给定大小为N的数组arr []和正整数K ，任务是找到大小为K的子数组，其元素可用于生成可被3整除的数字。如果不存在这样的子数组，则打印– 1 。
例子：
Input: arr[] = {84, 23, 45, 12 56, 82}, K = 3
Output: 12, 56, 82
Explanation:
Number formed by the subarray {12, 56, 82} is 125682, which is divisible by 3.
Input: arr[] = {84, 23, 45, 14 56, 82}, K = 3
Output : -1
天真的方法：最简单的方法是从给定的数组生成大小为K的所有可能的子数组，对于每个子数组，检查该子数组形成的数字是否可被3整除。
时间复杂度： O(N * K)
辅助空间： O(1)
高效方法：为了优化上述方法，该想法基于以下观察：
A number is divisible by 3 if and only if the summation of the digits of the number is divisible by 3.
请按照以下步骤解决问题：
- 将数组的前K个元素的总和存储在一个变量中，例如sum 。
- 遍历数组的其余元素
- 使用滑动窗口技术，从总和中减去子数组的第一个元素，然后将下一个数组元素添加到子数组中。
- 在每个步骤中，检查总和是否可被3整除。
- 如果发现为真，则打印当前的K大小子数组。
- 如果找不到这样的子数组，则打印-1。
下面是上述方法的实现。
C++
// C++ implementation of the above approach
#include
using namespace std;
// Function to find the
// K size subarray
void findSubArray(vector arr, int k)
{
pair ans;
int i, sum = 0;
// Check if the first K elements
// forms a number which is
// divisible by 3
for (i = 0; i < k; i++) {
sum += arr[i];
}
int found = 0;
if (sum % 3 == 0) {
ans = make_pair(0, i - 1);
found = 1;
}
// Using Sliding window technique
for (int j = i; j < arr.size(); j++) {
if (found == 1)
break;
// Calculate sum of next K
// size subarray
sum = sum + arr[j] - arr[j - k];
// Check if sum is divisible by 3
if (sum % 3 == 0) {
// Update the indices of
// the subarray
ans = make_pair(j - k + 1, j);
found = 1;
}
}
// If no such subarray is found
if (found == 0)
ans = make_pair(-1, 0);
if (ans.first == -1) {
cout << -1;
}
else {
// Print the subarray
for (i = ans.first; i <= ans.second;
i++) {
cout << arr[i] << " ";
}
}
}
// Driver's code
int main()
{
// Given array and K
vector arr = { 84, 23, 45,
12, 56, 82 };
int K = 3;
// Function Call
findSubArray(arr, K);
return 0;
} Java
// Java implementation of the above approach
import java.util.*;
import java.awt.Point;
class GFG{
// Function to find the
// K size subarray
public static void findSubArray(Vector arr,
int k)
{
Point ans = new Point(0, 0);
int i, sum = 0;
// Check if the first K elements
// forms a number which is
// divisible by 3
for(i = 0; i < k; i++)
{
sum += arr.get(i);
}
int found = 0;
if (sum % 3 == 0)
{
ans = new Point(0, i - 1);
found = 1;
}
// Using Sliding window technique
for(int j = i; j < arr.size(); j++)
{
if (found == 1)
break;
// Calculate sum of next K
// size subarray
sum = sum + arr.get(j) - arr.get(j - k);
// Check if sum is divisible by 3
if (sum % 3 == 0)
{
// Update the indices of
// the subarray
ans = new Point(j - k + 1, j);
found = 1;
}
}
// If no such subarray is found
if (found == 0)
ans = new Point(-1, 0);
if (ans.x == -1)
{
System.out.print(-1);
}
else
{
// Print the subarray
for(i = ans.x; i <= ans.y; i++)
{
System.out.print(arr.get(i) + " ");
}
}
}
// Driver code
public static void main(String[] args)
{
// Given array and K
Vector arr = new Vector();
arr.add(84);
arr.add(23);
arr.add(45);
arr.add(12);
arr.add(56);
arr.add(82);
int K = 3;
// Function call
findSubArray(arr, K);
}
}
// This code is contributed by divyeshrabadiya07 Python3
# Python3 implementation of the
# above approach
# Function to find the
# K size subarray
def findSubArray(arr, k):
ans = [(0, 0)]
sm = 0
i = 0
found = 0
# Check if the first K elements
# forms a number which is
# divisible by 3
while (i < k):
sm += arr[i]
i += 1
if (sm % 3 == 0):
ans = [(0, i - 1)]
found = 1
# Using Sliding window technique
for j in range(i, len(arr), 1):
if (found == 1):
break
# Calculate sm of next K
# size subarray
sm = sm + arr[j] - arr[j - k]
# Check if sm is divisible by 3
if (sm % 3 == 0):
# Update the indices of
# the subarray
ans = [(j - k + 1, j)]
found = 1
# If no such subarray is found
if (found == 0):
ans = [(-1, 0)]
if (ans[0][0] == -1):
print(-1)
else:
# Print the subarray
for i in range(ans[0][0],
ans[0][1] + 1, 1):
print(arr[i], end = " ")
# Driver code
if __name__ == '__main__':
# Given array and K
arr = [ 84, 23, 45, 12, 56, 82 ]
K = 3
# Function call
findSubArray(arr, K)
# This code is contributed by SURENDRA_GANGWARC#
// C# implementation of
// the above approach
using System;
using System.Collections.Generic;
class GFG{
class Point
{
public int x, y;
public Point(int first,
int second)
{
this.x = first;
this.y = second;
}
}
// Function to find the
// K size subarray
public static void findSubArray(List arr,
int k)
{
Point ans = new Point(0, 0);
int i, sum = 0;
// Check if the first K elements
// forms a number which is
// divisible by 3
for(i = 0; i < k; i++)
{
sum += arr[i];
}
int found = 0;
if (sum % 3 == 0)
{
ans = new Point(0, i - 1);
found = 1;
}
// Using Sliding window technique
for(int j = i; j < arr.Count; j++)
{
if (found == 1)
break;
// Calculate sum of next K
// size subarray
sum = sum + arr[j] -
arr[j - k];
// Check if sum is
// divisible by 3
if (sum % 3 == 0)
{
// Update the indices of
// the subarray
ans = new Point(j - k + 1, j);
found = 1;
}
}
// If no such subarray is found
if (found == 0)
ans = new Point(-1, 0);
if (ans.x == -1)
{
Console.Write(-1);
}
else
{
// Print the subarray
for(i = ans.x; i <= ans.y; i++)
{
Console.Write(arr[i] + " ");
}
}
}
// Driver code
public static void Main(String[] args)
{
// Given array and K
List arr = new List();
arr.Add(84);
arr.Add(23);
arr.Add(45);
arr.Add(12);
arr.Add(56);
arr.Add(82);
int K = 3;
// Function call
findSubArray(arr, K);
}
}
// This code is contributed by Rajput-Ji 输出：
12 56 82
时间复杂度： O(N)
辅助空间： O(1)