Find a sorted subsequence of size 3 three in linear time

最新推荐文章于 2023-02-25 23:03:36 发布

转载最新推荐文章于 2023-02-25 23:03:36 发布 · 381 阅读

Google面试专栏收录该内容

22 篇文章

订阅专栏

本文介绍了一种在整数数组中寻找满足特定条件的三元素的方法，即找到三个索引i、j、k，使得a[i]<a[j]<a[k]且i<j<k，并在O(n)时间内完成。文章提供了算法实现细节及示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given an array of n integers, find the 3 elements such that a[i] < a[j] < a[k] and i < j < k in 0(n) time. If there are multiple such triplets, then print any one of them.

Examples:

Input:  arr[] = {12, 11, 10, 5, 6, 2, 30}
Output: 5, 6, 30

Input:  arr[] = {1, 2, 3, 4}
Output: 1, 2, 3 OR 1, 2, 4 OR 2, 3, 4

Input:  arr[] = {4, 3, 2, 1}
Output: No such triplet

Solution:
1) Create an auxiliary array smaller[0..n-1]. smaller[i] should store the index of a number which is smaller than arr[i] and is on left side of arr[i]. smaller[i] should contain -1 if there is no such element.
2) Create another auxiliary array greater[0..n-1]. greater[i] should store the index of a number which is greater than arr[i] and is on right side of arr[i]. greater[i] should contain -1 if there is no such element.
3) Finally traverse both smaller[] and greater[] and find the index i for which both smaller[i] and greater[i] are not -1.

// A function to fund a sorted subsequence of size 3
void find3Numbers(int arr[], int n)
{
   int max = n-1; //Index of maximum element from right side
   int min = 0; //Index of minimum element from left side
   int i;

   // Create an array that will store index of a smaller
   // element on left side. If there is no smaller element
   // on left side, then smaller[i] will be -1.
   int *smaller = new int[n];
   smaller[0] = -1;  // first entry will always be -1
   for (i = 1; i < n; i++)
   {
       if (arr[i] <= arr[min])
       {
          min = i;
          smaller[i] = -1;
       }
       else
          smaller[i] = min;
   }

   // Create another array that will store index of a
   // greater element on right side. If there is no greater
   // element on right side, then greater[i] will be -1.
   int *greater = new int[n];
   greater[n-1] = -1;  // last entry will always be -1
   for (i = n-2; i >= 0; i--)
   {
       if (arr[i] >= arr[max])
       {
          max = i;
          greater[i] = -1;
       }
       else
          greater[i] = max;
   }

   // Now find a number which has both a greater number on
   // right side and smaller number on left side
   for (i = 0; i < n; i++)
   {
       if (smaller[i] != -1 && greater[i] != -1)
       {
          printf("%d %d %d", arr[smaller[i]],
                 arr[i], arr[greater[i]]);
          return;
       }
   }

   // If we reach number, then there are no such 3 numbers
   printf("No such triplet found");

   // Free the dynamically alloced memory to avoid memory leak
   delete [] smaller;
   delete [] greater;

   return;
}