给定长度为N的数组arr []使得(1 <= arr [i] <= N)，任务是通过仅插入范围[1，N]内的元素来修改数组，使得长度为K的所有子数组都相等。如果可能，打印修改后的数组。否则打印“不可能”。
例子：
Input: arr[] = {1, 2, 2, 1}, K = 2
Output: 1 2 1 2 1
Explanation:
Insert 1 between second and third element.
Now all subarray of length K has an equal sum of 3.
Input: arr[] = {1, 2, 3, 4}, K = 3
Output: Not possible
方法：
- 由于不允许删除元素，因此，只有当数组中最多有K个不同的元素时，才能实现这种修改后的数组。
- 因此，首先检查给定数组中是否有超过K个不同的元素。如果是，则打印“不可能”。
- 否则，将给定数组的所有不同元素存储在向量中。
- 如果不同元素的数量小于K，则在向量中添加/重复一些元素，并使向量的大小等于K。
- 现在，向量具有K个元素。现在以相同的顺序重复向量的元素，以显示修改后的数组。进行此重复以显示所有长度为K的子数组具有相同的总和。
下面是上述方法的实现
C++
// C++ implementation of above approach
#include
using namespace std;
// Function that prints another array
// whose all subarray of length k
// have an equal sum
void MakeArray(int a[], int n, int k)
{
unordered_map mp;
// Store all distinct elements in
// the unordered map
for (int i = 0; i < n; i++) {
if (mp.find(a[i]) == mp.end())
mp[a[i]] = 1;
}
// Condition if the number of
// distinct elements is greater
// than k
if (mp.size() > k) {
cout << "Not possible\n";
return;
}
vector ans;
// Push all distinct elements
// in a vector
for (auto i : mp) {
ans.push_back(i.first);
}
// Push 1 while the size of
// vector not equal to k
while (ans.size() < k) {
ans.push_back(1);
}
// Print the vector 2 times
for (int i = 0; i < 2; i++) {
for (int j = 0; j < k; j++)
cout << ans[j] << " ";
}
}
// Driver Code
int main()
{
int arr[] = { 1, 2, 2, 1 };
int K = 2;
int size = sizeof(arr) / sizeof(arr[0]);
MakeArray(arr, size, K);
return 0;
} Java
// Java implementation of above approach
import java.util.*;
class GFG{
// Function that prints another array
// whose all subarray of length k
// have an equal sum
static void MakeArray(int a[], int n, int k)
{
HashMap mp = new HashMap();
// Store all distinct elements in
// the unordered map
for (int i = 0; i < n; i++)
{
if (!mp.containsKey(a[i]))
mp.put(a[i], 1);
}
// Condition if the number of
// distinct elements is greater
// than k
if (mp.size() > k)
{
System.out.print("Not possible\n");
return;
}
Vector ans = new Vector();
// Push all distinct elements
// in a vector
for (Map.Entry i : mp.entrySet())
{
ans.add(i.getKey());
}
// Push 1 while the size of
// vector not equal to k
while (ans.size() < k)
{
ans.add(1);
}
// Print the vector 2 times
for (int i = 0; i < 2; i++)
{
for (int j = 0; j < k; j++)
System.out.print(ans.get(j) + " ");
}
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 1, 2, 2, 1 };
int K = 2;
int size = arr.length;
MakeArray(arr, size, K);
}
}
// This code is contributed by Rohit_ranjan Python3
# Python3 implementation of above approach
# Function that prints another array
# whose all subarray of length k
# have an equal sum
def MakeArray(a, n, k):
mp = dict()
# Store all distinct elements in
# the unordered map
for i in a:
if i not in mp:
mp[a[i]] = 1
# Condition if the number of
# distinct elements is greater
# than k
if(len(mp) > k):
print("Not possible")
return
ans = []
# Push all distinct elements
# in a vector
for i in mp:
ans.append(i)
# Push 1 while the size of
# vector not equal to k
while(len(ans) < k):
ans.append(1)
# Print the vector 2 times
for i in range(2):
for j in range(k):
print(ans[j], end = " ")
# Driver Code
if __name__ == '__main__':
arr = [ 1, 2, 2, 1 ]
K = 2
size = len(arr)
MakeArray(arr, size, K)
# This code is contributed by mohit kumar 29C#
// C# implementation of above approach
using System;
using System.Collections.Generic;
class GFG{
// Function that prints another array
// whose all subarray of length k
// have an equal sum
static void MakeArray(int []a, int n, int k)
{
Dictionary mp = new Dictionary();
// Store all distinct elements in
// the unordered map
for(int i = 0; i < n; i++)
{
if (!mp.ContainsKey(a[i]))
mp.Add(a[i], 1);
}
// Condition if the number of
// distinct elements is greater
// than k
if (mp.Count > k)
{
Console.Write("Not possible\n");
return;
}
List ans = new List();
// Push all distinct elements
// in a vector
foreach(KeyValuePair i in mp)
{
ans.Add(i.Key);
}
// Push 1 while the size of
// vector not equal to k
while (ans.Count < k)
{
ans.Add(1);
}
// Print the vector 2 times
for(int i = 0; i < 2; i++)
{
for(int j = 0; j < k; j++)
Console.Write(ans[j] + " ");
}
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 1, 2, 2, 1 };
int K = 2;
int size = arr.Length;
MakeArray(arr, size, K);
}
}
// This code is contributed by amal kumar choubey 输出：
2 1 2 1
如果您希望与行业专家一起参加现场课程，请参阅《 Geeks现场课程》和《 Geeks现场课程美国》。