Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Longest Decreasing Subsequence

  • Difficulty Level : Medium
  • Last Updated : 11 May, 2021

Given an array of N integers, find the length of the longest subsequence of a given sequence such that all elements of the subsequence are sorted in strictly decreasing order. 

Examples:  

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

Input: arr[] = [15, 27, 14, 38, 63, 55, 46, 65, 85] 
Output:
Explanation: The longest decreasing subsequence is {63, 55, 46} 
Input: arr[] = {50, 3, 10, 7, 40, 80} 
Output:
Explanation: The longest decreasing subsequence is {50, 10, 7} 



The problem can be solved using Dynamic Programming
Optimal Substructure: 
Let arr[0…n-1] be the input array and lds[i] be the length of the LDS ending at index i such that arr[i] is the last element of the LDS. 
Then, lds[i] can be recursively written as: 

lds[i] = 1 + max(lds[j]) where i > j > 0 and arr[j] > arr[i] or 
lds[i] = 1, if no such j exists.

To find the LDS for a given array, we need to return max(lds[i]) where n > i > 0.  

C++




// CPP program to find the length of the
// longest decreasing subsequence
#include <bits/stdc++.h>
using namespace std;
 
// Function that returns the length
// of the longest decreasing subsequence
int lds(int arr[], int n)
{
    int lds[n];
    int i, j, max = 0;
 
    // Initialize LDS with 1 for all index
    // The minimum LDS starting with any
    // element is always 1
    for (i = 0; i < n; i++)
        lds[i] = 1;
 
    // Compute LDS from every index
    // 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;
 
    // Select the maximum
    // of all the LDS values
    for (i = 0; i < n; i++)
        if (max < lds[i])
            max = lds[i];
 
    // returns the length of the LDS
    return max;
}
// Driver Code
int main()
{
    int arr[] = { 15, 27, 14, 38, 63, 55, 46, 65, 85 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "Length of LDS is " << lds(arr, n);
    return 0;
}

Java




// Java program to find the
// length of the longest
// decreasing subsequence
import java.io.*;
 
class GFG
{
 
// Function that returns the
// length of the longest
// decreasing subsequence
static int lds(int arr[], int n)
{
    int lds[] = new int[n];
    int i, j, max = 0;
 
    // Initialize LDS with 1
    // for all index. The minimum
    // LDS starting with any
    // element is always 1
    for (i = 0; i < n; i++)
        lds[i] = 1;
 
    // Compute LDS from every
    // index 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;
 
    // Select the maximum
    // of all the LDS values
    for (i = 0; i < n; i++)
        if (max < lds[i])
            max = lds[i];
 
    // returns the length
    // of the LDS
    return max;
}
// Driver Code
public static void main (String[] args)
{
    int arr[] = { 15, 27, 14, 38,
                  63, 55, 46, 65, 85 };
    int n = arr.length;
    System.out.print("Length of LDS is " +
                             lds(arr, n));
}
}
 
// This code is contributed by anuj_67.

Python 3




# Python 3 program to find the length of
# the longest decreasing subsequence
 
# Function that returns the length
# of the longest decreasing subsequence
def lds(arr, n):
 
    lds = [0] * n
    max = 0
 
    # Initialize LDS with 1 for all index
    # The minimum LDS starting with any
    # element is always 1
    for i in range(n):
        lds[i] = 1
 
    # Compute LDS from every index
    # in bottom up manner
    for i in range(1, n):
        for j in range(i):
            if (arr[i] < arr[j] and
                lds[i] < lds[j] + 1):
                lds[i] = lds[j] + 1
 
    # Select the maximum
    # of all the LDS values
    for i in range(n):
        if (max < lds[i]):
            max = lds[i]
 
    # returns the length of the LDS
    return max
 
# Driver Code
if __name__ == "__main__":
     
    arr = [ 15, 27, 14, 38,
            63, 55, 46, 65, 85 ]
    n = len(arr)
    print("Length of LDS is", lds(arr, n))
 
# This code is contributed by ita_c

C#




// C# program to find the
// length of the longest
// decreasing subsequence
using System;
 
class GFG
{
 
// Function that returns the
// length of the longest
// decreasing subsequence
static int lds(int []arr, int n)
{
    int []lds = new int[n];
    int i, j, max = 0;
 
    // Initialize LDS with 1
    // for all index. The minimum
    // LDS starting with any
    // element is always 1
    for (i = 0; i < n; i++)
        lds[i] = 1;
 
    // Compute LDS from every
    // index 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;
 
    // Select the maximum
    // of all the LDS values
    for (i = 0; i < n; i++)
        if (max < lds[i])
            max = lds[i];
 
    // returns the length
    // of the LDS
    return max;
}
// Driver Code
public static void Main ()
{
    int []arr = { 15, 27, 14, 38,
                63, 55, 46, 65, 85 };
    int n = arr.Length;
    Console.Write("Length of LDS is " +
                          lds(arr, n));
}
}
 
// This code is contributed by anuj_67.

PHP




<?php
// PHP program to find the
// length of the longest
// decreasing subsequence
 
 
// Function that returns the
// length of the longest
// decreasing subsequence
function lds($arr, $n)
{
    $lds = array();
    $i; $j; $max = 0;
 
    // Initialize LDS with 1
    // for all index The minimum
    // LDS starting with any
    // element is always 1
    for ($i = 0; $i < $n; $i++)
        $lds[$i] = 1;
 
    // Compute LDS from every
    // index in bottom up manner
    for ($i = 1; $i < $n; $i++)
        for ($j = 0; $j < $i; $j++)
            if ($arr[$i] < $arr[$j] and
                $lds[$i] < $lds[$j] + 1)
                {
                    $lds[$i] = $lds[$j] + 1;
                }
 
    // Select the maximum
    // of all the LDS values
    for ($i = 0; $i < $n; $i++)
        if ($max < $lds[$i])
            $max = $lds[$i];
 
    // returns the length
    // of the LDS
    return $max;
}
 
// Driver Code
$arr = array(15, 27, 14, 38, 63,
             55, 46, 65, 85);
$n = count($arr);
echo "Length of LDS is " ,
            lds($arr, $n);
 
// This code is contributed by anuj_67.
?>

Javascript




<script>
// Javascript program to find the
// length of the longest
// decreasing subsequence
     
    // Function that returns the
// length of the longest
// decreasing subsequence
    function lds(arr,n)
    {
        let lds = new Array(n);
    let i, j, max = 0;
   
    // Initialize LDS with 1
    // for all index. The minimum
    // LDS starting with any
    // element is always 1
    for (i = 0; i < n; i++)
        lds[i] = 1;
   
    // Compute LDS from every
    // index 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;
   
    // Select the maximum
    // of all the LDS values
    for (i = 0; i < n; i++)
        if (max < lds[i])
            max = lds[i];
   
    // returns the length
    // of the LDS
    return max;
    }
     
    // Driver Code
    let arr=[15, 27, 14, 38,
                  63, 55, 46, 65, 85 ];
     
    let n = arr.length;
    document.write("Length of LDS is " +
                             lds(arr, n));
     
     
    // This code is contributed by rag2127
</script>
Output : 
Length of LDS is 3

 

Time Complexity: O(n2
Auxiliary Space: O(n)
Related Article: https://www.geeksforgeeks.org/longest-increasing-subsequence/
 




My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!