给定大小为N的数组arr [] ,任务是计算其元素的连接被3整除的对,每个数组元素最多存在一对。
例子:
Input: arr[] = { 5, 3, 2, 8, 7 }
Output: 1
Explanation:
Possible pairs whose concatenation is divisible by 3 are { 27, 72, 78, 87 }, but the array element arr[4] will be present in at most one pair. Therefore, the required output is 1
Input: arr[] = { 10, 6, 3, 7, 2 }
Output: 2
天真的方法:解决此问题的最简单方法是遍历数组并生成给定数组的所有可能的对。对于每一对,检查该对中的元素的串联是否可被3整除。如果发现是正确的,则将两个元素都标记为false,以使该对中的两个元素不能出现在多个对中。
下面是幼稚方法的实现:
Java
// Java program to implement
// the above approach
import java.io.*;
import java.util.Arrays;
class GFG{
// Function to count pairs whose concatenation
// is divisible by 3 and each element can be
// present in at most one pair
public static int countDivBy3InArray(int[] arr)
{
// Stores count pairs whose concatenation
// is divisible by 3 and each element can
// be present in at most one pair
int ans = 0;
// Check if an element present
// in any pair or not
boolean[] taken = new boolean[arr.length];
Arrays.fill(taken, false);
// Generate all possible pairs
for(int i = 0; i < arr.length; i++)
{
// If the element already
// present in a pair
if (taken[i] == true)
{
continue;
}
for(int j = i + 1; j < arr.length; j++)
{
// If the element already
// present in a pair
if (taken[j] == true)
{
continue;
}
// If concatenation of elements
// is divisible by 3
if (Integer.parseInt(
Integer.toString(arr[i]) +
Integer.toString(arr[j])) % 3 == 0 ||
Integer.parseInt(
Integer.toString(arr[j]) +
Integer.toString(arr[i])) % 3 == 0)
{
// Update ans
ans += 1;
// Mark i is True
taken[i] = true;
// Mark j is True
taken[j] = true;
}
}
}
return ans;
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 5, 3, 2, 8, 7 };
// To display the result
System.out.println(countDivBy3InArray(arr));
}
}
// This code is contributed by aditya7409
Python3
# Python3 program to implement
# the above approach
# Function to count pairs whose concatenation is
# divisible by 3 and each element can be present
# in at most one pair
def countDivBy3InArray(arr):
# Stores count pairs whose concatenation is
# divisible by 3 and each element can be present
# in at most one pair
ans = 0
# Check if an element present
# in any pair or not
taken = [False] * len(arr)
# Generate all possible pairs
for i in range(len(arr)):
# If the element already
# present in a pair
if taken[i]:
continue
for j in range(i + 1, len(arr)):
# If the element already
# present in a pair
if taken[j]:
continue
# If concatenation of elements
# is divisible by 3
if (not int(str(arr[i])+str(arr[j])) % 3 or
not int(str(arr[j])+str(arr[i])) % 3):
# Update ans
ans += 1
# Mark i is True
taken[i] = True
# Mark j is True
taken[j] = True
return ans
# Driver Code
arr = [5, 3, 2, 8, 7]
# To display the result
print(countDivBy3InArray(arr))
C#
// C# program to implement
// the above approach
using System;
public class GFG
{
// Function to count pairs whose concatenation
// is divisible by 3 and each element can be
// present in at most one pair
public static int countDivBy3InArray(int[] arr)
{
// Stores count pairs whose concatenation
// is divisible by 3 and each element can
// be present in at most one pair
int ans = 0;
// Check if an element present
// in any pair or not
bool[] taken = new bool[arr.Length];
// Generate all possible pairs
for(int i = 0; i < arr.Length; i++)
{
// If the element already
// present in a pair
if (taken[i] == true)
{
continue;
}
for(int j = i + 1; j < arr.Length; j++)
{
// If the element already
// present in a pair
if (taken[j] == true)
{
continue;
}
// If concatenation of elements
// is divisible by 3
if (Int32.Parse(
(arr[i]).ToString() +
(arr[j]).ToString()) % 3 == 0 ||
Int32.Parse(
(arr[j]).ToString() +
(arr[i]).ToString()) % 3 == 0)
{
// Update ans
ans += 1;
// Mark i is True
taken[i] = true;
// Mark j is True
taken[j] = true;
}
}
}
return ans;
}
// Driver Code
public static void Main(String[] args)
{
int[] arr = { 5, 3, 2, 8, 7 };
// To display the result
Console.WriteLine(countDivBy3InArray(arr));
}
}
// This code is contributed by 29AjayKumar
Java
// Java program to implement
// the above approach
public class GFG
{
// Function to count pairs whose concatenation is
// divisible by 3 and each element can be present
// in at most one pair
static int countDiv(int[] arr)
{
// Stores count of array elements whose
// remainder is 0 by taking modulo by 3
int rem0 = 0;
// Stores count of array elements whose
// remainder is 1 by taking modulo by 3
int rem1 = 0;
// Stores count of array elements whose
// remainder is 2 by taking modulo by 3
int rem2 = 0;
// Traverse the array
for(int i : arr)
{
// Stores sum of digits
// of arr[i]
int digitSum = 0;
// Update digitSum
digitSum += i;
// If remainder of digitSum by
// by taking modulo 3 is 0
if(digitSum % 3 == 0)
{
// Update rem0
rem0 += 1;
}
// If remainder of digitSum by
// by taking modulo 3 is 1
else if(digitSum % 3 == 1)
{
// Update rem1
rem1 += 1;
}
else
{
// Update rem2
rem2 += 1;
}
}
return (rem0 / 2 + Math.min(rem1, rem2));
}
// Driver code
public static void main(String[] args) {
int[] arr = {5, 3, 2, 8, 7};
// To display the result
System.out.println(countDiv(arr));
}
}
// This code is contributed by divyesh072019.
Python3
# Python3 program to implement
# the above approach
# Function to count pairs whose concatenation is
# divisible by 3 and each element can be present
# in at most one pair
def countDiv(arr):
# Stores count of array elements whose
# remainder is 0 by taking modulo by 3
rem0 = 0
# Stores count of array elements whose
# remainder is 1 by taking modulo by 3
rem1 = 0
# Stores count of array elements whose
# remainder is 2 by taking modulo by 3
rem2 = 0
# Traverse the array
for i in arr:
# Stores sum of digits
# of arr[i]
digitSum = 0
for digit in str(i):
# Update digitSum
digitSum += int(digit)
# If remainder of digitSum by
# by taking modulo 3 is 0
if digitSum % 3 == 0:
# Update rem0
rem0 += 1
# If remainder of digitSum by
# by taking modulo 3 is 1
elif digitSum % 3 == 1:
# Update rem1
rem1 += 1
else:
# Update rem2
rem2 += 1
return (rem0 // 2 + min(rem1, rem2))
# Driver Code
arr = [5, 3, 2, 8, 7]
# To display the result
print(countDiv(arr))
C#
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG {
// Function to count pairs whose concatenation is
// divisible by 3 and each element can be present
// in at most one pair
static int countDiv(int[] arr)
{
// Stores count of array elements whose
// remainder is 0 by taking modulo by 3
int rem0 = 0;
// Stores count of array elements whose
// remainder is 1 by taking modulo by 3
int rem1 = 0;
// Stores count of array elements whose
// remainder is 2 by taking modulo by 3
int rem2 = 0;
// Traverse the array
foreach(int i in arr)
{
// Stores sum of digits
// of arr[i]
int digitSum = 0;
// Update digitSum
digitSum += i;
// If remainder of digitSum by
// by taking modulo 3 is 0
if(digitSum % 3 == 0)
{
// Update rem0
rem0 += 1;
}
// If remainder of digitSum by
// by taking modulo 3 is 1
else if(digitSum % 3 == 1)
{
// Update rem1
rem1 += 1;
}
else
{
// Update rem2
rem2 += 1;
}
}
return (rem0 / 2 + Math.Min(rem1, rem2));
}
// Driver code
static void Main() {
int[] arr = {5, 3, 2, 8, 7};
// To display the result
Console.Write(countDiv(arr));
}
}
// This code is contributed by divyeshrabadiya07.
Javascript
输出:
1
时间复杂度: O(N 2 )
辅助空间: O(N)
高效方法:可以使用检查数字是否可以被3整除的概念来优化上述方法。请按照以下步骤解决问题:
- 初始化三个变量,说rem0,REM1和REM2,存储阵列元件,其余数是0,1和2分别当除以3的计数。
- 遍历数组并检查以下条件:
- 如果arr [i]%3 == 0 ,则更新cnt0 + = 1 。
- 如果arr [i]%3 == 1 ,则更新cnt1 + = 1 。
- 如果arr [i]%3 == 2 ,则更新cnt2 + = 1 。
- 最后,打印对数,即(rem0 / 2 + min(rem1,rem2)) 。
下面是上述方法的实现:
Java
// Java program to implement
// the above approach
public class GFG
{
// Function to count pairs whose concatenation is
// divisible by 3 and each element can be present
// in at most one pair
static int countDiv(int[] arr)
{
// Stores count of array elements whose
// remainder is 0 by taking modulo by 3
int rem0 = 0;
// Stores count of array elements whose
// remainder is 1 by taking modulo by 3
int rem1 = 0;
// Stores count of array elements whose
// remainder is 2 by taking modulo by 3
int rem2 = 0;
// Traverse the array
for(int i : arr)
{
// Stores sum of digits
// of arr[i]
int digitSum = 0;
// Update digitSum
digitSum += i;
// If remainder of digitSum by
// by taking modulo 3 is 0
if(digitSum % 3 == 0)
{
// Update rem0
rem0 += 1;
}
// If remainder of digitSum by
// by taking modulo 3 is 1
else if(digitSum % 3 == 1)
{
// Update rem1
rem1 += 1;
}
else
{
// Update rem2
rem2 += 1;
}
}
return (rem0 / 2 + Math.min(rem1, rem2));
}
// Driver code
public static void main(String[] args) {
int[] arr = {5, 3, 2, 8, 7};
// To display the result
System.out.println(countDiv(arr));
}
}
// This code is contributed by divyesh072019.
Python3
# Python3 program to implement
# the above approach
# Function to count pairs whose concatenation is
# divisible by 3 and each element can be present
# in at most one pair
def countDiv(arr):
# Stores count of array elements whose
# remainder is 0 by taking modulo by 3
rem0 = 0
# Stores count of array elements whose
# remainder is 1 by taking modulo by 3
rem1 = 0
# Stores count of array elements whose
# remainder is 2 by taking modulo by 3
rem2 = 0
# Traverse the array
for i in arr:
# Stores sum of digits
# of arr[i]
digitSum = 0
for digit in str(i):
# Update digitSum
digitSum += int(digit)
# If remainder of digitSum by
# by taking modulo 3 is 0
if digitSum % 3 == 0:
# Update rem0
rem0 += 1
# If remainder of digitSum by
# by taking modulo 3 is 1
elif digitSum % 3 == 1:
# Update rem1
rem1 += 1
else:
# Update rem2
rem2 += 1
return (rem0 // 2 + min(rem1, rem2))
# Driver Code
arr = [5, 3, 2, 8, 7]
# To display the result
print(countDiv(arr))
C#
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG {
// Function to count pairs whose concatenation is
// divisible by 3 and each element can be present
// in at most one pair
static int countDiv(int[] arr)
{
// Stores count of array elements whose
// remainder is 0 by taking modulo by 3
int rem0 = 0;
// Stores count of array elements whose
// remainder is 1 by taking modulo by 3
int rem1 = 0;
// Stores count of array elements whose
// remainder is 2 by taking modulo by 3
int rem2 = 0;
// Traverse the array
foreach(int i in arr)
{
// Stores sum of digits
// of arr[i]
int digitSum = 0;
// Update digitSum
digitSum += i;
// If remainder of digitSum by
// by taking modulo 3 is 0
if(digitSum % 3 == 0)
{
// Update rem0
rem0 += 1;
}
// If remainder of digitSum by
// by taking modulo 3 is 1
else if(digitSum % 3 == 1)
{
// Update rem1
rem1 += 1;
}
else
{
// Update rem2
rem2 += 1;
}
}
return (rem0 / 2 + Math.Min(rem1, rem2));
}
// Driver code
static void Main() {
int[] arr = {5, 3, 2, 8, 7};
// To display the result
Console.Write(countDiv(arr));
}
}
// This code is contributed by divyeshrabadiya07.
Java脚本
输出:
1
时间复杂度: O(N)
辅助空间: O()