给定一个数组arr[]具有N 个不同的正元素,任务是生成另一个数组B[]使得对于数组中的每个第i个索引, arr[] , B[i]是从arr 中丢失的最小正数[]不包括arr[i] 。
例子:
Input: arr[] = {2, 1, 5, 3}
Output: B[] = {2, 1, 4, 3}
Explanation: After excluding the arr[0], the array is {1, 5, 3}, and the minimum positive number which is not present in this array is 2. Therefore, B[0] = 2. Similarly, after excluding arr[1], arr[2], arr[3], the minimum positive numbers which are not present in the array are 1, 4 and 3, respectively. Hence, B[1] = 1, B[2] = 4, B[3] = 3.
Input: arr[] = {1, 9, 2, 4}
Output: B[] = {1, 3, 2, 3}
朴素的方法:解决这个问题最简单的方法是遍历数组arr[]并为每个索引i初始化一个数组hash[]并为每个索引j (其中j ≠ i )更新hash[arr[j]] =1 。现在从索引1遍历数组hash[]并找到hash[k] = 0的最小索引k并更新B[i] = k 。最后,完成上述步骤后打印数组B[] 。
时间复杂度: O(N 2 ) 其中 N 是给定数组的长度。
辅助空间: O(N)
高效方法:为了优化上述方法,思想是计算数组arr[] 的MEX并遍历数组arr[] 。如果ARR [i]是小于阵列ARR的MEX []然后MEX不含此元件将是改编[I]本身,如果ARR [i]为比阵列A的MEX []大于所述阵列的MEX不会排除此元素后更改。
请按照以下步骤解决问题:
- 初始化一个数组,比如hash[] ,以存储值i是否存在于数组arr[] 中。如果i存在hash[i] = 1否则hash[i] = 0。
- 初始化变量MexOfArr以存储数组arr[] 的MEX并从1遍历数组hash[]以找到hash[j] = 0的最小索引j ,这意味着数组arr[] 中不存在值j并存储MexOfArr = j 。
- 现在遍历数组arr[]并且如果arr[i]小于MexOfArr,则存储B[i] = arr[i] else B[i] = MexOfArr 。
- 完成上述步骤后,将数组B[] 的元素打印为所需答案。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
#define MAXN 100001
// Function to construct array B[] that
// stores MEX of array A[] excluding A[i]
void constructMEX(int arr[], int N)
{
// Stores elements present in arr[]
int hash[MAXN] = { 0 };
// Mark all values 1, if present
for (int i = 0; i < N; i++) {
hash[arr[i]] = 1;
}
// Initialize variable to store MEX
int MexOfArr;
// Find MEX of arr[]
for (int i = 1; i < MAXN; i++) {
if (hash[i] == 0) {
MexOfArr = i;
break;
}
}
// Stores MEX for all indices
int B[N];
// Traverse the given array
for (int i = 0; i < N; i++) {
// Update MEX
if (arr[i] < MexOfArr)
B[i] = arr[i];
// MEX default
else
B[i] = MexOfArr;
}
// Print the array B
for (int i = 0; i < N; i++)
cout << B[i] << ' ';
}
// Driver Code
int main()
{
// Given array
int arr[] = { 2, 1, 5, 3 };
// Given size
int N = sizeof(arr)
/ sizeof(arr[0]);
// Function call
constructMEX(arr, N);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
static int MAXN = 100001;
// Function to construct array
// B[] that stores MEX of array
// A[] excluding A[i]
static void constructMEX(int arr[],
int N)
{
// Stores elements present
// in arr[]
int hash[] = new int[MAXN];
for (int i = 0; i < N; i++)
{
hash[i] = 0;
}
// Mark all values 1, if
// present
for (int i = 0; i < N; i++)
{
hash[arr[i]] = 1;
}
// Initialize variable to
// store MEX
int MexOfArr = 0 ;
// Find MEX of arr[]
for (int i = 1; i < MAXN; i++)
{
if (hash[i] == 0)
{
MexOfArr = i;
break;
}
}
// Stores MEX for all
// indices
int B[] = new int[N];
// Traverse the given array
for (int i = 0; i < N; i++)
{
// Update MEX
if (arr[i] < MexOfArr)
B[i] = arr[i];
// MEX default
else
B[i] = MexOfArr;
}
// Print the array B
for (int i = 0; i < N; i++)
System.out.print(B[i] + " ");
}
// Driver Code
public static void main(String[] args)
{
// Given array arr[]
int arr[] = {2, 1, 5, 3};
// Size of array
int N = arr.length;
// Function call
constructMEX(arr, N);
}
}
// This code is contributed by sanjoy_62
Python3
# Python3 program for the
# above approach
MAXN = 100001
# Function to construct
# array B[] that stores
# MEX of array A[] excluding
# A[i]
def constructMEX(arr, N):
# Stores elements present
# in arr[]
hash = [0] * MAXN
# Mark all values 1,
# if present
for i in range(N):
hash[arr[i]] = 1
# Initialize variable to
# store MEX
MexOfArr = 0
# Find MEX of arr[]
for i in range(1, MAXN):
if (hash[i] == 0):
MexOfArr = i
break
# Stores MEX for all
# indices
B = [0] * N
# Traverse the given array
for i in range(N):
# Update MEX
if (arr[i] < MexOfArr):
B[i] = arr[i]
# MEX default
else:
B[i] = MexOfArr
# Prthe array B
for i in range(N):
print(B[i], end = " ")
# Driver Code
if __name__ == '__main__':
# Given array
arr = [2, 1, 5, 3]
# Given size
N = len(arr)
# Function call
constructMEX(arr, N)
# This code is contributed by Mohit Kumar 29
C#
// C# program for the above approach
using System;
class GFG{
static int MAXN = 100001;
// Function to construct array
// B[] that stores MEX of array
// A[] excluding A[i]
static void constructMEX(int[] arr,
int N)
{
// Stores elements present
// in arr[]
int[] hash = new int[MAXN];
for(int i = 0; i < N; i++)
{
hash[i] = 0;
}
// Mark all values 1, if
// present
for(int i = 0; i < N; i++)
{
hash[arr[i]] = 1;
}
// Initialize variable to
// store MEX
int MexOfArr = 0;
// Find MEX of arr[]
for(int i = 1; i < MAXN; i++)
{
if (hash[i] == 0)
{
MexOfArr = i;
break;
}
}
// Stores MEX for all
// indices
int[] B = new int[N];
// Traverse the given array
for(int i = 0; i < N; i++)
{
// Update MEX
if (arr[i] < MexOfArr)
B[i] = arr[i];
// MEX default
else
B[i] = MexOfArr;
}
// Print the array B
for(int i = 0; i < N; i++)
Console.Write(B[i] + " ");
}
// Driver Code
public static void Main()
{
// Given array arr[]
int[] arr = { 2, 1, 5, 3 };
// Size of array
int N = arr.Length;
// Function call
constructMEX(arr, N);
}
}
// This code is contributed by code_hunt
Javascript
2 1 4 3
时间复杂度: O(N)
辅助空间: O(N)
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