给定数组中可被 M 整除的最大 K 位整数
给定一个大小为N的数组arr[]和 2 个整数K和M ，任务是从数组arr[]中找到可被M整除的K位的最大整数。如果不可能找到任何这样的整数，则打印-1 。
Note: The value of K is such that the found integer is always less than or equal to INT_MAX.
例子：
Input: arr[] = {1, 2, 3, 4, 5, 6}, N=6, K=3, M=4
Output: 652
Explanation: The largest of 3 digits is formed by the digits 6, 5, 2 with their indices 1. 4, and 5.
Input: arr[] = {4, 5, 4}, N=3, K=3, M=50
Output: -1
Explanation :- There exists no integer from the array which is divisible by 50 as there is no 0.
朴素方法：这可以通过查找所有大小为 K 的子序列然后检查条件以找到最大的子序列来完成。
时间复杂度： O(2 N )
辅助空间：O(1)
有效方法：这个想法是检查数组中M的倍数中的每个数字是否有可能形成该数字。如果可能，请更新答案。请按照以下步骤解决问题：
- 如果K大于N，则打印-1并返回。
- 用值0初始化大小为10的向量freq[]以存储数组arr[] 的所有元素的频率。
- 遍历范围[0, N)并将每个元素的频率存储在数组freq[] 中。
- 按升序对数组arr[]进行排序。
- 初始化变量X并将其值设置为数组arr[]中最大的K位数作为可能的最大数。
- 如果X小于M则打印-1并返回。
- 将变量answer初始化为-1以存储答案。
- 使用变量i迭代范围[M, X]并执行以下步骤：
- 将变量y初始化为i。
- 用值0初始化大小为10的向量v[]以存储数字y 的所有数字的频率。
- 遍历向量freq[]和v[] ，如果对于所有位置freq[j]大于等于v[j]，则将答案更新为 answer或y 的最大值。
- 执行上述步骤后，打印变量answer作为答案。
下面是上述方法的实现。
C++14
// C++ program for the above approach
#include
using namespace std;
// Function to find the largest integer
// of K digits divisible by M
void find(vector& arr, int N, int K, int M)
{
// Base Case, as there is no sub-sequence
// of size greater than N is possible
if (K > N) {
cout << "-1\n";
return;
}
// Vector to store the frequency of each
// number from the array
vector freq(10, 0);
// Calculate the frequency of each from
// the array
for (int i = 0; i < N; i++) {
freq[arr[i]]++;
}
// Sort the array in ascending order
sort(arr.begin(), arr.end());
// Variable to store the maximum number
// possible
int X = 0;
// Traverse and store the largest K digits
for (int i = N - 1; K > 0; i--) {
X = X * 10 + arr[i];
K--;
}
// Base Case
if (X < M) {
cout << "-1\n";
return;
}
// Variable to store the answer
int answer = -1;
// Traverse and check for each number
for (int i = M; i <= X; i = i + M) {
int y = i;
vector v(10, 0);
// Calculate the frequency of each number
while (y > 0) {
v[y % 10]++;
y = y / 10;
}
bool flag = true;
// If frequency of any digit in y is less than
// that in freq[], then it's not possible.
for (int j = 0; j <= 9; j++) {
if (v[j] > freq[j]) {
flag = false;
break;
}
}
if (flag)
answer = max(answer, i);
}
// Print the answer
cout << answer << endl;
}
// Driver Code
int main()
{
vector arr = { 1, 2, 3, 4, 5, 6 };
int N = 6, K = 3, M = 4;
find(arr, N, K, M);
return 0;
} Java
// Java code for the above approach
import java.io.*;
import java.util.Arrays;;
class GFG
{
// Function to find the largest integer
// of K digits divisible by M
static void find(int[] arr, int N, int K, int M)
{
// Base Case, as there is no sub-sequence
// of size greater than N is possible
if (K > N) {
System.out.println("-1\n");
return;
}
// Vector to store the frequency of each
// number from the array
int freq[] = new int[10];
// Calculate the frequency of each from
// the array
for (int i = 0; i < N; i++) {
freq[arr[i]]++;
}
// Sort the array in ascending order
Arrays.sort(arr);
// Variable to store the maximum number
// possible
int X = 0;
// Traverse and store the largest K digits
for (int i = N - 1; K > 0; i--) {
X = X * 10 + arr[i];
K--;
}
// Base Case
if (X < M) {
System.out.println("-1");
return;
}
// Variable to store the answer
int answer = -1;
// Traverse and check for each number
for (int i = M; i <= X; i = i + M) {
int y = i;
int v[] = new int[10];
// Calculate the frequency of each number
while (y > 0) {
v[y % 10]++;
y = y / 10;
}
boolean flag = true;
// If frequency of any digit in y is less than
// that in freq[], then it's not possible.
for (int j = 0; j <= 9; j++) {
if (v[j] > freq[j]) {
flag = false;
break;
}
}
if (flag)
answer = Math.max(answer, i);
}
// Print the answer
System.out.println(answer);
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 1, 2, 3, 4, 5, 6 };
int N = 6, K = 3, M = 4;
find(arr, N, K, M);
}
}
// This code is contributed by Potta LokeshPython3
# Python program for the above approach
# Function to find the largest integer
# of K digits divisible by M
def find(arr, N, K, M):
# Base Case, as there is no sub-sequence
# of size greater than N is possible
if (K > N):
print("-1")
return
# Vector to store the frequency of each
# number from the array
freq = [0] * 10
# Calculate the frequency of each from
# the array
for i in range(N):
freq[arr[i]] += 1
# Sort the array in ascending order
arr.sort()
# Variable to store the maximum number
# possible
X = 0
# Traverse and store the largest K digits
for i in range(N - 1, 2, -1):
X = X * 10 + arr[i]
K -= 1
# Base Case
if (X < M):
print("-1")
return
# Variable to store the answer
answer = -1
# Traverse and check for each number
for i in range(M, X + 1, M):
y = i
v = [0] * 10
# Calculate the frequency of each number
while (y > 0):
v[y % 10] += 1
y = y // 10
flag = True
# If frequency of any digit in y is less than
# that in freq[], then it's not possible.
for j in range(10):
if (v[j] > freq[j]):
flag = False
break
if (flag):
answer = max(answer, i)
# Print the answer
print(answer)
# Driver Code
arr = [1, 2, 3, 4, 5, 6]
N = 6
K = 3
M = 4
find(arr, N, K, M)
# This code is contributed by gfgking.C#
// C# code for the above approach
using System;
public class GFG
{
// Function to find the largest integer
// of K digits divisible by M
static void find(int[] arr, int N, int K, int M)
{
// Base Case, as there is no sub-sequence
// of size greater than N is possible
if (K > N) {
Console.WriteLine("-1\n");
return;
}
// Vector to store the frequency of each
// number from the array
int []freq = new int[10];
// Calculate the frequency of each from
// the array
for (int i = 0; i < N; i++) {
freq[arr[i]]++;
}
// Sort the array in ascending order
Array.Sort(arr);
// Variable to store the maximum number
// possible
int X = 0;
// Traverse and store the largest K digits
for (int i = N - 1; K > 0; i--) {
X = X * 10 + arr[i];
K--;
}
// Base Case
if (X < M) {
Console.WriteLine("-1");
return;
}
// Variable to store the answer
int answer = -1;
// Traverse and check for each number
for (int i = M; i <= X; i = i + M) {
int y = i;
int []v = new int[10];
// Calculate the frequency of each number
while (y > 0) {
v[y % 10]++;
y = y / 10;
}
bool flag = true;
// If frequency of any digit in y is less than
// that in freq[], then it's not possible.
for (int j = 0; j <= 9; j++) {
if (v[j] > freq[j]) {
flag = false;
break;
}
}
if (flag)
answer = Math.Max(answer, i);
}
// Print the answer
Console.WriteLine(answer);
}
// Driver Code
public static void Main(string[] args)
{
int[] arr = { 1, 2, 3, 4, 5, 6 };
int N = 6, K = 3, M = 4;
find(arr, N, K, M);
}
}
// This code is contributed by AnkThonJavascript
输出
652时间复杂度：O(max(M, N))
辅助空间：O(1)