检查是否存在三元组 (i, j, k) 使得 arr[i] < arr[k] < arr[j] for i < j < k

给定一个数组arr[] ，任务是检查是否存在三元组(i, j, k)使得arr[i]然后打印Yes否则打印No 。

例子：

Input: arr[] = {1, 2, 3, 4, 5}
Output: No
Explanation:
There is no such sub-sequence such that arr[i] < arr[k] < arr[j]

Input: arr[] = {3, 1, 5, 0, 4}
Output: Yes
Explanation:
There exist a triplet (3, 5, 4) which is arr[i] < arr[k] < arr[j]

朴素的方法：这个想法是生成所有可能的三元组，如果有任何三元组满足给定的条件，则打印Yes else 打印No 。
时间复杂度： O(N ³ )
辅助空间： O(1)
高效的方法：优化上述方法的想法是使用堆栈来跟踪数组arr[]中每个元素右侧的较小元素。以下是步骤：

• 从末尾遍历数组并维护一个以降序存储元素的堆栈。
• 要按降序维护堆栈，请弹出小于当前元素的元素。然后将弹出的元素标记为三元组的第三个元素。
• 如果任何元素小于最后弹出的元素（标记为三元组的第三个元素），则反向遍历数组。然后它们存在一个满足给定条件的三元组并打印Yes 。
• 否则打印No 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to check if there exist
// triplet in the array such that
// i < j < k and arr[i] < arr[k] < arr[j]
bool findTriplet(vector& arr)
{
    int n = arr.size();
    stack st;
 
    // Initialize the heights of h1 and h2
    // to INT_MAX and INT_MIN respectively
    int h3 = INT_MIN, h1 = INT_MAX;
    for (int i = n - 1; i >= 0; i--) {
 
        // Store the current element as h1
        h1 = arr[i];
 
        // If the element at top of stack
        // is less than the current element
        // then pop the stack top
        // and keep updating the value of h3
        while (!st.empty()
            && st.top() < arr[i]) {
 
            h3 = st.top();
            st.pop();
        }
 
        // Push the current element
        // on the stack
        st.push(arr[i]);
 
        // If current element is less
        // than h3, then we found such
        // triplet and return true
        if (h1 < h3) {
            return true;
        }
    }
 
    // No triplet found, hence return false
    return false;
}
 
// Driver Code
int main()
{
    // Given array
    vector arr = { 4, 7, 5, 6 };
 
    // Function Call
    if (findTriplet(arr)) {
        cout << " Yes" << '\n';
    }
    else {
        cout << " No" << '\n';
    }
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
     
// Function to check if there exist
// triplet in the array such that
// i < j < k and arr[i] < arr[k] < arr[j]
public static boolean findTriplet(int[] arr)
{
    int n = arr.length;
    Stack st = new Stack<>();
 
    // Initialize the heights of h1 and h2
    // to INT_MAX and INT_MIN respectively
    int h3 = Integer.MIN_VALUE;
    int h1 = Integer.MAX_VALUE;
 
    for(int i = n - 1; i >= 0; i--)
    {
         
        // Store the current element as h1
        h1 = arr[i];
 
        // If the element at top of stack
        // is less than the current element
        // then pop the stack top
        // and keep updating the value of h3
        while (!st.empty() && st.peek() < arr[i])
        {
            h3 = st.peek();
            st.pop();
        }
 
        // Push the current element
        // on the stack
        st.push(arr[i]);
 
        // If current element is less
        // than h3, then we found such
        // triplet and return true
        if (h1 < h3)
        {
            return true;
        }
    }
 
    // No triplet found, hence return false
    return false;
}
 
// Driver code
public static void main(String[] args)
{
     
    // Given array
    int arr[] = { 4, 7, 5, 6 };
 
    // Function call
    if (findTriplet(arr))
    {
        System.out.println("Yes");
    }
    else
    {
        System.out.println("No");
    }
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 program for the above approach
import sys
 
# Function to check if there exist
# triplet in the array such that
# i < j < k and arr[i] < arr[k] < arr[j]
def findTriplet(arr):
    n = len(arr)
    st = []
 
    # Initialize the heights of h1 and h3
    # to INT_MAX and INT_MIN respectively
    h3 = -sys.maxsize - 1
    h1 = sys.maxsize
     
    for i in range(n - 1, -1, -1):
 
        # Store the current element as h1
        h1 = arr[i]
 
        # If the element at top of stack
        # is less than the current element
        # then pop the stack top
        # and keep updating the value of h3
        while (len(st) > 0 and st[-1] < arr[i]):
            h3 = st[-1]
            del st[-1]
 
        # Push the current element
        # on the stack
        st.append(arr[i])
 
        # If current element is less
        # than h3, then we found such
        # triplet and return true
        if (h1 < h3):
            return True
         
    # No triplet found, hence
    # return false
    return False
 
# Driver Code
if __name__ == '__main__':
 
    # Given array
    arr = [ 4, 7, 5, 6 ]
 
    # Function Call
    if (findTriplet(arr)):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
     
// Function to check if there exist
// triplet in the array such that
// i < j < k and arr[i] < arr[k] < arr[j]
public static bool findTriplet(int[] arr)
{
    int n = arr.Length;
    Stack st = new Stack();
 
    // Initialize the heights of h1 and h2
    // to INT_MAX and INT_MIN respectively
    int h3 = int.MinValue;
    int h1 = int.MaxValue;
 
    for(int i = n - 1; i >= 0; i--)
    {
         
        // Store the current element as h1
        h1 = arr[i];
 
        // If the element at top of stack
        // is less than the current element
        // then pop the stack top
        // and keep updating the value of h3
        while (st.Count != 0 && st.Peek() < arr[i])
        {
            h3 = st.Peek();
            st.Pop();
        }
 
        // Push the current element
        // on the stack
        st.Push(arr[i]);
 
        // If current element is less
        // than h3, then we found such
        // triplet and return true
        if (h1 < h3)
        {
            return true;
        }
    }
 
    // No triplet found, hence return false
    return false;
}
 
// Driver code
public static void Main(String[] args)
{
     
    // Given array
    int []arr = { 4, 7, 5, 6 };
 
    // Function call
    if (findTriplet(arr))
    {
        Console.WriteLine("Yes");
    }
    else
    {
        Console.WriteLine("No");
    }
}
}
 
// This code is contributed by PrinciRaj1992

Javascript

 
// JavaScript program for the above approach
 
// Function to check if there exist
// triplet in the array such that
// i < j < k and arr[i] < arr[k] < arr[j]
function findTriplet(arr)
{
    var n = arr.length;
    var st = [];
 
    // Initialize the heights of h1 and h2
    // to INT_MAX and INT_MIN respectively
    var h3 = -1000000000, h1 = 1000000000;
    for (var i = n - 1; i >= 0; i--) {
 
        // Store the current element as h1
        h1 = arr[i];
 
        // If the element at top of stack
        // is less than the current element
        // then pop the stack top
        // and keep updating the value of h3
        while (st.length!=0
            && st[st.length-1] < arr[i]) {
 
            h3 = st[st.length-1];
            st.pop();
        }
 
        // Push the current element
        // on the stack
        st.push(arr[i]);
 
        // If current element is less
        // than h3, then we found such
        // triplet and return true
        if (h1 < h3) {
            return true;
        }
    }
 
    // No triplet found, hence return false
    return false;
}
 
// Driver Code
// Given array
var arr = [4, 7, 5, 6 ];
// Function Call
if (findTriplet(arr)) {
    document.write( " Yes");
}
else {
    document.write( " No" );
}
 
 

输出：

Yes

时间复杂度： O(N)
辅助空间： O(N)

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。