Longest Decreasing Subsequence
You are given an array/list ARR consisting of N integers. Your task is to find the length of the longest decreasing subsequence.

A subsequence is a sequence of numbers obtained by deleting zero or more elements from the array/list, keeping the relative positions of elements same as it was in the initial sequence. A decreasing subsequence is a subsequence in which every element is strictly less than the previous number.

Note:

There can be more than one subsequences with the longest length.

For example:-

For the given array [5, 0, 3, 2, 9], the longest decreasing subsequence is of length 3, i.e. [5, 3, 2]

Note:

Try to solve the problem in O(N log N) time complexity.

Input Format :

The first line of input contains an integer 'T' representing the number of the test case. Then the test case follows. The first line of each test case contains an integer ‘N’ representing the size of the array/list. The second line of each test case contains N single space-separated integers representing the array/list elements.

Output Format :

For each test case, print the integer denoting the length of the longest decreasing subsequence. Print the output of each test case in a separate line.

Note:

You do not need to print anything; it has already been taken care of. Just implement the given function.

Constraints:

1

Question

Longest Decreasing Subsequence

You are given an array/list ARR consisting of N integers. Your task is to find the length of the longest decreasing subsequence.

A subsequence is a sequence of numbers obtained by deleting zero or more elements from the array/list, keeping the relative positions of elements same as it was in the initial sequence. A decreasing subsequence is a subsequence in which every element is strictly less than the previous number.

Note:

There can be more than one subsequences with the longest length.

For example:-

For the given array [5, 0, 3, 2, 9], the longest decreasing subsequence is of length 3, i.e. [5, 3, 2]

Note:

Try to solve the problem in O(N log N) time complexity.

Input Format :

The first line of input contains an integer 'T' representing the number of the test case. Then the test case follows.

The first line of each test case contains an integer ‘N’ representing the size of the array/list.

The second line of each test case contains N single space-separated integers representing the array/list elements.

Output Format :

For each test case, print the integer denoting the length of the longest decreasing subsequence.

Print the output of each test case in a separate line.

Note:

You do not need to print anything; it has already been taken care of. Just implement the given function.

Constraints:

1 <= T <= 50 
1 <= N <= 5000
1 <= ARR[i] <= 10^9

Time Limit: 1 sec

Accepted Answer

Recursive

We can use recursion to solve this problem.

For each element, there are two possibilities.
1. We include the current item in LDS if it is smaller than the previous element and recurse for remaining items.
2. We exclude the current item from LDS and recurse for remaining items.
Finally, we return the length of LDS we get by including or excluding the current item.
The base case of the recursion would be to return 0 when no items are left.

Space Complexity: O(n)Explanation:

O(N), where N is the length of the array.

As stack space of size N required for recursive function calls.Hence the overall complexity will be O(N).

Time Complexity: O(2^n)Explanation:

O(2^N), where N is the length of the array.

We are exploring all possible subsets of the array and there are 2^N subsets.

Accepted Answer

Memoization

In the recursive approach, many recursive calls had to be made again and again with the same set of parameters. This redundancy can be eliminated by storing the results obtained for a particular call in memory (as an array, MEMO) so that whenever we need to call the function again if we have stored its value earlier.

This can be done as:

For each element, there are two possibilities.
1. We include the current item in LDS if it is smaller than the previous element and recurse for remaining items. For this, we first check if we have value for LDS stored in MEMO.
  - If MEMO[currentElement] > -1, then return MEMO
  - Else calculate the value for currentElement and store it in MEMO[currentElement] and return it.
2. We exclude the current item from LDS and recurse for remaining items.
Finally, we return the length of the LDS we get by including or excluding the current item.
The base case of the recursion would be to return 0 when no items are left.

Space Complexity: O(n^2)Explanation:

O(N^2), where N is the length of the array.

We’ll be creating an auxiliary array of size N^2 to store the length of LDS for every index. So the overall complexity will be O(N^2).

Time Complexity: O(n^2)Explanation:

O(N^2), where N is the length of the array.

As because size of recursion tree can go up to N^2. So the overall complexity will be O(N^2).

Python (3.5)

'''

	Time Complexity: O(N^2)

	Space Complexity: O(N^2)

	

	where N is the length of array

'''





def LDSHelper(arr, prev, curr, memo):

    # Base case.

    if curr == len(arr):

        return 0



    if memo[prev + 1][curr] >= 0:

        return memo[prev + 1][curr]



    taken = 0



    if prev < 0 or arr[curr] < arr[prev]:



        taken = 1 + LDSHelper(arr, curr, curr + 1, memo)



    # Exclude the current item from LDS and recurse for remaining items.

    nottaken = LDSHelper(arr, prev, curr + 1, memo)



    memo[prev + 1][curr] = max(taken, nottaken)



    return memo[prev + 1][curr]





def LDS(arr):



    # Initialising memoization table.

    memo = [[-1 for i in range(len(arr) + 1)] for j in range(len(arr))]



    return LDSHelper(arr, -1, 0, memo)

Java (SE 1.8)

/*

    Time complexity: O(N^2)

    Space complexity: O(N)



    Where N is the length of array.

*/



public class Solution {

    public static int LDS(int[] arr) {

            int n = arr.length;



            // Initialising memoization table.

            int[][]memo = new int[n + 1][];



            for (int i = 0; i <= n; i++){

                memo[i] = new int[n + 1];



                for (int j = 0; j <= n; j++){

                    memo[i][j] = -1;

                }

            }

            return LDSHelper(arr, -1, 0, memo);

        }





    static int LDSHelper(int[] arr, int prev, int curr, int[][]memo){

        // Base case.

        if (curr == arr.length){

            return 0;

        }



        if (memo[prev + 1][curr] >= 0){

            return memo[prev + 1][curr];

        }



        int taken = 0;



        if (prev < 0 || arr[curr] < arr[prev]){

            taken = 1 + LDSHelper(arr, curr, curr + 1, memo);

        }



        // Exclude the current item from LDS and recurse for remaining items.

        int nottaken = LDSHelper(arr, prev, curr + 1, memo);

        memo[prev + 1][curr] = Math.max(taken, nottaken);

        return memo[prev + 1][curr];

    }



}

C++ (g++ 5.4)

/*

	Time Complexity: O(N^2)

	Space Complexity: O(N^2)

	

	where N is the length of array.

*/



int LDSHelper(vector arr, int prev, int curr, int **memo){



	// Base case.

    if (curr == arr.size()){

        return 0;

    }



    if (memo[prev + 1][curr] >= 0){

        return memo[prev + 1][curr];

    }



    int taken = 0;



    if (prev < 0 || arr[curr] < arr[prev]){

        taken = 1 + LDSHelper(arr, curr, curr + 1, memo);

    }

    

	// Exclude the current item from LDS and recurse for remaining items.

    int nottaken = LDSHelper(arr, prev, curr + 1, memo);

    memo[prev + 1][curr] = max(taken, nottaken);

    return memo[prev + 1][curr];

}



int LDS(vector arr){



    int n = arr.size();



	// Initialising memoization table.

    int **memo = new int *[n + 1];



    for (int i = 0; i <= n; i++){

        memo[i] = new int[n + 1];



        for (int j = 0; j <= n; j++){

            memo[i][j] = -1;

        }

    }

    return LDSHelper(arr, -1, 0, memo);

}

Accepted Answer

Dynamic Programming

The problem has optimal substructure. That means the problem can be broken down into smaller sub-problems, which can further be divided until the solution becomes trivial.

The solution also exhibits overlapping sub-problems. If we draw the recursion tree of the solution, we can see that the same sub-problems are getting computed again and again.

We know that problems having optimal substructure and overlapping subproblems can be solved by using Dynamic Programming.

We’ll solve this problem in a bottom-up manner. In this bottom-up approach, we will solve smaller subproblems first, then larger sub-problems from them.

We will compute L[i], for each 0 <= i < N, which stores the length of the longest decreasing subsequence of the subarray arr[0..i] that ends with arr[i].
We will initialise L[i] = 1, for each 0 <= i < N because the longest decreasing subsequence ending with ARR[i] has length 1.
To calculate L[i], for each 1 <= i < N, we iterate over L[j], for each 0 <= j < i.
1. If the value at index j is greater than the value at index i, then we can append the current element (at i-th index) to the LDS ending at index j. Hence, we update L[i] = max(L[i], 1 + L[j]).
Return the maximum value of all L[i], for 0 <= i

Space Complexity: O(n)Explanation:

O(N), where N is the length of the array.

We’ll be creating an auxiliary array to store the length of LDS for every index. Hence the overall complexity will be O(N).

Time Complexity: O(n^2)Explanation:

O(N^2), where N is the length of the array.

We traverse the array once in the outer loop, yielding a complexity of O(N). Then for every element ARR[i], we make a linear search of all elements from 0 up to i. Hence the overall complexity is O(N^2).

Java (SE 1.8)

/*

	Time Complexity: O(N^2)

	Space Complexity: O(N)



	where N is the length of array

*/



public class Solution {

	public static int LDS(int[] arr) {

		int n = arr.length;

		int lds[] = new int[n];

		int i, j, max = 0;



		// Initialize LDS values for all indexes .

		for (i = 0; i < n; i++) {

			lds[i] = 1;

		}



		// Compute optimized LDS values in bottom up manner.

		for (i = 1; i < n; i++) {

			for (j = 0; j < i; j++) {

				if (arr[i] < arr[j] && lds[i] < lds[j] + 1) {

					lds[i] = lds[j] + 1;

				}

			}

		}



		// Pick maximum of all LDS values.

		for (i = 0; i < n; i++) {

			if (max < lds[i]) {

				max = lds[i];

			}

		}



		return max;

	}

}

Python (3.5)

'''

	Time Complexity: O(N^2)

	Space Complexity: O(N)



	where N is the length of array

'''



def LDS(arr):



    n = len(arr)

    lds = [0] * n

    i = 0

    j = 0

    mx = 0



    # Initialize LDS values for all indexes.

    for i in range(n):

        lds[i] = 1



    # Compute optimized LDS values in bottom up manner.

    for i in range(1, n):

        for j in range(0, i):

            if arr[i] < arr[j] and lds[i] < lds[j] + 1:

                lds[i] = lds[j] + 1



    # Pick maximum of all LDS values.

    for i in range(n):

        if mx < lds[i]:

            mx = lds[i]



    return mx

C++ (g++ 5.4)

/*

    Time complexity: O(N^2)

    Space complexity: O(N)



    where N is the length of array.

*/



int LDS(vector arr){

	int n = arr.size();

	int lds[n];

	int i, j, max = 0;



	// Initialize LDS values for all indexes. 

	for (i = 0; i < n; i++) {

		lds[i] = 1;

	}



	// Compute optimized LDS values in bottom up manner. 

	for (i = 1; i < n; i++) {

		for (j = 0; j < i; j++) {

			if (arr[i] < arr[j] && lds[i] < lds[j] + 1) {

				lds[i] = lds[j] + 1;

			}

		}

	}



	// Pick maximum of all LDS values.

	for (i = 0; i < n; i++) {

		if (max < lds[i]) {

			max = lds[i];

		}

	}



	return max;

}

Accepted Answer

Efficient approach

The idea here is to maintain lists of decreasing sequences. In general, we have a set of active lists of varying length. We are adding an element A[i] to these lists. We can always add A[i] at a point just larger than it and we can show that this doesn't hamper the LDS. We scan the lists (for end elements) in decreasing order of their length. We will verify the end elements of all the lists to find a list whose end element is smaller than A[i] (floor value).

The basic conditions to create these lists will be:

If A[i] is largest among all end candidates of active lists, we will start a new active list of length 1.
If A[i] is smallest among all end candidates of active lists, we will clone the largest active list, and extend it by A[i].
If A[i] is in between, we will find a list with the smallest end element that is larger than A[i]. Clone and extend this list by A[i].

We will discard all other lists of the same length as that of this modified list.

Note: At any instance during our construction of active lists, the following condition is maintained.

“The end element of a smaller list is smaller than the end elements of larger lists”.

Space Complexity: O(n)Explanation:

O(N), where ‘N’ is the length of the array.

This is because we are using an auxiliary array to keep end elements. The maximum length of this array is that of input. In the worst case, the array is divided into N lists of size one.

Hence the overall space complexity is O(N).

Time Complexity: O(nlogn)Explanation:

O(N log (N)), where N is the length of the array.

As we are performing a Binary Search on the array of size ‘N’. Hence the overall complexity will be O(N log(N)).

Java (SE 1.8)

/*

    Time complexity: O(N log(N))

    Space complexity: O(N)



    Where N is the length of array.

*/

public class Solution {

	public static int LDS(int[] arr) {



		// Add boundary case, when array size is one.

		int size = arr.length;

		int[] tailTable = new int[size];



        // Always points empty slot.

		int len; 



		tailTable[0] = arr[0];

		len = 1;

		for (int i = 1; i < size; i++) {

			if (arr[i] > tailTable[0]) {



				// New greatest value.

				tailTable[0] = arr[i];

			}



			else if (arr[i] < tailTable[len - 1]) {



				// 'arr[i]' wants to extend largest subsequence.

				tailTable[len++] = arr[i];

			}



			else {



                /*

                    'arr[i]' wants to be current end candidate of an existing

				    subsequence. It will replace ceiling value in tailTable.

                */

				tailTable[binSearch(tailTable, -1, len - 1, arr[i])] = arr[i];

			}

		}



		return len;



	}



	static int binSearch(int tailTable[], int l, int r, int key) {

		while (r - l > 1) {

			int m = l + (r - l) / 2;

			if (tailTable[m] <= key) {

				r = m;

			} 

            

            else {

				l = m;

			}

		}



		return r;

	}



}

Python (3.5)

'''

	Time complexity: O(N log N)

	Space complexity: O(N)



	where N is the length of array

'''



def binarySearch(tailTable, left, right, key):

    while right - left > 1:

        mid = left + (right - left) // 2



        if tailTable[mid] <= key:

            right = mid



        else:

            left = mid



    return right





def LDS(arr):



    # Add boundary case, when array size is one.

    size = len(arr)

    tailTable = [0] * size



    # Always points empty slot.

    l = 0



    tailTable[0] = arr[0]

    l = 1



    for i in range(1, size):



        if arr[i] > tailTable[0]:



            # New greatest value.

            tailTable[0] = arr[i]



        elif arr[i] < tailTable[l - 1]:

            # arr[i] wants to extend largest subsequence.

            tailTable[l] = arr[i]

            l += 1



        else:

            

            '''

                arr[i] wants to be current end candidate of an existing subsequence.

                It will replace ceiling value in tailTable.

            '''

            tailTable[binarySearch(tailTable, -1, l - 1, arr[i])] = arr[i]



    return l

C++ (g++ 5.4)

/*

    Time complexity: O(N log(N))

    Space complexity: O(N)



    where N is the length of array.

*/



int binarySearch(int tailTable[], int l, int r, int key){

    while (r - l > 1){

        int m = l + (r - l) / 2;



        if (tailTable[m] <= key){

            r = m;

        }



        else{

            l = m;

        }

    }

    return r;

}



int LDS(vector arr){



    // Add boundary case, when array size is one.

    int size = arr.size();



    int tailTable[size];



    // Always points empty slot.

    int len; 

    tailTable[0] = arr[0];

    len = 1;



    for (int i = 1; i < size; i++){



        if (arr[i] > tailTable[0]){



            // New greatest value.

            tailTable[0] = arr[i];

        }



        else if (arr[i] < tailTable[len - 1]){



            // arr[i] wants to extend largest subsequence.

            tailTable[len++] = arr[i];

        }



        else{



			/*

            	arr[i] wants to be current end candidate of an existing subsequence.

            	It will replace ceiling value in tailTable.

			*/

            tailTable[binarySearch(tailTable, -1, len - 1, arr[i])] = arr[i];

        }

    }

    return len;

}

You are given an array/list ARR consisting of N integers. Your task is to find the length of the longest decreasing subsequence.

A subsequence is a sequence of numbers obtained by deleting zero or more elements from the array/list, keeping the relative positions of elements same as it was in the initial sequence. A decreasing subsequence is a subsequence in which every element is strictly less than the previous number.

Note:

For example:-

Note:

Input Format :

Output Format :

Note:

Constraints:

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