Compress a matrix into a single number using given operations
Last Updated :
27 Oct, 2021
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 Program for the above approach
import 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.
# Python Program for the above approach
# Function to compress an
# array to a single number
def 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 number
def 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 Code
mat = [ [ 3, 2, 4], [5, 6, 1], [8, 1, 3] ]
print( getResult(mat) )
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to compress an
// array to a single number
static 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 number
static 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 Code
public 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
<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>
Time Complexity: O(N*M)
Auxiliary Space: O(N)
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