AND of arr[i] and arr[j] is mat[i][j].
Examples:
Input: mat[][] = {{-1, 0, 1, 1, 1}, {0, -1, 2, 0, 2}, {1, 2, -1, 1, 3}, {1, 0, 1, -1, -1}, {1, 2, 3, 1, -1}}
Output: 10
Explanation:
The array that satisfy the above criteria is {1, 2, 3, 1, 3}.
The sum of array elements is 1 + 2 + 3 + 1 + 3 = 10, which is minimum.Input: mat[][] = {{-1, 2, 3}, {2, -1, 7}, {3, 7, -1}}
Output: 17
Approach: The given problem can be solved based on the following observations:
- The binary representation of the Bitwise AND of two numbers have only those bit set that is set in both the numbers.
- So from the property of Bitwise AND it can be observed that a number in the resultant array must have all the bit set that is set in at least one of the matrix elements in the ith row.
- Therefore, the number at ith position in the resultant array is Bitwise OR of all the elements of the matrix at the ith row.
Follow the steps below to solve the problem:
- Initialize a variable say sum to 0 that stores the minimum possible sum of the array.
- Iterate over the range [0, N – 1] and add the value of Bitwise OR of all the elements of the matrix at the ith row except mat[i][i] to the variable sum.
- After completing the above steps, print the value of the sum as the resultant possible sum.
Below is the implementation of the above approach:
C++
// C++ program of the above approach#include <bits/stdc++.h>using namespace std;// Function to find the minimum sum// of the array such that Bitwise// AND of arr[i] ana arr[j] is mat[i][j]int findMinSum(vector<vector<int> > mat, int N){ // Stores the minimum possible sum int sum = 0; // Traverse the range [0, N - 1] for (int i = 0; i < N; i++) { // Stores the Bitwise OR of // all the element of a row int res = 0; // Traverse the range [0, N - 1] for (int j = 0; j < N; j++) { // If i not equal to j if (i != j) { // Update the value of // res res |= mat[i][j]; } } // Increment the sum by res sum += res; } // Return minimum possible sum return sum;}// Driver Codeint main(){ vector<vector<int> > mat = { { -1, 2, 3 }, { 9, -1, 7 }, { 4, 5, -1 } }; int N = mat.size(); cout << findMinSum(mat, N); return 0;} |
Java
// Java program for the above approachimport java.io.*;import java.lang.*;import java.util.*;class GFG{// Function to find the minimum sum// of the array such that Bitwise// AND of arr[i] ana arr[j] is mat[i][j]static int findMinSum(int mat[][], int N){ // Stores the minimum possible sum int sum = 0; // Traverse the range [0, N - 1] for(int i = 0; i < N; i++) { // Stores the Bitwise OR of // all the element of a row int res = 0; // Traverse the range [0, N - 1] for(int j = 0; j < N; j++) { // If i not equal to j if (i != j) { // Update the value of // res res |= mat[i][j]; } } // Increment the sum by res sum += res; } // Return minimum possible sum return sum;}// Driver Codepublic static void main(String[] args){ int mat[][] = { { -1, 2, 3 }, { 9, -1, 7 }, { 4, 5, -1 } }; int N = mat.length; System.out.println(findMinSum(mat, N));}}// This code is contributed by Kingash |
Python3
# Python 3 program of the above approach# Function to find the minimum sum# of the array such that Bitwise# AND of arr[i] ana arr[j] is mat[i][j]def findMinSum(mat, N): # Stores the minimum possible sum sum1 = 0 # Traverse the range [0, N - 1] for i in range(N): # Stores the Bitwise OR of # all the element of a row res = 0 # Traverse the range [0, N - 1] for j in range(N): # If i not equal to j if (i != j): # Update the value of # res res |= mat[i][j] # Increment the sum by res sum1 += res # Return minimum possible sum return sum1 # Driver codeif __name__ == '__main__': mat = [[-1, 2, 3], [9, -1, 7], [4, 5,-1]] N = 3 print(findMinSum(mat, N)) # This code is contributed by ipg2016107. |
C#
// C# program for the above approachusing System;class GFG{// Function to find the minimum sum// of the array such that Bitwise// AND of arr[i] ana arr[j] is mat[i][j]static int findMinSum(int[,] mat, int N){ // Stores the minimum possible sum int sum = 0; // Traverse the range [0, N - 1] for(int i = 0; i < N; i++) { // Stores the Bitwise OR of // all the element of a row int res = 0; // Traverse the range [0, N - 1] for(int j = 0; j < N; j++) { // If i not equal to j if (i != j) { // Update the value of // res res |= mat[i, j]; } } // Increment the sum by res sum += res; } // Return minimum possible sum return sum;}// Driver Codepublic static void Main(string[] args){ int[,] mat = { { -1, 2, 3 }, { 9, -1, 7 }, { 4, 5, -1 } }; int N = mat.GetLength(0); Console.WriteLine(findMinSum(mat, N));}}// This code is contributed by ukasp |
Javascript
<script>// Javascript program for the above approach // Function to find the minimum sum// of the array such that Bitwise// AND of arr[i] ana arr[j] is mat[i][j]function findMinSum(mat, N){ // Stores the minimum possible sum var sum = 0; // Traverse the range [0, N - 1] for(var i = 0; i < N; i++) { // Stores the Bitwise OR of // all the element of a row var res = 0; // Traverse the range [0, N - 1] for(var j = 0; j < N; j++) { // If i not equal to j if (i != j) { // Update the value of // res res |= mat[i][j]; } } // Increment the sum by res sum += res; } // Return minimum possible sum return sum;}// Driver Codevar mat = [ [ -1, 2, 3 ], [ 9, -1, 7 ], [ 4, 5, -1 ] ];var N = mat.length;document.write(findMinSum(mat, N));// This code is contributed by Ankita saini</script> |
Output:
23
Time Complexity: O(N2)
Auxiliary Space: O(1)
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