给定一个维度为M * N的矩阵mat[][] ，任务是首先压缩它以获得一个数组，然后使用以下操作再次压缩它以获得单个整数：
- 当矩阵被压缩时，其值的二进制表示被压缩。
- 因此，每个位都被考虑，如果一个位的位置有S 个设置位和NS 个未设置位，那么如果S > NS则为该位置设置该位，否则取消设置。
- 压缩每一列以将矩阵转换为数组，然后将该数组进一步压缩为单个数字。
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.
例子：
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
方法：想法是计算每个位置的设置位数。请按照以下步骤解决问题：
- 遍历矩阵每一列的每个元素，并将每一列压缩为一个数字。
- 计算每个位置的设置位数。
- 为设置位数超过未设置位数的位置设置位。
- 决定是否为每个位置设置位后计算值。
- 获得数组后，应用相同的步骤来获得所需的整数。
下面是上述方法的实现：
C++
// c++ Program for the above approach
#include
using namespace std;
// Function to compress an
// array to a single number
vector append(vector arr, int x)
{
// create a new ArrayList
arr.push_back(x);
return arr;
}
int compress(vector 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> mat)
{
// Stores compressed array
vector compressedArr;
int len = mat.size();
int len2 = mat[0].size();
for (int i = 0; i < len; i++)
{
vector 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> 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 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 arrlist
= new ArrayList(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 contibuted by rohitsingh07052. Python3
# 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#
// 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 arrlist = new List(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 输出：
1时间复杂度： O(N*M)
辅助空间： O(N)
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