Given a matrix mat[][] of dimension M * N, the task is to first compress it to obtain an array and, then compress it again to obtain a single integer using following operations:
- When a matrix is compressed, its value’s binary representation gets compressed.
- Therefore, each bit is considered, and if a position of a bit has S set bits and NS non-set bits, then the bit is set for the position if S > NS and unset otherwise.
- Each column is compressed to turn the matrix into an array, then that array is compressed further into a single number.
For example, if 5, 2, 3, and 1 gets compressed then their binary representations (101)2, (010)2, (011)2, and (001)2 gets compressed then for the 0th and 1st positions, S ? NS and for the 2nd position S > NS, then the number modifies to (001)2 = 1.
Examples:
Input: arr[][] ={{ 3, 2, 4}, {5, 6, 1}, {8, 1, 3}}
Output: 1
Explanation: The array obtained after compressing the given matrix from top is {1, 2, 1 }.Then, the obtained array is compressed to 1.Input: arr[][] = {{ 5, 3}, {6, 7}}
Output: 0
Approach: The idea is to count the number of set bits for each position. Follow the steps below to solve the problem:
- Traverse through each element of each column of a matrix and compress each column to a single number.
- Count the number of set bits for each position.
- Set the bit for the positions with number of set bits exceeding the number of unset bits.
- Calculate the value after deciding whether to set or not to set the bit for each position.
- After obtaining an array, apply the same steps to obtain the required integer.
Below is the implementation of the above approach:
C++
// c++ Program for the above approach#include<bits/stdc++.h>using namespace std; // Function to compress an // array to a single number vector<int> append(vector<int> arr, int x) { // create a new ArrayList arr.push_back(x); return arr; } int compress(vector<int> arr) { // Stores the required integer int ans = 0; int getBit = 1; // Checking for each position for (int i = 0; i < 32; i++) { int S = 0; int NS = 0; for (int j = 0; j < arr.size(); j++) { // Count set and unset bit int temp = getBit & arr[j]; if (temp == 1) { S += 1; } else { NS += 1; } // If count of set bits exceeds // count of unset bits if (S > NS) { // Add value of set bits to ans ans += pow(2, i); } getBit <<= 1; } } return ans; } // Function to compress // matrix to a single number int getResult(vector<vector<int>> mat) { // Stores compressed array vector<int> compressedArr; int len = mat.size(); int len2 = mat[0].size(); for (int i = 0; i < len; i++) { vector<int> col; for (int j = 0; j < len2; j++) { col = append(col, mat[j][i]); } // Compress all columns // to a single number compressedArr = append(compressedArr, compress(col)); } return compress(compressedArr); } // Driver Code int main() { vector<vector<int>> mat{{3, 2, 4 },{5, 6, 1},{8, 1, 3}}; cout<<(getResult(mat)); }// THIS CODE IS CONTRIBUTED BY SURENDRA_GANGWAR. |
Java
// Java Program for the above approachimport java.io.*;import java.lang.*;import java.lang.Math;import java.util.*;class GFG { // Function to compress an // array to a single number static Integer[] append(Integer arr[], int x) { // create a new ArrayList List<Integer> arrlist = new ArrayList<Integer>(Arrays.asList(arr)); // Add the new element arrlist.add(x); // Convert the Arraylist to array arr = arrlist.toArray(arr); // return the array return arr; } static int compress(Integer[] arr) { // Stores the required integer int ans = 0; int getBit = 1; // Checking for each position for (int i = 0; i < 32; i++) { int S = 0; int NS = 0; for (int j = 0; j < arr.length; j++) { // Count set and unset bit int and = getBit & arr[j]; if (and == 1) { S += 1; } else { NS += 1; } // If count of set bits exceeds // count of unset bits if (S > NS) { // Add value of set bits to ans ans += Math.pow(2, i); } getBit <<= 1; } } return ans; } // Function to compress // matrix to a single number static int getResult(Integer[][] mat) { // Stores compressed array Integer[] compressedArr = {}; int len = mat.length; int len2 = mat[0].length; for (int i = 0; i < len; i++) { Integer[] col = {}; for (int j = 0; j < len2; j++) { col = append(col, mat[j][i]); } // Compress all columns // to a single number compressedArr = append(compressedArr, compress(col)); } return compress(compressedArr); } // Driver Code public static void main(String[] args) { Integer[][] mat = { { 3, 2, 4 }, { 5, 6, 1 }, { 8, 1, 3 } }; System.out.println(getResult(mat)); }}// This code is contributed by rohitsingh07052. |
Python3
# Python Program for the above approach# Function to compress an# array to a single numberdef compress(arr): # Stores the required integer ans = 0 getBit = 1 # Checking for each position for i in range(32): S = 0 NS = 0 for j in arr: # Count set and unset bits if getBit&j: S += 1 else: NS += 1 # If count of set bits exceeds # count of unset bits if S > NS: # Add value of set bits to ans ans += 2**i getBit <<= 1 return ans# Function to compress# matrix to a single numberdef getResult(mat): # Stores compressed array compressedArr = [] for i in range(len(mat)): col = [] for j in range(len(mat[0])): col.append(mat[j][i]) # Compress all columns # to a single number compressedArr.append(compress(col)) return compress(compressedArr)# Driver Codemat = [ [ 3, 2, 4], [5, 6, 1], [8, 1, 3] ]print( getResult(mat) ) |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{// Function to compress an// array to a single numberstatic int[] append(int []arr, int x){ // create a new List List<int> arrlist = new List<int>(arr); // Add the new element arrlist.Add(x); // Convert the Arraylist to array arr = arrlist.ToArray(); // return the array return arr;}static int compress(int[] arr){ // Stores the required integer int ans = 0; int getBit = 1; // Checking for each position for(int i = 0; i < 32; i++) { int S = 0; int NS = 0; for(int j = 0; j < arr.Length; j++) { // Count set and unset bit int and = getBit & arr[j]; if (and == 1) { S += 1; } else { NS += 1; } // If count of set bits exceeds // count of unset bits if (S > NS) { // Add value of set bits to ans ans += (int)Math.Pow(2, i); } getBit <<= 1; } } return ans;}// Function to compress// matrix to a single numberstatic int getResult(int[,] mat){ // Stores compressed array int[] compressedArr = {}; int len = mat.GetLength(0); int len2 = mat.GetLength(1); for(int i = 0; i < len; i++) { int[] col = {}; for (int j = 0; j < len2; j++) { col = append(col, mat[j,i]); } // Compress all columns // to a single number compressedArr = append(compressedArr, compress(col)); } return compress(compressedArr);}// Driver Codepublic static void Main(String[] args){ int[,] mat = { { 3, 2, 4 }, { 5, 6, 1 }, { 8, 1, 3 } }; Console.WriteLine(getResult(mat));}}// This code is contributed by shikhasingrajput |
Javascript
<script> // Javascript Program for the above approach // Function to compress an // array to a single number function append(arr, x) { // create a new ArrayList arr.push(x); return arr; } function compress(arr) { // Stores the required integer let ans = 0; let getBit = 1; // Checking for each position for (let i = 0; i < 32; i++) { let S = 0; let NS = 0; for (let j = 0; j < arr.length; j++) { // Count set and unset bit let temp = getBit & arr[j]; if (temp == 1) { S += 1; } else { NS += 1; } // If count of set bits exceeds // count of unset bits if (S > NS) { // Add value of set bits to ans ans += Math.pow(2, i); } getBit <<= 1; } } return ans; } // Function to compress // matrix to a single number function getResult(mat) { // Stores compressed array let compressedArr = []; let len = mat.length; let len2 = mat[0].length; for (let i = 0; i < len; i++) { let col = []; for (let j = 0; j < len2; j++) { col = append(col, mat[j][i]); } // Compress all columns // to a single number compressedArr = append(compressedArr, compress(col)); } return compress(compressedArr); } let mat = [[3, 2, 4 ],[5, 6, 1],[8, 1, 3]]; document.write(getResult(mat));// This code is contributed by mukesh07.</script> |
Output:
1
Time Complexity: O(N*M)
Auxiliary Space: O(N)
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