Find a sorted subsequence of size 3 in linear time | GeeksforGeeks
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.
Read full article from Find a sorted subsequence of size 3 in linear time | GeeksforGeeks
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.
Key Point: Pre Smaller(Smallest), Next Larger(laegest)
-- Smaller/Greater Array. minSoFas, maxSoFar.
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.
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.
Time Complexity: O(n) Auxliary Space: O(n)
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;}
Actually we don't have to compare at last round, just check index value is not equal itself.
https://chesterli0130.wordpress.com/2012/09/27/find-a-sorted-subsequence-of-size-3-in-linear-time/
http://algorithms.tutorialhorizon.com/find-increasing-triplet-sub-sequence/
http://algorithms.tutorialhorizon.com/find-increasing-triplet-sub-sequence/
Exercise:
1. Find a subsequence of size 3 such that arr[i] < arr[j] > arr[k].
2. Find a sorted subsequence of size 4 in linear time
http://stackoverflow.com/questions/17654673/find-a-sorted-subsequence-of-size-4-in-an-array-in-linear-timevoid find4Numbers(int arr[], int n)
{
int max = n-1; //Index of maximum element from right side
int min = 0, second = -1; //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];
int[] betweenSmallerAndCurrent = new int[n];
smaller[0] = -1; // first entry will always be -1
betweenSmallerAndCurrent[0] = -1;
for (i = 1; i < n; i++)
{
if (arr[i] <= arr[min])
{
min = i;
smaller[i] = -1;
betweenSmallerAndCurrent[i] = -1;
}
else
{
smaller[i] = min;
if (second != -1 && arr[second] < arr[i])
betweenSmallerAndCurrent[i] = second;
else
betweenSmallerAndCurrent[i] = -1;
if (second == -1 || arr[i] < arr[second])
second = i;
}
}
// 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;
}
.....
}