给定两个序列，一个是递增序列a[]另一个是正常序列b[] ，在递增序列中找到给定序列中不存在的第 K 个缺失元素。如果没有第 k 个缺失元素，则输出 -1
例子：
Input: a[] = {0, 2, 4, 6, 8, 10, 12, 14, 15};
b[] = {4, 10, 6, 8, 12};
k = 3
Output: 14
Explanation : The numbers from increasing sequence that
are not present in the given sequence are 0, 2, 14, 15.
The 3rd missing number is 14. n1递增序列 a[] 上的元素数。
n2给定序列 b[] 中的元素数。
一种天真的方法是迭代递增序列中的每个元素并检查它是否存在于给定序列中，并保留不存在元素的计数器，并打印第 k 个不存在元素。这将不够有效，因为它有两个嵌套的 for 循环，需要 O(n2)。
时间复杂度：O(n1 * n2)
辅助空间：O(1)
一种有效的方法是使用散列。我们将给定序列的所有元素存储在一个哈希表中。然后我们遍历所有递增序列的元素。对于每个元素，我们在哈希表中搜索它。如果元素不在哈希表中，那么我们增加丢失元素的计数。如果 count 变为 k，我们返回缺失的元素。
下面是上述方法的实现
C++
// C++ program to find the k-th missing element
// in a given sequence
#include
using namespace std;
// Returns k-th missing element. It returns -1 if
// no k is more than number of missing elements.
int find(int a[], int b[], int k, int n1, int n2)
{
// Insert all elements of givens sequence b[].
unordered_set s;
for (int i = 0; i < n2; i++)
s.insert(b[i]);
// Traverse through increasing sequence and
// keep track of count of missing numbers.
int missing = 0;
for (int i = 0; i < n1; i++) {
if (s.find(a[i]) == s.end())
missing++;
if (missing == k)
return a[i];
}
return -1;
}
// driver program to test the above function
int main()
{
int a[] = { 0, 2, 4, 6, 8, 10, 12, 14, 15 };
int b[] = { 4, 10, 6, 8, 12 };
int n1 = sizeof(a) / sizeof(a[0]);
int n2 = sizeof(b) / sizeof(b[0]);
int k = 3;
cout << find(a, b, k, n1, n2);
return 0;
} Java
// Java program to find the k-th missing element
// in a given sequence
import java.util.*;
class GFG
{
// Returns k-th missing element. It returns -1 if
// no k is more than number of missing elements.
static int find(int a[], int b[], int k, int n1, int n2)
{
// Insert all elements of givens sequence b[].
LinkedHashSet s = new LinkedHashSet<>();
for (int i = 0; i < n2; i++)
s.add(b[i]);
// Traverse through increasing sequence and
// keep track of count of missing numbers.
int missing = 0;
for (int i = 0; i < n1; i++)
{
if(!s.contains(a[i]) )
missing++;
if (missing == k)
return a[i];
}
return -1;
}
// Driver code
public static void main(String[] args)
{
int a[] = { 0, 2, 4, 6, 8, 10, 12, 14, 15 };
int b[] = { 4, 10, 6, 8, 12 };
int n1 = a.length;
int n2 = b.length;
int k = 3;
System.out.println(find(a, b, k, n1, n2));
}
}
// This code has been contributed by 29AjayKumar Python3
# Python3 program to find the k-th
# missing element in a given sequence
# Returns k-th missing element. It returns -1 if
# no k is more than number of missing elements.
def find(a, b, k, n1, n2):
# insert all elements of
# given sequence b[].
s = set()
for i in range(n2):
s.add(b[i])
# Traverse through increasing sequence and
# keep track of count of missing numbers.
missing = 0
for i in range(n1):
if a[i] not in s:
missing += 1
if missing == k:
return a[i]
return -1
# Driver code
a = [0, 2, 4, 6, 8, 10, 12, 14, 15]
b = [4, 10, 6, 8, 12]
n1 = len(a)
n2 = len(b)
k = 3
print(find(a, b, k, n1, n2))
# This code is contributed by Shrikant13C#
// C# program to find the k-th missing element
// in a given sequence
using System;
using System.Collections.Generic;
class GFG
{
// Returns k-th missing element. It returns -1 if
// no k is more than number of missing elements.
static int find(int []a, int []b, int k, int n1, int n2)
{
// Insert all elements of givens sequence b[].
HashSet s = new HashSet();
for (int i = 0; i < n2; i++)
s.Add(b[i]);
// Traverse through increasing sequence and
// keep track of count of missing numbers.
int missing = 0;
for (int i = 0; i < n1; i++)
{
if(!s.Contains(a[i]) )
missing++;
if (missing == k)
return a[i];
}
return -1;
}
// Driver code
public static void Main(String[] args)
{
int []a = { 0, 2, 4, 6, 8, 10, 12, 14, 15 };
int []b = { 4, 10, 6, 8, 12 };
int n1 = a.Length;
int n2 = b.Length;
int k = 3;
Console.WriteLine(find(a, b, k, n1, n2));
}
}
/* This code contributed by PrinciRaj1992 */ Javascript
输出：
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