给出了一个包含正元素的数组。 “ A”和“ B”是定义范围的两个数字。编写函数以检查数组是否包含给定范围内的所有元素。
例子 ：
Input : arr[] = {1 4 5 2 7 8 3}
A : 2, B : 5
Output : Yes
Input : arr[] = {1 4 5 2 7 8 3}
A : 2, B : 6
Output : No
方法1 ：(直观)
最直观的方法是对数组进行排序，然后从大于“ A”的元素到大于“ B”的元素进行检查。如果这些元素按连续顺序排列，则范围中的所有元素都存在于数组中。
时间复杂度：O(n log n)
辅助空间：O(1)
方法2 ：(散列)
我们可以维护一个count数组或一个散列表，其中存储该数组中A…B范围内所有数字的计数。然后，我们可以简单地检查每个数字是否至少出现过一次。
时间复杂度：O(n)
辅助空间：O(max_element)
方法3 ：(最佳)
对数组进行线性遍历。如果找到的元素使得| arr [i] | > = A和| arr [i] |
C++
#include
using namespace std;
// Function to check the array for elements in
// given range
bool check_elements(int arr[], int n, int A, int B)
{
// Range is the no. of elements that are
// to be checked
int range = B - A;
// Traversing the array
for (int i = 0; i < n; i++) {
// If an element is in range
if (abs(arr[i]) >= A && abs(arr[i]) <= B) {
// Negating at index ‘element – A’
int z = abs(arr[i]) - A;
if (arr[z] > 0) {
arr[z] = arr[z] * -1;
}
}
}
// Checking whether elements in range 0-range
// are negative
int count=0;
for (int i = 0; i <= range && i 0)
return false;
else
count++;
}
if(count!= (range+1))
return false;
// All range elements are present
return true;
}
// Driver code
int main()
{
// Defining Array and size
int arr[] = { 1, 4, 5, 2, 7, 8, 3 };
int n = sizeof(arr) / sizeof(arr[0]);
// A is lower limit and B is the upper limit
// of range
int A = 2, B = 5;
// True denotes all elements were present
if (check_elements(arr, n, A, B))
cout << "Yes";
// False denotes any element was not present
else
cout << "No";
return 0;
} Java
// JAVA Code for Check if an array contains
// all elements of a given range
import java.util.*;
class GFG {
// Function to check the array for elements in
// given range
public static boolean check_elements(int arr[], int n,
int A, int B)
{
// Range is the no. of elements that are
// to be checked
int range = B - A;
// Traversing the array
for (int i = 0; i < n; i++) {
// If an element is in range
if (Math.abs(arr[i]) >= A &&
Math.abs(arr[i]) <= B) {
int z = Math.abs(arr[i]) - A;
if (arr[z] > 0) {
arr[z] = arr[z] * -1;
}
}
}
// Checking whether elements in range 0-range
// are negative
int count=0;
for (int i = 0; i <= range && i 0)
return false;
else
count++;
}
if(count!= (range+1))
return false;
// All range elements are present
return true;
}
/* Driver program to test above function */
public static void main(String[] args)
{
// Defining Array and size
int arr[] = { 1, 4, 5, 2, 7, 8, 3 };
int n = arr.length;
// A is lower limit and B is the upper limit
// of range
int A = 2, B = 5;
// True denotes all elements were present
if (check_elements(arr, n, A, B))
System.out.println("Yes");
// False denotes any element was not present
else
System.out.println("No");
}
}
// This code is contributed by Arnav Kr. Mandal. Python3
# Function to check the array for
# elements in given range
def check_elements(arr, n, A, B) :
# Range is the no. of elements
# that are to be checked
rangeV = B - A
# Traversing the array
for i in range(0, n):
# If an element is in range
if (abs(arr[i]) >= A and
abs(arr[i]) <= B) :
# Negating at index ‘element – A’
z = abs(arr[i]) - A
if (arr[z] > 0) :
arr[z] = arr[z] * -1
# Checking whether elements in
# range 0-range are negative
count = 0
for i in range(0, rangeV + 1):
if i >= n:
break
# Element from range is
# missing from array
if (arr[i] > 0):
return False
else:
count = count + 1
if(count != (rangeV + 1)):
return False
# All range elements are present
return True
# Driver code
# Defining Array and size
arr = [ 1, 4, 5, 2, 7, 8, 3 ]
n = len(arr)
# A is lower limit and B is
# the upper limit of range
A = 2
B = 5
# True denotes all elements
# were present
if (check_elements(arr, n, A, B)) :
print("Yes")
# False denotes any element
# was not present
else:
print("No")
# This code is contributed
# by Yatin GuptaC#
// C# Code for Check if an array contains
// all elements of a given range
using System;
class GFG {
// Function to check the array for
// elements in given range
public static bool check_elements(int []arr, int n,
int A, int B)
{
// Range is the no. of elements
// that are to be checked
int range = B - A;
// Traversing the array
for (int i = 0; i < n; i++)
{
// If an element is in range
if (Math.Abs(arr[i]) >= A &&
Math.Abs(arr[i]) <= B)
{
int z = Math.Abs(arr[i]) - A;
if (arr[z] > 0)
{
arr[z] = arr[z] * - 1;
}
}
}
// Checking whether elements in
// range 0-range are negative
int count=0;
for (int i = 0; i <= range
&& i < n; i++)
{
// Element from range is
// missing from array
if (arr[i] > 0)
return false;
else
count++;
}
if(count != (range + 1))
return false;
// All range elements are present
return true;
}
// Driver Code
public static void Main(String []args)
{
// Defining Array and size
int []arr = {1, 4, 5, 2, 7, 8, 3};
int n = arr.Length;
// A is lower limit and B is
// the upper limit of range
int A = 2, B = 5;
// True denotes all elements were present
if (check_elements(arr, n, A, B))
Console.WriteLine("Yes");
// False denotes any element was not present
else
Console.WriteLine("No");
}
}
// This code is contributed by vt_m.PHP
= $A &&
abs($arr[$i]) <= $B)
{
// Negating at index
// ‘element – A’
$z = abs($arr[$i]) - $A;
if ($arr[$z] > 0)
{
$arr[$z] = $arr[$z] * -1;
}
}
}
// Checking whether elements
// in range 0-range are negative
$count = 0;
for ($i = 0; $i <= $range &&
$i< $n; $i++)
{
// Element from range is
// missing from array
if ($arr[$i] > 0)
return -1;
else
$count++;
}
if($count!= ($range + 1))
return -1;
// All range elements
// are present
return true;
}
// Driver code
// Defining Array and size
$arr = array(1, 4, 5, 2,
7, 8, 3);
$n = sizeof($arr);
// A is lower limit and
// B is the upper limit
// of range
$A = 2; $B = 5;
// True denotes all
// elements were present
if ((check_elements($arr, $n,
$A, $B)) == true)
echo "Yes";
// False denotes any
// element was not present
else
echo "No";
// This code is contributed by aj_36
?>输出 ：
Yes
时间复杂度： O(n)
辅助空间： O(1)