找到所有元素都小于它的元素，然后所有元素都大于的元素

给定一个数组，找到一个元素，在该元素之前所有元素都小于该元素，之后所有元素都大于该元素。如果存在这样的元素，则返回该元素的索引，否则返回-1。

例子：

Input: arr[] = {5, 1, 4, 3, 6, 8, 10, 7, 9}; 
Output: 4 
Explanation: All elements on left of arr[4] are smaller than it 
and all elements on right are greater.

Input: arr[] = {5, 1, 4, 4}; 
Output: -1 
Explanation : No such index exits.

预期时间复杂度：O(n)。

一个简单的解决方案是逐个考虑每个元素。对于每个元素，将其与左侧的所有元素和右侧的所有元素进行比较。该解决方案的时间复杂度为O(n ² )。

一个有效的解决方案可以使用O(n)的额外空间在O(n)的时间内解决此问题。以下是详细的解决方案。

1. 创建两个数组leftMax []和rightMin []。
2. 从左到右遍历输入数组，并填充leftMax []，以使leftMax [i]在输入数组中包含从0到i-1的最大元素。
3. 从右到左遍历输入数组，并填充rightMin []，以使rightMin [i]在输入数组中包含从到n-1到i + 1的最小元素。
4. 遍历输入数组。对于每个元素arr [i]，检查arr [i]是否大于leftMax [i]并小于rightMin [i]。如果是，请返回i。

对上述方法的进一步优化是仅使用一个额外的数组，并且仅遍历两次输入数组。第一次遍历与上面相同，并填充leftMax []。下一个遍历从右侧遍历并跟踪最小值。第二次遍历还会找到所需的元素。

下图是上述方法的模拟：

下面是上述方法的实现。

C++

// C++ program to find the element which is greater than
// all left elements and smaller than all right elements.
#include 
using namespace std;
 
// Function to return the index of the element which is greater than
// all left elements and smaller than all right elements.
int findElement(int arr[], int n)
{
    // leftMax[i] stores maximum of arr[0..i-1]
    int leftMax[n];
    leftMax[0] = INT_MIN;
 
    // Fill leftMax[]1..n-1]
    for (int i = 1; i < n; i++)
        leftMax[i] = max(leftMax[i-1], arr[i-1]);
 
    // Initialize minimum from right
    int rightMin = INT_MAX;
 
    // Traverse array from right
    for (int i=n-1; i>=0; i--)
    {
        // Check if we found a required element
        if (leftMax[i] < arr[i] && rightMin > arr[i])
             return i;
 
        // Update right minimum
        rightMin = min(rightMin, arr[i]);
    }
 
    // If there was no element matching criteria
    return -1;
}
 
// Driver program
int main()
{
    int arr[] = {5, 1, 4, 3, 6, 8, 10, 7, 9};
    int n = sizeof arr / sizeof arr[0];
    cout << "Index of the element is " << findElement(arr, n);
    return 0;
}

Java

// Java program to find the element which is greater than
// all left elements and smaller than all right elements.
import java.io.*;
import java.util.*;
 
public class GFG {
       static int findElement(int[] arr, int n)
       {
              // leftMax[i] stores maximum of arr[0..i-1]
              int[] leftMax = new int[n];
              leftMax[0] = Integer.MIN_VALUE;
 
              // Fill leftMax[]1..n-1]
              for (int i = 1; i < n; i++)
                   leftMax[i] = Math.max(leftMax[i - 1], arr[i - 1]);
                    
              // Initialize minimum from right
              int rightMin = Integer.MAX_VALUE;
 
              // Traverse array from right
              for (int i = n - 1; i >= 0; i--)
              {
                   // Check if we found a required element
                   if (leftMax[i] < arr[i] && rightMin > arr[i])
                       return i;
 
                   // Update right minimum
                   rightMin = Math.min(rightMin, arr[i]);
              }
                
              // If there was no element matching criteria
              return -1;
 
               
       }
 
       // Driver code
       public static void main(String args[])
       {
              int[] arr = {5, 1, 4, 3, 6, 8, 10, 7, 9};
              int n = arr.length;
              System.out.println("Index of the element is " +
              findElement(arr, n));
       }
 
       // This code is contributed
       // by rachana soma
}

Python3

# Python3 program to find the element which is greater than
# all left elements and smaller than all right elements.
 
def findElement(arr, n):
  
    # leftMax[i] stores maximum of arr[0..i-1]
    leftMax = [None] * n
    leftMax[0] = float('-inf')
 
    # Fill leftMax[]1..n-1]
    for i in range(1, n):
        leftMax[i] = max(leftMax[i-1], arr[i-1])
 
    # Initialize minimum from right
    rightMin = float('inf')
 
    # Traverse array from right
    for i in range(n-1, -1, -1):
      
        # Check if we found a required element
        if leftMax[i] < arr[i] and rightMin > arr[i]:
            return i
 
        # Update right minimum
        rightMin = min(rightMin, arr[i])
      
    # If there was no element matching criteria
    return -1
 
# Driver program
if __name__ == "__main__":
 
    arr = [5, 1, 4, 3, 6, 8, 10, 7, 9]
    n = len(arr)
    print("Index of the element is", findElement(arr, n))
  
# This code is contributed by Rituraj Jain

C#

// C# program to find the element which is greater than
// all left elements and smaller than all right elements.
using System;
 
class GFG
{
static int findElement(int[] arr, int n)
{
    // leftMax[i] stores maximum of arr[0..i-1]
    int[] leftMax = new int[n];
    leftMax[0] = int.MinValue;
 
    // Fill leftMax[]1..n-1]
    for (int i = 1; i < n; i++)
        leftMax[i] = Math.Max(leftMax[i - 1], arr[i - 1]);
 
    // Initialize minimum from right
    int rightMin = int.MaxValue;
 
    // Traverse array from right
    for (int i=n-1; i>=0; i--)
    {
        // Check if we found a required element
        if (leftMax[i] < arr[i] && rightMin > arr[i])
            return i;
 
        // Update right minimum
        rightMin = Math.Min(rightMin, arr[i]);
    }
 
    // If there was no element matching criteria
    return -1;
}
 
// Driver program
public static void Main()
{
    int[] arr = {5, 1, 4, 3, 6, 8, 10, 7, 9};
    int n = arr.Length;
    Console.Write( "Index of the element is " + findElement(arr, n));
}
}
 
// This code is contributed
// by Akanksha Rai(Abby_akku)

PHP

= 0; $i--)
    {
        // Check if we found a required
        // element
        if ($leftMax[$i] < $arr[$i] &&
            $rightMin > $arr[$i])
            return $i;
 
        // Update right minimum
        $rightMin = min($rightMin, $arr[$i]);
    }
 
    // If there was no element
    // matching criteria
    return -1;
}
 
// Driver Code
$arr = array(5, 1, 4, 3, 6, 8, 10, 7, 9);
$n = count($arr);
echo "Index of the element is " ,
           findElement($arr, $n);
 
// This code is contributed
// by Sach_Code
?>

Javascript

 
// Javascript program to find the element
// which is greater than all left elements
// and smaller than all right elements.
 
// Function to return the index of the
// element which is greater than all
// left elements and smaller than all right elements.
function findElement(arr, n)
{
     
    // leftMax[i] stores maximum of arr[0..i-1]
    var leftMax = Array(n).fill(0);
    leftMax[0] = Number.MIN_VALUE;
 
    // Fill leftMax1..n-1]
    for(i = 1; i < n; i++)
        leftMax[i] = Math.max(leftMax[i - 1],
                                  arr[i - 1]);
 
    // Initialize minimum from right
    var rightMin = Number.MAX_VALUE;
 
    // Traverse array from right
    for(i = n - 1; i >= 0; i--)
    {
         
        // Check if we found a required element
        if (leftMax[i] < arr[i] && 
              rightMin > arr[i])
            return i;
 
        // Update right minimum
        rightMin = Math.min(rightMin, arr[i]);
    }
 
    // If there was no element
    // matching criteria
    return -1;
}
 
// Driver code
var arr = [ 5, 1, 4, 3, 6, 8, 10, 7, 9 ];
var n = arr.length;
 
document.write("Index of the element is " +
               findElement(arr, n));
 
// This code is contributed by aashish1995
 

输出：

Index of the element is 4

时间复杂度： O(n)
辅助空间： O(n)
感谢Gaurav Ahirwar提供上述解决方案。

空间优化方法：

C++

// C++ program to find the element which is greater than
// all left elements and smaller than all right elements.
#include 
using namespace std;
 
int findElement(int a[], int n)
{
    // Base case
    if (n == 1 || n == 2) {
        return -1;
    }
 
    // 1.element is the possible candidate for the solution
    // of the problem.
      // 2.idx is the index of the possible
    // candidate.
      // 3.maxx is the value which is maximum on the
    // left side of the array.
      // 4.bit tell whether the loop is
    // terminated from the if condition or from the else
    // condition when loop do not satisfied the condition.
    // 5.check is the variable which tell whether the
    // element is updated or not
 
    int element = a[0], maxx = a[0], bit = -1, check = 0;
    int idx = -1;
 
    // The extreme two of the array can not be the solution
    // Therefore iterate the loop from i = 1 to < n-1
    for (int i = 1; i < (n - 1);) {
 
        // here we find the possible candidate where Element
        // with left side smaller and right side greater.
        // when the if condition fail we check and update in
        // else condition.
 
        if (a[i] < maxx && i < (n - 1)) {
            i++;
            bit = 0;
        }
 
        // here we update the possible element if the
        // element is greater than the maxx (maximum element
        // so far). In while loop we sur-pass the value which
        // is greater than the element
        else {
            if (a[i] >= maxx) {
                element = a[i];
                idx = i;
                check = 1;
                maxx = a[i];
            }
            if (check == 1) {
                i++;
            }
            bit = 1;
            while (a[i] >= element && i < (n - 1)) {
                if (a[i] > maxx) {
                    maxx = a[i];
                }
                i++;
            }
            check = 0;
        }
    }
 
    // checking for the last value and whether the loop is
    // terminated from else or if block.
 
    if (element <= a[n - 1] && bit == 1) {
        return idx;
    }
    else {
        return -1;
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 5, 1, 4, 3, 6, 8, 10, 7, 9 };
    int n = sizeof arr / sizeof arr[0];
   
      // Function Call
    cout << "Index of the element is "
         << findElement(arr, n);
    return 0;
}

Java

// Java program to find the element
// which is greater than all left
// elements and smaller than all
// right elements.
class GFG{
     
static int findElement(int []a, int n)
{
     
    // Base case
    if (n == 1 || n == 2)
    {
        return -1;
    }
     
    // 1.element is the possible candidate for
    // the solution of the problem.
    // 2.idx is the index of the possible
    // candidate.
    // 3.maxx is the value which is maximum on the
    // left side of the array.
    // 4.bit tell whether the loop is
    // terminated from the if condition or from
    // the else condition when loop do not
    // satisfied the condition.
    // 5.check is the variable which tell whether the
    // element is updated or not
    int element = a[0], maxx = a[0],
            bit = -1, check = 0;
    int idx = -1;
     
    // The extreme two of the array can
    // not be the solution. Therefore
    // iterate the loop from i = 1 to < n-1
    for(int i = 1; i < (n - 1);)
    {
         
        // Here we find the possible candidate
        // where Element with left side smaller
        // and right side greater. When the if
        // condition fail we check and update in
        // else condition.
        if (a[i] < maxx && i < (n - 1))
        {
            i++;
            bit = 0;
        }
         
        // Here we update the possible element
        // if the element is greater than the
        // maxx (maximum element so far). In
        // while loop we sur-pass the value which
        // is greater than the element
        else
        {
            if (a[i] >= maxx)
            {
                element = a[i];
                idx = i;
                check = 1;
                maxx = a[i];
            }
            if (check == 1)
            {
                i++;
            }
            bit = 1;
              
            while (a[i] >= element && i < (n - 1))
            {
                if (a[i] > maxx)
                {
                    maxx = a[i];
                }
                i++;
            }
            check = 0;
        }
    }
     
    // Checking for the last value and whether
    // the loop is terminated from else or
    // if block.
    if (element <= a[n - 1] && bit == 1)
    {
        return idx;
    }
    else
    {
        return -1;
    }
         
}
 
// Driver code
public static void main(String []args)
{
    int []arr = { 5, 1, 4, 3, 6, 8, 10, 7, 9 };
    int n = arr.length;
     
    System.out.println("Index of the element is " +
                       findElement(arr, n));
}
}
 
// This code is contributed by avanitrachhadiya2155

Python3

# Python3 program to find the element which
# is greater than all left elements and
# smaller than all right elements.
def findElement (a, n):
 
    # Base case
    if (n == 1 or n == 2):
        return -1
 
    # 1. element is the possible candidate
    # for the solution of the problem
    # 2. idx is the index of the
    # possible candidate
    # 3. maxx is the value which is maximum
    # on the left side of the array
    # 4. bit tell whether the loop is
    # terminated from the if condition or
    # from the else condition when loop do
    # not satisfied the condition.
    # 5. check is the variable which tell
    # whether the element is updated or not
    element, maxx, bit = a[0], a[0], -1
    check = 0
    idx = -1
 
    # The extreme of the array can't be
    # the solution Therefore iterate
    # the loop from i = 1 to < n-1
    i = 1
    while (i < (n - 1)):
 
        # Here we find the possible candidate
        # where element with left side smaller
        # and right side greater. when the if
        # condition fail we check and update
        # in else condition
        if (a[i] < maxx and i < (n - 1)):
            i += 1
            bit = 0
 
        # Here we update the possible element
        # if the element is greater than the
        # maxx (maximum element so far). In
        # while loop we sur-pass the value
        # which is greater than the element
        else:
            if (a[i] >= maxx):
                element = a[i]
                idx = i
                check = 1
                maxx = a[i]
 
            if (check == 1):
                i += 1
 
            bit = 1
            while (a[i] >= element and i < (n - 1)):
                if (a[i] > maxx):
                    maxx = a[i]
 
                i += 1
 
            check = 0
 
    # Checking for the last value and whether
    # the loop is terminated from else or
    # if block
    if (element <= a[n - 1] and bit == 1):
        return idx
    else:
        return -1
 
# Driver Code
if __name__ == '__main__':
     
    arr = [ 5, 1, 4, 3,
            6, 8, 10, 7, 9 ]
    n = len(arr)
 
    # Function call
    print("Index of the element is",
           findElement(arr, n))
 
# This code is contributed by himanshu77

C＃

// C# program to find the element
// which is greater than all left
// elements and smaller than all
// right elements.
using System;
  
class GFG{
  
static int findElement(int []a, int n)
{
     
    // Base case
    if (n == 1 || n == 2)
    {
        return -1;
    }
     
    // 1.element is the possible candidate for
    // the solution of the problem.
    // 2.idx is the index of the possible
    // candidate.
    // 3.maxx is the value which is maximum on the
    // left side of the array.
    // 4.bit tell whether the loop is
    // terminated from the if condition or from
    // the else condition when loop do not
    // satisfied the condition.
    // 5.check is the variable which tell whether the
    // element is updated or not
    int element = a[0], maxx = a[0],
        bit = -1, check = 0;
    int idx = -1;
  
    // The extreme two of the array can
    // not be the solution. Therefore
    // iterate the loop from i = 1 to < n-1
    for(int i = 1; i < (n - 1);)
    {
         
        // Here we find the possible candidate
        // where Element with left side smaller
        // and right side greater. When the if
        // condition fail we check and update in
        // else condition.
        if (a[i] < maxx && i < (n - 1))
        {
            i++;
            bit = 0;
        }
         
        // Here we update the possible element
        // if the element is greater than the
        // maxx (maximum element so far). In
        // while loop we sur-pass the value which
        // is greater than the element
        else
        {
            if (a[i] >= maxx)
            {
                element = a[i];
                idx = i;
                check = 1;
                maxx = a[i];
            }
            if (check == 1)
            {
                i++;
            }
            bit = 1;
             
            while (a[i] >= element && i < (n - 1))
            {
                if (a[i] > maxx)
                {
                    maxx = a[i];
                }
                i++;
            }
            check = 0;
        }
    }
  
    // Checking for the last value and whether
    // the loop is terminated from else or
    // if block.
    if (element <= a[n - 1] && bit == 1)
    {
        return idx;
    }
    else
    {
        return -1;
    }
}
 
// Driver code
public static void Main(string[] args)
{
    int []arr = { 5, 1, 4, 3, 6, 8, 10, 7, 9 };
    int n = arr.Length;
 
    // Function Call
    Console.Write("Index of the element is " +
                   findElement(arr, n));
}
}
 
// This code is contributed by rutvik_56

Java脚本

// javascript program to find the element
// which is greater than all left
// elements and smaller than all
// right elements.   
function findElement(a , n)
{
 
        // Base case
        if (n == 1 || n == 2)
        {
            return -1;
        }
 
        // 1.element is the possible candidate for
        // the solution of the problem.
        // 2.idx is the index of the possible
        // candidate.
        // 3.maxx is the value which is maximum on the
        // left side of the array.
        // 4.bit tell whether the loop is
        // terminated from the if condition or from
        // the else condition when loop do not
        // satisfied the condition.
        // 5.check is the variable which tell whether the
        // element is updated or not
        var element = a[0], maxx = a[0], bit = -1, check = 0;
        var idx = -1;
 
        // The extreme two of the array can
        // not be the solution. Therefore
        // iterate the loop from i = 1 to < n-1
        for (i = 1; i < (n - 1);) {
 
            // Here we find the possible candidate
            // where Element with left side smaller
            // and right side greater. When the if
            // condition fail we check and update in
            // else condition.
            if (a[i] < maxx && i < (n - 1)) {
                i++;
                bit = 0;
            }
 
            // Here we update the possible element
            // if the element is greater than the
            // maxx (maximum element so far). In
            // while loop we sur-pass the value which
            // is greater than the element
            else {
                if (a[i] >= maxx) {
                    element = a[i];
                    idx = i;
                    check = 1;
                    maxx = a[i];
                }
                if (check == 1) {
                    i++;
                }
                bit = 1;
 
                while (a[i] >= element && i < (n - 1)) {
                    if (a[i] > maxx) {
                        maxx = a[i];
                    }
                    i++;
                }
                check = 0;
            }
        }
 
        // Checking for the last value and whether
        // the loop is terminated from else or
        // if block.
        if (element <= a[n - 1] && bit == 1)
        {
            return idx;
        }
        else
        {
            return -1;
        }
 
    }
 
    // Driver code
        var arr = [ 5, 1, 4, 3, 6, 8, 10, 7, 9 ];
        var n = arr.length;
        document.write("Index of the element is " + findElement(arr, n));
 
// This code is contributed by gauravrajput1

输出

Index of the element is 4

时间复杂度： O(n)
辅助空间： O(1)