在数组元素的二进制表示中查找给定二进制模式的出现

给定一个由N个正整数组成的数组arr[]和一个由集合{0, 1}中的字符组成的字符串patt ，任务是在给定数组中每个整数的二进制表示中找到patt的出现次数。
例子：

Input: arr[] = {5, 106, 7, 8}, patt = “10” 
Output: 1 3 0 1 
binary(5) = 101 -> occurrence(10) = 1 
binary(106) = 1101010 -> occurrence(10) = 3 
binary(7) = 111 -> occurrence(10) = 0 
binary(8) = 1000 -> occurrence(10) = 1
Input: arr[] = {1, 1, 1, 1}, patt = “10” 
Output: 0 0 0 0 
 

方法：找到本文中讨论的每个数组元素的二进制表示。
现在，在之前找到的二进制表示中找到给定模式的出现。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the binary
// representation of n
string decToBinary(int n)
{
    // Array to store binary representation
    int binaryNum[32];
 
    // Counter for binary array
    int i = 0;
    while (n > 0) {
 
        // Storing remainder in binary array
        binaryNum[i] = n % 2;
        n = n / 2;
        i++;
    }
 
    // To store the binary representation
    // as a string
    string binary = "";
    for (int j = i - 1; j >= 0; j--)
        binary += to_string(binaryNum[j]);
 
    return binary;
}
 
// Function to return the frequency of
// pat in the given string txt
int countFreq(string& pat, string& txt)
{
    int M = pat.length();
    int N = txt.length();
    int res = 0;
 
    /* A loop to slide pat[] one by one */
    for (int i = 0; i <= N - M; i++) {
        /* For current index i, check for 
           pattern match */
        int j;
        for (j = 0; j < M; j++)
            if (txt[i + j] != pat[j])
                break;
 
        // If pat[0...M-1] = txt[i, i+1, ...i+M-1]
        if (j == M) {
            res++;
            j = 0;
        }
    }
    return res;
}
 
// Function to find the occurrence of
// the given pattern in the binary
// representation of elements of arr[]
void findOccurrence(int arr[], int n, string pattern)
{
 
    // For every element of the array
    for (int i = 0; i < n; i++) {
 
        // Find its binary representation
        string binary = decToBinary(arr[i]);
 
        // Print the occurrence of the given pattern
        // in its binary representation
        cout << countFreq(pattern, binary) << " ";
    }
}
 
// Driver code
int main()
{
    int arr[] = { 5, 106, 7, 8 };
    string pattern = "10";
    int n = sizeof(arr) / sizeof(arr[0]);
 
    findOccurrence(arr, n, pattern);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
    // Function to return the binary
    // representation of n
    static String decToBinary(int n)
    {
        // Array to store binary representation
        int[] binaryNum = new int[32];
 
        // Counter for binary array
        int i = 0;
        while (n > 0)
        {
 
            // Storing remainder in binary array
            binaryNum[i] = n % 2;
            n = n / 2;
            i++;
        }
 
        // To store the binary representation
        // as a string
        String binary = "";
        for (int j = i - 1; j >= 0; j--)
        {
            binary += String.valueOf(binaryNum[j]);
        }
        return binary;
    }
 
    // Function to return the frequency of
    // pat in the given string txt
    static int countFreq(String pat, String txt)
    {
        int M = pat.length();
        int N = txt.length();
        int res = 0;
 
        /* A loop to slide pat[] one by one */
        for (int i = 0; i <= N - M; i++)
        {
            /* For current index i, check for
            pattern match */
            int j;
            for (j = 0; j < M; j++)
            {
                if (txt.charAt(i + j) != pat.charAt(j))
                {
                    break;
                }
            }
 
            // If pat[0...M-1] = txt[i, i+1, ...i+M-1]
            if (j == M)
            {
                res++;
                j = 0;
            }
        }
        return res;
    }
 
    // Function to find the occurrence of
    // the given pattern in the binary
    // representation of elements of arr[]
    static void findOccurrence(int arr[], int n,
                               String pattern)
    {
 
        // For every element of the array
        for (int i = 0; i < n; i++)
        {
 
            // Find its binary representation
            String binary = decToBinary(arr[i]);
 
            // Print the occurrence of the given pattern
            // in its binary representation
            System.out.print(countFreq(pattern,
                                       binary) + " ");
        }
    }
     
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = {5, 106, 7, 8};
        String pattern = "10";
        int n = arr.length;
        findOccurrence(arr, n, pattern);
    }
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 implementation of the approach
 
# Function to return the binary
# representation of n
def decToBinary(n):
     
    # Array to store binary representation
    binaryNum = [0 for i in range(32)]
 
    # Counter for binary array
    i = 0
    while (n > 0):
 
        # Storing remainder in binary array
        binaryNum[i] = n % 2
        n = n // 2
        i += 1
 
    # To store the binary representation
    # as a string
    binary = ""
    for j in range(i - 1, -1, -1):
        binary += str(binaryNum[j])
 
    return binary
 
# Function to return the frequency of
# pat in the given txt
def countFreq(pat, txt):
    M = len(pat)
    N = len(txt)
    res = 0
 
    # A loop to slide pat[] one by one
    for i in range(N - M + 1):
         
        # For current index i, check for
        # pattern match
        j = 0
        while(j < M):
            if (txt[i + j] != pat[j]):
                break
            j += 1
 
        # If pat[0...M-1] = txt[i, i+1, ...i+M-1]
        if (j == M):
            res += 1
            j = 0
 
    return res
 
# Function to find the occurrence of
# the given pattern in the binary
# representation of elements of arr[]
def findOccurrence(arr, n, pattern):
 
    # For every element of the array
    for i in range(n):
 
        # Find its binary representation
        binary = decToBinary(arr[i])
 
        # Print the occurrence of the given pattern
        # in its binary representation
        print(countFreq(pattern, binary), end = " ")
 
# Driver code
arr = [5, 106, 7, 8]
pattern = "10"
n = len(arr)
 
findOccurrence(arr, n, pattern)
 
# This code is contributed by Mohit Kumar

C#

// C# code implementation for above approach
using System;
 
class GFG
{
 
    // Function to return the binary
    // representation of n
    static String decToBinary(int n)
    {
        // Array to store binary representation
        int[] binaryNum = new int[32];
 
        // Counter for binary array
        int i = 0;
        while (n > 0)
        {
 
            // Storing remainder in binary array
            binaryNum[i] = n % 2;
            n = n / 2;
            i++;
        }
 
        // To store the binary representation
        // as a string
        String binary = "";
        for (int j = i - 1; j >= 0; j--)
        {
            binary += String.Join("", binaryNum[j]);
        }
        return binary;
    }
 
    // Function to return the frequency of
    // pat in the given string txt
    static int countFreq(String pat, String txt)
    {
        int M = pat.Length;
        int N = txt.Length;
        int res = 0;
 
        /* A loop to slide pat[] one by one */
        for (int i = 0; i <= N - M; i++)
        {
            /* For current index i, check for
            pattern match */
            int j;
            for (j = 0; j < M; j++)
            {
                if (txt[i + j] != pat[j])
                {
                    break;
                }
            }
 
            // If pat[0...M-1] = txt[i, i+1, ...i+M-1]
            if (j == M)
            {
                res++;
                j = 0;
            }
        }
        return res;
    }
 
    // Function to find the occurrence of
    // the given pattern in the binary
    // representation of elements of arr[]
    static void findOccurrence(int []arr, int n,
                               String pattern)
    {
 
        // For every element of the array
        for (int i = 0; i < n; i++)
        {
 
            // Find its binary representation
            String binary = decToBinary(arr[i]);
 
            // Print the occurrence of the given pattern
            // in its binary representation
            Console.Write(countFreq(pattern,
                                    binary) + " ");
        }
    }
     
    // Driver code
    public static void Main(String[] args)
    {
        int []arr = {5, 106, 7, 8};
        String pattern = "10";
        int n = arr.Length;
        findOccurrence(arr, n, pattern);
    }
}
     
// This code is contributed by PrinciRaj1992

Javascript

 
// Javascript implementation of the approach
 
// Function to return the binary
// representation of n
function decToBinary(n)
{
    // Array to store binary representation
    var binaryNum = Array(32);
 
    // Counter for binary array
    var i = 0;
    while (n > 0) {
 
        // Storing remainder in binary array
        binaryNum[i] = n % 2;
        n = parseInt(n / 2);
        i++;
    }
 
    // To store the binary representation
    // as a string
    var binary = "";
    for (var j = i - 1; j >= 0; j--)
        binary += (binaryNum[j].toString());
 
    return binary;
}
 
// Function to return the frequency of
// pat in the given string txt
function countFreq(pat, txt)
{
    var M = pat.length;
    var N = txt.length;
    var res = 0;
 
    /* A loop to slide pat[] one by one */
    for (var i = 0; i <= N - M; i++) {
        /* For current index i, check for 
           pattern match */
        var j;
        for (j = 0; j < M; j++)
            if (txt[i + j] != pat[j])
                break;
 
        // If pat[0...M-1] = txt[i, i+1, ...i+M-1]
        if (j == M) {
            res++;
            j = 0;
        }
    }
    return res;
}
 
// Function to find the occurrence of
// the given pattern in the binary
// representation of elements of arr[]
function findOccurrence(arr, n, pattern)
{
 
    // For every element of the array
    for (var i = 0; i < n; i++) {
 
        // Find its binary representation
        var binary = decToBinary(arr[i]);
 
        // Print the occurrence of the given pattern
        // in its binary representation
        document.write( countFreq(pattern, binary) + " ");
    }
}
 
// Driver code
var arr = [ 5, 106, 7, 8 ];
var pattern = "10";
var n = arr.length;
findOccurrence(arr, n, pattern);
 

输出：

1 3 0 1