# Longest Increasing Subsequence (LIS)

Given an array arr[] of size N, the task is to find the length of the Longest Increasing Subsequence (LIS) i.e., the longest possible subsequence in which the elements of the subsequence are sorted in increasing order.

Longest Increasing Subsequence

Examples:

Input: arr[] = {3, 10, 2, 1, 20}
Output: 3
Explanation: The longest increasing subsequence is 3, 10, 20

Input: arr[] = {3, 2}
Output:1
Explanation: The longest increasing subsequences are {3} and {2}

Input: arr[] = {50, 3, 10, 7, 40, 80}
Output: 4
Explanation: The longest increasing subsequence is {3, 7, 40, 80}

## Longest Increasing Sequence using Recursion:

The problem can be solved based on the following idea:

Let L(i) be the length of the LIS ending at index i such that arr[i] is the last element of the LIS. Then, L(i) can be recursively written as:

• L(i) = 1 + max(L(j) ) where 0 < j < i and arr[j] < arr[i]; or
• L(i) = 1, if no such j exists.

Formally, the length of LIS ending at index i, is 1 greater than the maximum of lengths of all LIS ending at some index j such that arr[j] < arr[i] where j < i.

We can see that the above recurrence relation follows the optimal substructure property.

Illustration:

Follow the below illustration for a better understanding:

Consider arr[] = {3, 10, 2, 11}

L(i): Denotes LIS of subarray ending at position ‘i’

Recursion Tree

Follow the steps mentioned below to implement the above idea:

• Create a recursive function.
• For each recursive call, Iterate from the i = 1 to the current position and do the following:
• Find the possible length of the longest increasing subsequence ending at the current position if the previous sequence ended at i.
• Update the maximum possible length accordingly.
• Repeat this for all indices and find the answer

Below is the implementation of the recursive approach:

## C++

 `// A Naive C++ recursive implementation` `// of LIS problem` `#include ` `using` `namespace` `std;`   `// To make use of recursive calls, this` `// function must return two things:` `// 1) Length of LIS ending with element` `// arr[n-1].` `// We use max_ending_here for this purpose` `// 2) Overall maximum as the LIS may end` `// with an element before arr[n-1] max_ref` `// is used this purpose.` `// The value of LIS of full array of size` `// n is stored in *max_ref which is` `// our final result` `int` `_lis(``int` `arr[], ``int` `n, ``int``* max_ref)` `{`   `    ``// Base case` `    ``if` `(n == 1)` `        ``return` `1;`   `    ``// 'max_ending_here' is length of` `    ``// LIS ending with arr[n-1]` `    ``int` `res, max_ending_here = 1;`   `    ``// Recursively get all LIS ending with` `    ``// arr[0], arr[1] ... arr[n-2]. If` `    ``// arr[i-1] is smaller than arr[n-1],` `    ``// and max ending with arr[n-1] needs` `    ``// to be updated, then update it` `    ``for` `(``int` `i = 1; i < n; i++) {` `        ``res = _lis(arr, i, max_ref);` `        ``if` `(arr[i - 1] < arr[n - 1]` `            ``&& res + 1 > max_ending_here)` `            ``max_ending_here = res + 1;` `    ``}`   `    ``// Compare max_ending_here with the` `    ``// overall max. And update the` `    ``// overall max if needed` `    ``if` `(*max_ref < max_ending_here)` `        ``*max_ref = max_ending_here;`   `    ``// Return length of LIS ending` `    ``// with arr[n-1]` `    ``return` `max_ending_here;` `}`   `// The wrapper function for _lis()` `int` `lis(``int` `arr[], ``int` `n)` `{`   `    ``// The max variable holds the result` `    ``int` `max = 1;`   `    ``// The function _lis() stores its` `    ``// result in max` `    ``_lis(arr, n, &max);`   `    ``// Returns max` `    ``return` `max;` `}`   `// Driver program to test above function` `int` `main()` `{` `    ``int` `arr[] = { 10, 22, 9, 33, 21, 50, 41, 60 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);`   `    ``// Function call` `    ``cout << ``"Length of lis is "` `<< lis(arr, n);` `    ``return` `0;` `}`

## C

 `// A Naive C recursive implementation` `// of LIS problem` `#include ` `#include `   `// To make use of recursive calls, this` `// function must return two things:` `// 1) Length of LIS ending with element arr[n-1].` `// We use max_ending_here for this purpose` `// 2) Overall maximum as the LIS may end with` `// an element before arr[n-1] max_ref is` `//  used this purpose.` `// The value of LIS of full array of size n` `// is stored in *max_ref which is our final result` `int` `_lis(``int` `arr[], ``int` `n, ``int``* max_ref)` `{` `    ``// Base case` `    ``if` `(n == 1)` `        ``return` `1;`   `    ``// 'max_ending_here' is length of LIS` `    ``// ending with arr[n-1]` `    ``int` `res, max_ending_here = 1;`   `    ``// Recursively get all LIS ending with arr[0],` `    ``// arr[1] ... arr[n-2]. If arr[i-1] is smaller` `    ``// than arr[n-1], and max ending with arr[n-1]` `    ``// needs to be updated, then update it` `    ``for` `(``int` `i = 1; i < n; i++) {` `        ``res = _lis(arr, i, max_ref);` `        ``if` `(arr[i - 1] < arr[n - 1]` `            ``&& res + 1 > max_ending_here)` `            ``max_ending_here = res + 1;` `    ``}`   `    ``// Compare max_ending_here with the overall` `    ``// max. And update the overall max if needed` `    ``if` `(*max_ref < max_ending_here)` `        ``*max_ref = max_ending_here;`   `    ``// Return length of LIS ending with arr[n-1]` `    ``return` `max_ending_here;` `}`   `// The wrapper function for _lis()` `int` `lis(``int` `arr[], ``int` `n)` `{` `    ``// The max variable holds the result` `    ``int` `max = 1;`   `    ``// The function _lis() stores its result in max` `    ``_lis(arr, n, &max);`   `    ``// returns max` `    ``return` `max;` `}`   `// Driver program to test above function` `int` `main()` `{` `    ``int` `arr[] = { 10, 22, 9, 33, 21, 50, 41, 60 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);`   `    ``// Function call` `    ``printf``(``"Length of lis is %d"``, lis(arr, n));` `    ``return` `0;` `}`

## Java

 `// A Naive Java Program for LIS Implementation` `import` `java.io.*;` `import` `java.util.*;`   `class` `LIS {`   `    ``// Stores the LIS` `    ``static` `int` `max_ref;`   `    ``// To make use of recursive calls, this function must` `    ``// return two things: 1) Length of LIS ending with` `    ``// element arr[n-1]. We use max_ending_here for this` `    ``// purpose 2) Overall maximum as the LIS may end with an` `    ``// element before arr[n-1] max_ref is used this purpose.` `    ``// The value of LIS of full array of size n is stored in` `    ``// *max_ref which is our final result` `    ``static` `int` `_lis(``int` `arr[], ``int` `n)` `    ``{` `        ``// Base case` `        ``if` `(n == ``1``)` `            ``return` `1``;`   `        ``// 'max_ending_here' is length of LIS ending with` `        ``// arr[n-1]` `        ``int` `res, max_ending_here = ``1``;`   `        ``// Recursively get all LIS ending with arr[0],` `        ``// arr[1] ... arr[n-2]. If   arr[i-1] is smaller` `        ``// than arr[n-1], and max ending with arr[n-1] needs` `        ``// to be updated, then update it` `        ``for` `(``int` `i = ``1``; i < n; i++) {` `            ``res = _lis(arr, i);` `            ``if` `(arr[i - ``1``] < arr[n - ``1``]` `                ``&& res + ``1` `> max_ending_here)` `                ``max_ending_here = res + ``1``;` `        ``}`   `        ``// Compare max_ending_here with the overall max. And` `        ``// update the overall max if needed` `        ``if` `(max_ref < max_ending_here)` `            ``max_ref = max_ending_here;`   `        ``// Return length of LIS ending with arr[n-1]` `        ``return` `max_ending_here;` `    ``}`   `    ``// The wrapper function for _lis()` `    ``static` `int` `lis(``int` `arr[], ``int` `n)` `    ``{` `        ``// The max variable holds the result` `        ``max_ref = ``1``;`   `        ``// The function _lis() stores its result in max` `        ``_lis(arr, n);`   `        ``// Returns max` `        ``return` `max_ref;` `    ``}`   `    ``// Driver program to test above functions` `    ``public` `static` `void` `main(String args[])` `    ``{` `        ``int` `arr[] = { ``10``, ``22``, ``9``, ``33``, ``21``, ``50``, ``41``, ``60` `};` `        ``int` `n = arr.length;`   `        ``// Function call` `        ``System.out.println(``"Length of lis is "` `                           ``+ lis(arr, n));` `    ``}` `}` `// This code is contributed by Rajat Mishra`

## Python3

 `# A naive Python implementation of LIS problem`     `# Global variable to store the maximum` `global` `maximum`     `# To make use of recursive calls, this function must return` `# two things:` `# 1) Length of LIS ending with element arr[n-1]. We use` `# max_ending_here for this purpose` `# 2) Overall maximum as the LIS may end with an element` `# before arr[n-1] max_ref is used this purpose.` `# The value of LIS of full array of size n is stored in` `# *max_ref which is our final result` `def` `_lis(arr, n):`   `    ``# To allow the access of global variable` `    ``global` `maximum`   `    ``# Base Case` `    ``if` `n ``=``=` `1``:` `        ``return` `1`   `    ``# maxEndingHere is the length of LIS ending with arr[n-1]` `    ``maxEndingHere ``=` `1`   `    ``# Recursively get all LIS ending with` `    ``# arr[0], arr[1]..arr[n-2]` `    ``# If arr[i-1] is smaller than arr[n-1], and` `    ``# max ending with arr[n-1] needs to be updated,` `    ``# then update it` `    ``for` `i ``in` `range``(``1``, n):` `        ``res ``=` `_lis(arr, i)` `        ``if` `arr[i``-``1``] < arr[n``-``1``] ``and` `res``+``1` `> maxEndingHere:` `            ``maxEndingHere ``=` `res ``+` `1`   `    ``# Compare maxEndingHere with overall maximum. And` `    ``# update the overall maximum if needed` `    ``maximum ``=` `max``(maximum, maxEndingHere)`   `    ``return` `maxEndingHere`     `def` `lis(arr):`   `    ``# To allow the access of global variable` `    ``global` `maximum`   `    ``# Length of arr` `    ``n ``=` `len``(arr)`   `    ``# Maximum variable holds the result` `    ``maximum ``=` `1`   `    ``# The function _lis() stores its result in maximum` `    ``_lis(arr, n)` `    ``return` `maximum`     `# Driver program to test the above function` `if` `__name__ ``=``=` `'__main__'``:` `    ``arr ``=` `[``10``, ``22``, ``9``, ``33``, ``21``, ``50``, ``41``, ``60``]` `    ``n ``=` `len``(arr)`   `    ``# Function call` `    ``print``(``"Length of lis is"``, lis(arr))`   `# This code is contributed by NIKHIL KUMAR SINGH`

## C#

 `using` `System;`   `// A Naive C# Program for LIS Implementation` `class` `LIS {`   `    ``// Stores the LIS` `    ``static` `int` `max_ref;`   `    ``// To make use of recursive calls, this function must` `    ``// return two things: 1) Length of LIS ending with` `    ``// element arr[n-1]. We use max_ending_here for this` `    ``// purpose 2) Overall maximum as the LIS may end with an` `    ``// element before arr[n-1] max_ref is used this purpose.` `    ``// The value of LIS of full array of size n is stored in` `    ``// *max_ref which is our final result` `    ``static` `int` `_lis(``int``[] arr, ``int` `n)` `    ``{` `        ``// Base case` `        ``if` `(n == 1)` `            ``return` `1;`   `        ``// 'max_ending_here' is length of LIS ending with` `        ``// arr[n-1]` `        ``int` `res, max_ending_here = 1;`   `        ``// Recursively get all LIS ending with arr[0],` `        ``// arr[1] ... arr[n-2]. If   arr[i-1] is smaller` `        ``// than arr[n-1], and max ending with arr[n-1] needs` `        ``// to be updated, then update it` `        ``for` `(``int` `i = 1; i < n; i++) {` `            ``res = _lis(arr, i);` `            ``if` `(arr[i - 1] < arr[n - 1]` `                ``&& res + 1 > max_ending_here)` `                ``max_ending_here = res + 1;` `        ``}`   `        ``// Compare max_ending_here with the overall max` `        ``// and update the overall max if needed` `        ``if` `(max_ref < max_ending_here)` `            ``max_ref = max_ending_here;`   `        ``// Return length of LIS ending with arr[n-1]` `        ``return` `max_ending_here;` `    ``}`   `    ``// The wrapper function for _lis()` `    ``static` `int` `lis(``int``[] arr, ``int` `n)` `    ``{` `        ``// The max variable holds the result` `        ``max_ref = 1;`   `        ``// The function _lis() stores its result in max` `        ``_lis(arr, n);`   `        ``// Returns max` `        ``return` `max_ref;` `    ``}`   `    ``// Driver program to test above functions` `    ``public` `static` `void` `Main()` `    ``{` `        ``int``[] arr = { 10, 22, 9, 33, 21, 50, 41, 60 };` `        ``int` `n = arr.Length;`   `        ``// Function call` `        ``Console.Write(``"Length of lis is "` `+ lis(arr, n)` `                      ``+ ``"\n"``);` `    ``}` `}`

## Javascript

 ``

Output

`Length of lis is 5`

Time Complexity: O(2n) The time complexity of this recursive approach is exponential as there is a case of overlapping subproblems as explained in the recursive tree diagram above.
Auxiliary Space: O(1). No external space is used for storing values apart from the internal stack space.

## Longest Increasing Subsequence using Memoization:

If noticed carefully, we can see that the above recursive solution also follows the overlapping subproblems property i.e., same substructure solved again and again in different recursion call paths. We can avoid this using the memoization approach.

We can see that each state can be uniquely identified using two parameters:

• Current index (denotes the last index of the LIS) and
• Previous index (denotes the ending index of the previous LIS behind which the arr[i] is being concatenated).

Below is the implementation of the above approach.

## C++

 `// C++ code of memoization approach for LIS` `#include ` `using` `namespace` `std;`   `// To make use of recursive calls, this` `// function must return two things:` `// 1) Length of LIS ending with element` `// arr[n-1].` `// We use max_ending_here for this purpose` `// Overall maximum as the LIS may end with` `// an element before arr[n-1] max_ref is` `// used this purpose.` `// The value of LIS of full array of size` `// n is stored in *max_ref which is` `// our final result` `int` `f(``int` `idx, ``int` `prev_idx, ``int` `n, ``int` `a[],` `      ``vector >& dp)` `{` `    ``if` `(idx == n) {` `        ``return` `0;` `    ``}`   `    ``if` `(dp[idx][prev_idx + 1] != -1) {` `        ``return` `dp[idx][prev_idx + 1];` `    ``}`   `    ``int` `notTake = 0 + f(idx + 1, prev_idx, n, a, dp);` `    ``int` `take = INT_MIN;` `    ``if` `(prev_idx == -1 || a[idx] > a[prev_idx]) {` `        ``take = 1 + f(idx + 1, idx, n, a, dp);` `    ``}`   `    ``return` `dp[idx][prev_idx + 1] = max(take, notTake);` `}`   `// Function to find length of` `// longest increasing subsequence` `int` `longestSubsequence(``int` `n, ``int` `a[])` `{` `    ``vector > dp(n + 1, vector<``int``>(n + 1, -1));` `    ``return` `f(0, -1, n, a, dp);` `}`   `// Driver program to test above function` `int` `main()` `{` `    ``int` `a[] = { 3, 10, 2, 1, 20 };` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]);`   `    ``// Function call` `    ``cout << ``"Length of lis is "` `<< longestSubsequence(n, a);` `    ``return` `0;` `}`

## Java

 `// A Memoization Java Program for LIS Implementation`   `import` `java.lang.*;` `import` `java.util.Arrays;`   `class` `LIS {`   `    ``// To make use of recursive calls, this function must` `    ``// return two things: 1) Length of LIS ending with` `    ``// element arr[n-1]. We use max_ending_here for this` `    ``// purpose 2) Overall maximum as the LIS may end with an` `    ``// element before arr[n-1] max_ref is used this purpose.` `    ``// The value of LIS of full array of size n is stored in` `    ``// *max_ref which is our final result` `    ``static` `int` `f(``int` `idx, ``int` `prev_idx, ``int` `n, ``int` `a[],` `                 ``int``[][] dp)` `    ``{` `        ``if` `(idx == n) {` `            ``return` `0``;` `        ``}`   `        ``if` `(dp[idx][prev_idx + ``1``] != -``1``) {` `            ``return` `dp[idx][prev_idx + ``1``];` `        ``}`   `        ``int` `notTake = ``0` `+ f(idx + ``1``, prev_idx, n, a, dp);` `        ``int` `take = Integer.MIN_VALUE;` `        ``if` `(prev_idx == -``1` `|| a[idx] > a[prev_idx]) {` `            ``take = ``1` `+ f(idx + ``1``, idx, n, a, dp);` `        ``}`   `        ``return` `dp[idx][prev_idx + ``1``]` `            ``= Math.max(take, notTake);` `    ``}`   `    ``// The wrapper function for _lis()` `    ``static` `int` `lis(``int` `arr[], ``int` `n)` `    ``{` `        ``// The function _lis() stores its result in max` `        ``int` `dp[][] = ``new` `int``[n + ``1``][n + ``1``];` `        ``for` `(``int` `row[] : dp)` `            ``Arrays.fill(row, -``1``);`   `        ``return` `f(``0``, -``1``, n, arr, dp);` `    ``}`   `    ``// Driver program to test above functions` `    ``public` `static` `void` `main(String args[])` `    ``{` `        ``int` `a[] = { ``3``, ``10``, ``2``, ``1``, ``20` `};` `        ``int` `n = a.length;`   `        ``// Function call` `        ``System.out.println(``"Length of lis is "` `+ lis(a, n));` `    ``}` `}`   `// This code is contributed by Sanskar.`

## Python3

 `# A Naive Python recursive implementation` `# of LIS problem`     `import` `sys`   `# To make use of recursive calls, this` `# function must return two things:` `# 1) Length of LIS ending with element arr[n-1].` `#     We use max_ending_here for this purpose` `# 2) Overall maximum as the LIS may end with` `#     an element before arr[n-1] max_ref is` `#     used this purpose.` `# The value of LIS of full array of size n` `# is stored in *max_ref which is our final result`     `def` `f(idx, prev_idx, n, a, dp):`   `    ``if` `(idx ``=``=` `n):` `        ``return` `0`   `    ``if` `(dp[idx][prev_idx ``+` `1``] !``=` `-``1``):` `        ``return` `dp[idx][prev_idx ``+` `1``]`   `    ``notTake ``=` `0` `+` `f(idx ``+` `1``, prev_idx, n, a, dp)` `    ``take ``=` `-``sys.maxsize ``-` `1` `    ``if` `(prev_idx ``=``=` `-``1` `or` `a[idx] > a[prev_idx]):` `        ``take ``=` `1` `+` `f(idx ``+` `1``, idx, n, a, dp)`   `    ``dp[idx][prev_idx ``+` `1``] ``=` `max``(take, notTake)` `    ``return` `dp[idx][prev_idx ``+` `1``]`   `# Function to find length of longest increasing` `# subsequence.`     `def` `longestSubsequence(n, a):`   `    ``dp ``=` `[[``-``1` `for` `i ``in` `range``(n ``+` `1``)]``for` `j ``in` `range``(n ``+` `1``)]` `    ``return` `f(``0``, ``-``1``, n, a, dp)`     `# Driver program to test above function` `if` `__name__ ``=``=` `'__main__'``:` `    ``a ``=` `[``3``, ``10``, ``2``, ``1``, ``20``]` `    ``n ``=` `len``(a)`   `    ``# Function call` `    ``print``(``"Length of lis is"``, longestSubsequence(n, a))`   `# This code is contributed by shinjanpatra`

## C#

 `// C# approach to implementation the memoization approach`   `using` `System;`   `class` `GFG {`   `    ``// To make use of recursive calls, this` `    ``// function must return two things:` `    ``// 1) Length of LIS ending with element arr[n-1].` `    ``// We use max_ending_here for this purpose` `    ``// 2) Overall maximum as the LIS may end with` `    ``// an element before arr[n-1] max_ref is` `    ``//  used this purpose.` `    ``// The value of LIS of full array of size n` `    ``// is stored in *max_ref which is our final result` `    ``public` `static` `int` `INT_MIN = -2147483648;`   `    ``public` `static` `int` `f(``int` `idx, ``int` `prev_idx, ``int` `n,` `                        ``int``[] a, ``int``[, ] dp)` `    ``{` `        ``if` `(idx == n) {` `            ``return` `0;` `        ``}` `        ``if` `(dp[idx, prev_idx + 1] != -1) {` `            ``return` `dp[idx, prev_idx + 1];` `        ``}` `        ``int` `notTake = 0 + f(idx + 1, prev_idx, n, a, dp);` `        ``int` `take = INT_MIN;` `        ``if` `(prev_idx == -1 || a[idx] > a[prev_idx]) {` `            ``take = 1 + f(idx + 1, idx, n, a, dp);` `        ``}`   `        ``return` `dp[idx, prev_idx + 1]` `            ``= Math.Max(take, notTake);` `    ``}`   `    ``// Function to find length of longest increasing` `    ``// subsequence.` `    ``public` `static` `int` `longestSubsequence(``int` `n, ``int``[] a)` `    ``{` `        ``int``[, ] dp = ``new` `int``[n + 1, n + 1];`   `        ``for` `(``int` `i = 0; i < n + 1; i++) {` `            ``for` `(``int` `j = 0; j < n + 1; j++) {` `                ``dp[i, j] = -1;` `            ``}` `        ``}` `        ``return` `f(0, -1, n, a, dp);` `    ``}`   `    ``// Driver code` `    ``static` `void` `Main()` `    ``{` `        ``int``[] a = { 3, 10, 2, 1, 20 };` `        ``int` `n = a.Length;` `        ``Console.WriteLine(``"Length of lis is "` `                          ``+ longestSubsequence(n, a));` `    ``}` `}`   `// The code is contributed by Nidhi goel.`

## Javascript

 `/* A Naive Javascript recursive implementation` `      ``of LIS problem */`   `      ``/* To make use of recursive calls, this` `      ``function must return two things:` `      ``1) Length of LIS ending with element arr[n-1].` `          ``We use max_ending_here for this purpose` `      ``2) Overall maximum as the LIS may end with` `          ``an element before arr[n-1] max_ref is` `          ``used this purpose.` `      ``The value of LIS of full array of size n` `      ``is stored in *max_ref which is our final result` `      ``*/`   `      ``function` `f(idx, prev_idx, n, a, dp) {` `        ``if` `(idx == n) {` `          ``return` `0;` `        ``}`   `        ``if` `(dp[idx][prev_idx + 1] != -1) {` `          ``return` `dp[idx][prev_idx + 1];` `        ``}`   `        ``var` `notTake = 0 + f(idx + 1, prev_idx, n, a, dp);` `        ``var` `take = Number.MIN_VALUE;` `        ``if` `(prev_idx == -1 || a[idx] > a[prev_idx]) {` `          ``take = 1 + f(idx + 1, idx, n, a, dp);` `        ``}`   `        ``return` `(dp[idx][prev_idx + 1] = Math.max(take, notTake));` `      ``}` `      ``// Function to find length of longest increasing` `      ``// subsequence.` `      ``function` `longestSubsequence(n, a) {` `        ``var` `dp = Array(n + 1)` `          ``.fill()` `          ``.map(() => Array(n + 1).fill(-1));` `        ``return` `f(0, -1, n, a, dp);` `      ``}`   `      ``/* Driver program to test above function */`   `      ``var` `a = [3, 10, 2, 1, 20];` `      ``var` `n = 5;` `      ``console.log(``"Length of lis is "` `+ longestSubsequence(n, a));` `      `  `      ``// This code is contributed by satwiksuman.`

Output

`Length of lis is 3`

Time Complexity: O(N2)
Auxiliary Space: O(N2)

## Longest Increasing Subsequence using Dynamic Programming:

Due to optimal substructure and overlapping subproblem property, we can also utilise Dynamic programming to solve the problem. Instead of memoization, we can use the nested loop to implement the recursive relation.

The outer loop will run from i = 1 to N and the inner loop will run from j = 0 to i and use the recurrence relation to solve the problem.

Below is the implementation of the above approach:

## C++

 `// Dynamic Programming C++ implementation` `// of LIS problem` `#include ` `using` `namespace` `std;`   `// lis() returns the length of the longest` `// increasing subsequence in arr[] of size n` `int` `lis(``int` `arr[], ``int` `n)` `{` `    ``int` `lis[n];`   `    ``lis[0] = 1;`   `    ``// Compute optimized LIS values in` `    ``// bottom up manner` `    ``for` `(``int` `i = 1; i < n; i++) {` `        ``lis[i] = 1;` `        ``for` `(``int` `j = 0; j < i; j++)` `            ``if` `(arr[i] > arr[j] && lis[i] < lis[j] + 1)` `                ``lis[i] = lis[j] + 1;` `    ``}`   `    ``// Return maximum value in lis[]` `    ``return` `*max_element(lis, lis + n);` `}`   `// Driver program to test above function` `int` `main()` `{` `    ``int` `arr[] = { 10, 22, 9, 33, 21, 50, 41, 60 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);`   `    ``// Function call` `    ``printf``(``"Length of lis is %d\n"``, lis(arr, n));` `    ``return` `0;` `}`

## Java

 `// Dynamic Programming Java implementation` `// of LIS problem`   `import` `java.lang.*;`   `class` `LIS {`   `    ``// lis() returns the length of the longest` `    ``// increasing subsequence in arr[] of size n` `    ``static` `int` `lis(``int` `arr[], ``int` `n)` `    ``{` `        ``int` `lis[] = ``new` `int``[n];` `        ``int` `i, j, max = ``0``;`   `        ``// Initialize LIS values for all indexes` `        ``for` `(i = ``0``; i < n; i++)` `            ``lis[i] = ``1``;`   `        ``// Compute optimized LIS values in` `        ``// bottom up manner` `        ``for` `(i = ``1``; i < n; i++)` `            ``for` `(j = ``0``; j < i; j++)` `                ``if` `(arr[i] > arr[j] && lis[i] < lis[j] + ``1``)` `                    ``lis[i] = lis[j] + ``1``;`   `        ``// Pick maximum of all LIS values` `        ``for` `(i = ``0``; i < n; i++)` `            ``if` `(max < lis[i])` `                ``max = lis[i];`   `        ``return` `max;` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String args[])` `    ``{` `        ``int` `arr[] = { ``10``, ``22``, ``9``, ``33``, ``21``, ``50``, ``41``, ``60` `};` `        ``int` `n = arr.length;`   `        ``// Function call` `        ``System.out.println(``"Length of lis is "` `                           ``+ lis(arr, n));` `    ``}` `}`   `// This code is contributed by Rajat Mishra`

## Python3

 `# Dynamic programming Python implementation` `# of LIS problem`     `# lis returns length of the longest` `# increasing subsequence in arr of size n` `def` `lis(arr):` `    ``n ``=` `len``(arr)`   `    ``# Declare the list (array) for LIS and` `    ``# initialize LIS values for all indexes` `    ``lis ``=` `[``1``]``*``n`   `    ``# Compute optimized LIS values in bottom up manner` `    ``for` `i ``in` `range``(``1``, n):` `        ``for` `j ``in` `range``(``0``, i):` `            ``if` `arr[i] > arr[j] ``and` `lis[i] < lis[j] ``+` `1``:` `                ``lis[i] ``=` `lis[j]``+``1`   `    ``# Initialize maximum to 0 to get` `    ``# the maximum of all LIS` `    ``maximum ``=` `0`   `    ``# Pick maximum of all LIS values` `    ``for` `i ``in` `range``(n):` `        ``maximum ``=` `max``(maximum, lis[i])`   `    ``return` `maximum`     `# Driver program to test above function` `if` `__name__ ``=``=` `'__main__'``:` `    ``arr ``=` `[``10``, ``22``, ``9``, ``33``, ``21``, ``50``, ``41``, ``60``]` `    ``print``(``"Length of lis is"``, lis(arr))`     `# This code is contributed by Nikhil Kumar Singh`

## C#

 `// Dynamic Programming C# implementation of LIS problem`   `using` `System;` `class` `LIS {`   `    ``// lis() returns the length of the longest increasing` `    ``// subsequence in arr[] of size n` `    ``static` `int` `lis(``int``[] arr, ``int` `n)` `    ``{` `        ``int``[] lis = ``new` `int``[n];` `        ``int` `i, j, max = 0;`   `        ``// Initialize LIS values for all indexes` `        ``for` `(i = 0; i < n; i++)` `            ``lis[i] = 1;`   `        ``// Compute optimized LIS values in bottom up manner` `        ``for` `(i = 1; i < n; i++)` `            ``for` `(j = 0; j < i; j++)` `                ``if` `(arr[i] > arr[j] && lis[i] < lis[j] + 1)` `                    ``lis[i] = lis[j] + 1;`   `        ``// Pick maximum of all LIS values` `        ``for` `(i = 0; i < n; i++)` `            ``if` `(max < lis[i])` `                ``max = lis[i];`   `        ``return` `max;` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `Main()` `    ``{` `        ``int``[] arr = { 10, 22, 9, 33, 21, 50, 41, 60 };` `        ``int` `n = arr.Length;`   `        ``// Function call` `        ``Console.WriteLine(``"Length of lis is "` `                          ``+ lis(arr, n));` `    ``}` `}`   `// This code is contributed by Ryuga`

## Javascript

 ``

Output

```Length of lis is 5
```

Time Complexity: O(N2) As a nested loop is used.
Auxiliary Space: O(N) Use of any array to store LIS values at each index.

Note: The time complexity of the above Dynamic Programming (DP) solution is O(n^2), but there is an O(N* logN) solution for the LIS problem. We have not discussed the O(N log N) solution here.
Refer: Longest Increasing Subsequence Size (N * logN) for the mentioned approach.

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