给定一个大小为N的整数数组arr和一个整数K ，任务是使用以下操作使数组的所有元素等于K ：
- 选择输入数组的任意子数组[l….r]
- 替换此子数组的所有值等于已排序子数组 [l…r] 中的第[((r – l) + 2) / 2]个值
例子：
Input: arr [ ] = {5, 4, 3, 1, 2, 6, 7, 8, 9, 10}, K = 3
Output: YES
Explanation:
Choose first five elements and replace all elements with 3: arr = {3, 3, 3, 3, 3, 6, 7, 8, 9, 10}
Now take forth to sixth elements and replace all elements with 3: arr = {3, 3, 3, 3, 3, 3, 7, 8, 9, 10}
By appling same operations we can make all elements of arr equal to K: arr = {3, 3, 3, 3, 3, 3, 3, 3, 3, 3}
Input: arr [ ] = {3, 1, 2, 3}, K = 4
Output: NO
方法：
- 我们可以观察到，只有满足以下两个条件，才有可能使数组的所有元素都等于K ：
- 必须至少有一个元素等于K 。
- 必须存在一个连续的三元组，使得该三元组的任意两个值都大于或等于 K。
- 为了解决这个问题，我们需要创建一个辅助数组，比如aux[] ，它包含三个值 0、1、2。
- 最后的任务是检查是否有可能使aux数组的所有元素都等于 1。如果aux[]中的三个连续元素中有两个大于 0，那么我们可以取一个大小为 3 的子数组，并使所有元素都该子数组的值等于 1。然后我们将这个逻辑扩展到整个数组。
下面是上述方法的实现：
C++
// C++ implementation of above approach
#include
using namespace std;
// Function that prints
// whether is to possible
// to make all elements
// of the array equal to K
void makeAllK(int a[], int k, int n)
{
vector aux;
bool one_found = false;
// Fill vector aux
// according to the
// above approach
for (int i = 0; i < n; i++) {
if (a[i] < k)
aux.push_back(0);
else if (a[i] == k) {
aux.push_back(1);
one_found = true;
}
else
aux.push_back(2);
}
// Condition if K
// does not exist in
// the given array
if (one_found == false) {
cout << "NO"
<< "\n";
return;
}
bool ans = false;
if (n == 1
&& aux[0] == 1)
ans = true;
if (n == 2
&& aux[0] > 0
&& aux[1] > 1)
ans = true;
for (int i = 0; i < n - 2; i++) {
// Condition for minimum
// two elements is
// greater than 0 in
// pair of three elements
if (aux[i] > 0
&& aux[i + 1] > 0) {
ans = true;
break;
}
else if (aux[i] > 0
&& aux[i + 2] > 0) {
ans = true;
break;
}
else if (aux[i + 2] > 0
&& aux[i + 1] > 0) {
ans = true;
break;
}
}
if (ans == true)
cout << "YES"
<< "\n";
else
cout << "NO"
<< "\n";
}
// Driver Code
int main()
{
int arr[]
= { 1, 2, 3,
4, 5, 6,
7, 8, 9, 10 };
int K = 3;
int size = sizeof(arr)
/ sizeof(arr[0]);
makeAllK(arr, K, size);
return 0;
} Java
// Java implementation of the above approach
import java.util.*;
class GFG{
// Function that prints
// whether is to possible
// to make all elements
// of the array equal to K
static void makeAllK(int a[], int k, int n)
{
Vector aux = new Vector();
boolean one_found = false;
// Fill vector aux according
// to the above approach
for(int i = 0; i < n; i++)
{
if (a[i] < k)
aux.add(0);
else if (a[i] == k)
{
aux.add(1);
one_found = true;
}
else
aux.add(2);
}
// Condition if K does not
// exist in the given array
if (one_found == false)
{
System.out.print("NO" + "\n");
return;
}
boolean ans = false;
if (n == 1 && aux.get(0) == 1)
ans = true;
if (n == 2 && aux.get(0) > 0 &&
aux.get(1) > 1)
ans = true;
for(int i = 0; i < n - 2; i++)
{
// Condition for minimum
// two elements is
// greater than 0 in
// pair of three elements
if (aux.get(i) > 0 &&
aux.get(i + 1) > 0)
{
ans = true;
break;
}
else if (aux.get(i) > 0 &&
aux.get(i + 2) > 0)
{
ans = true;
break;
}
else if (aux.get(i + 2) > 0 &&
aux.get(i + 1) > 0)
{
ans = true;
break;
}
}
if (ans == true)
System.out.print("YES" + "\n");
else
System.out.print("NO" + "\n");
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 1, 2, 3, 4, 5,
6, 7, 8, 9, 10 };
int K = 3;
int size = arr.length;
makeAllK(arr, K, size);
}
}
// This code is contributed by amal kumar choubey Python3
# Python3 implementation of above approach
# Function that prints whether is
# to possible to make all elements
# of the array equal to K
def makeAllK(a, k, n):
aux = []
one_found = False
# Fill vector aux according
# to the above approach
for i in range(n):
if (a[i] < k):
aux.append(0)
elif (a[i] == k):
aux.append(1)
one_found = True
else:
aux.append(2)
# Condition if K does
# not exist in the given
# array
if (one_found == False):
print("NO")
return
ans = False
if (n == 1 and aux[0] == 1):
ans = True
if (n == 2 and aux[0] > 0 and aux[1] > 1):
ans = True
for i in range(n - 2):
# Condition for minimum two
# elements is greater than
# 0 in pair of three elements
if (aux[i] > 0 and aux[i + 1] > 0):
ans = True
break
elif (aux[i] > 0 and aux[i + 2] > 0):
ans = True
break
elif (aux[i + 2] > 0 and aux[i + 1] > 0):
ans = True
break
if (ans == True):
print("YES")
else:
print("NO")
# Driver Code
if __name__ == '__main__':
arr = [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ]
K = 3
size = len(arr)
makeAllK(arr, K, size)
# This code is contributed by Surendra_GangwarC#
// C# implementation of the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function that prints
// whether is to possible
// to make all elements
// of the array equal to K
static void makeAllK(int []a, int k, int n)
{
List aux = new List();
bool one_found = false;
// Fill vector aux according
// to the above approach
for(int i = 0; i < n; i++)
{
if (a[i] < k)
{
aux.Add(0);
}
else if (a[i] == k)
{
aux.Add(1);
one_found = true;
}
else
aux.Add(2);
}
// Condition if K does not
// exist in the given array
if (one_found == false)
{
Console.Write("NO" + "\n");
return;
}
bool ans = false;
if (n == 1 && aux[0] == 1)
ans = true;
if (n == 2 && aux[0] > 0 &&
aux[1] > 1)
ans = true;
for(int i = 0; i < n - 2; i++)
{
// Condition for minimum
// two elements is
// greater than 0 in
// pair of three elements
if (aux[i] > 0 &&
aux[i + 1] > 0)
{
ans = true;
break;
}
else if (aux[i] > 0 &&
aux[i + 2] > 0)
{
ans = true;
break;
}
else if (aux[i + 2] > 0 &&
aux[i + 1] > 0)
{
ans = true;
break;
}
}
if (ans == true)
Console.Write("YES" + "\n");
else
Console.Write("NO" + "\n");
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 1, 2, 3, 4, 5,
6, 7, 8, 9, 10 };
int K = 3;
int size = arr.Length;
makeAllK(arr, K, size);
}
}
// This code is contributed by amal kumar choubey Javascript
输出：
YES时间复杂度：O(N)
辅助空间：O(N)
如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live