给定大小为N的数组arr []和整数X。任务是找到数组中不存在的大于X的最小元素。
例子：
Input: arr[] = {1, 5, 10, 4, 7}, X = 4
Output: 6
6 is the samllest number greater than 4
which is not present in the array.
Input: arr[] = {1, 5, 10, 6, 11}, X = 10
Output: 12
方法：一种有效的解决方案基于二进制搜索。首先对数组进行排序。低取为零，高取为N – 1 。并搜索元素X +1。如果mid的元素( (low + high)/ 2)小于或等于search元素，则将low设为mid +1 。否则，取高到中– 1 。如果在中间的元素获得等于搜索元素，然后通过一个递增的搜索元素，使高N – 1。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the smallest element greater
// than x which is not present in a[]
int Next_greater(int a[], int n, int x)
{
// Sort the array
sort(a, a + n);
int low = 0, high = n - 1, ans = x + 1;
// Continue until low is less
// than or equals to high
while (low <= high) {
// Find mid
int mid = (low + high) / 2;
// If element at mid is less than
// or equals to searching element
if (a[mid] <= ans) {
// If mid is equals
// to searching element
if (a[mid] == ans) {
// Increment searching element
ans++;
// Make high as N - 1
high = n - 1;
}
// Make low as mid + 1
low = mid + 1;
}
// Make high as mid - 1
else
high = mid - 1;
}
// Return the next greater element
return ans;
}
// Driver code
int main()
{
int a[] = { 1, 5, 10, 4, 7 }, x = 4;
int n = sizeof(a) / sizeof(a[0]);
cout << Next_greater(a, n, x);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function to return the smallest element greater
// than x which is not present in a[]
static int Next_greater(int a[], int n, int x)
{
// Sort the array
Arrays.sort(a);
int low = 0, high = n - 1, ans = x + 1;
// Continue until low is less
// than or equals to high
while (low <= high)
{
// Find mid
int mid = (low + high) / 2;
// If element at mid is less than
// or equals to searching element
if (a[mid] <= ans)
{
// If mid is equals
// to searching element
if (a[mid] == ans)
{
// Increment searching element
ans++;
// Make high as N - 1
high = n - 1;
}
// Make low as mid + 1
low = mid + 1;
}
// Make high as mid - 1
else
high = mid - 1;
}
// Return the next greater element
return ans;
}
// Driver code
public static void main(String[] args)
{
int a[] = { 1, 5, 10, 4, 7 }, x = 4;
int n = a.length;
System.out.println(Next_greater(a, n, x));
}
}
// This code is contributed by PrinciRaj1992Python3
# Python3 implementation of the approach
# Function to return the smallest element
# greater than x which is not present in a[]
def Next_greater(a, n, x):
# Sort the array
a = sorted(a)
low, high, ans = 0, n - 1, x + 1
# Continue until low is less
# than or equals to high
while (low <= high):
# Find mid
mid = (low + high) // 2
# If element at mid is less than
# or equals to searching element
if (a[mid] <= ans):
# If mid is equals
# to searching element
if (a[mid] == ans):
# Increment searching element
ans += 1
# Make high as N - 1
high = n - 1
# Make low as mid + 1
low = mid + 1
# Make high as mid - 1
else:
high = mid - 1
# Return the next greater element
return ans
# Driver code
a = [1, 5, 10, 4, 7]
x = 4
n = len(a)
print(Next_greater(a, n, x))
# This code is contributed
# by Mohit KumarC#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the smallest element greater
// than x which is not present in a[]
static int Next_greater(int []a, int n, int x)
{
// Sort the array
Array.Sort(a);
int low = 0, high = n - 1, ans = x + 1;
// Continue until low is less
// than or equals to high
while (low <= high)
{
// Find mid
int mid = (low + high) / 2;
// If element at mid is less than
// or equals to searching element
if (a[mid] <= ans)
{
// If mid is equals
// to searching element
if (a[mid] == ans)
{
// Increment searching element
ans++;
// Make high as N - 1
high = n - 1;
}
// Make low as mid + 1
low = mid + 1;
}
// Make high as mid - 1
else
high = mid - 1;
}
// Return the next greater element
return ans;
}
// Driver code
public static void Main(String[] args)
{
int []a = { 1, 5, 10, 4, 7 };
int x = 4;
int n = a.Length;
Console.WriteLine(Next_greater(a, n, x));
}
}
// This code is contributed by Princi SinghPHP
输出：
6