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Size of array after repeated deletion of LIS
• Difficulty Level : Hard
• Last Updated : 01 Dec, 2020

Given an array arr[0..n-1] of the positive element. The task is to print the remaining elements of arr[] after repeated deletion of LIS (of size greater than 1). If there are multiple LIS with the same length, we need to choose the LIS that ends first.
Examples:

```Input : arr[] = {1, 2, 5, 3, 6, 4, 1}
Output : 1
Explanation :
{1, 2, 5, 3, 6, 4, 1} - {1, 2, 5, 6} = {3, 4, 1}
{3, 4, 1} - {3, 4} = {1}

Input : arr[] = {1, 2, 3, 1, 5, 2}
Output : -1
Explanation :
{1, 2, 3, 1, 5, 2} - {1, 2, 3, 5} = {1, 2}
{1, 2} - {1, 2} = {}

Input : arr[] = {5, 3, 2}
Output : 3```

We repeatedly find LIS and remove its elements from an array.

```// input vector array = arr[]
// max-sum LIS vector array = maxArray[]
while (arr.size())
{
// find LIS
lis = findLIS(arr, arr.size());
if (lis.size() < 2)
break;

// Remove lis elements from current array. Note
// that both lis[] and arr[] are sorted in
// increasing order.
for (i=0; i<arr.size() && lis.size()>0; i++)
{
if (arr[i] == lis[0])
{
// Remove lis element from arr[]
arr.erase(arr.begin()+i) ;
i--;

// erase the element from lis[]. This is
// needed to make sure that next element
// to be removed is first element of lis[]
lis.erase(lis.begin()) ;
}
}
}
// print remaining element of array
for (i=0; i<arr.size(); i++)
cout  << arr[i] << " ";
if (i == 0)
cout << "-1";```

C++

 `/* C++ program to find size of array after repeated` `  ``deletion of LIS */` `#include ` `using` `namespace` `std;`   `// Function to construct Maximum Sum LIS` `vector<``int``> findLIS(vector<``int``> arr, ``int` `n)` `{` `    ``// L[i] - The Maximum Sum Increasing` `    ``// Subsequence that ends with arr[i]` `    ``vector > L(n);`   `    ``// L[0] is equal to arr[0]` `    ``L[0].push_back(arr[0]);`   `    ``// start from index 1` `    ``for` `(``int` `i = 1; i < n; i++)` `    ``{` `        ``// for every j less than i` `        ``for` `(``int` `j = 0; j < i; j++)` `        ``{` `            ``/* L[i] = {MaxSum(L[j])} + arr[i]` `            ``where j < i and arr[j] < arr[i] */` `            ``if` `(arr[i] > arr[j] && (L[i].size() < L[j].size()))` `                ``L[i] = L[j];` `        ``}`   `        ``// L[i] ends with arr[i]` `        ``L[i].push_back(arr[i]);` `    ``}`   `    ``// set lis =  LIS` `    ``// whose size is max among all` `    ``int` `maxSize = 1;` `    ``vector<``int``> lis;` `    ``for` `(vector<``int``> x : L)` `    ``{` `        ``// The > sign makes sure that the LIS` `        ``// ending first is chose.    ` `        ``if` `(x.size() > maxSize)` `        ``{` `            ``lis = x;` `            ``maxSize = x.size();` `        ``}` `    ``}`   `    ``return` `lis;` `}`   `// Function to minimize array` `void` `minimize(``int` `input[], ``int` `n)` `{` `    ``vector<``int``> arr(input, input + n);`   `    ``while` `(arr.size())` `    ``{` `        ``// Find LIS of current array` `        ``vector<``int``> lis = findLIS(arr, arr.size());`   `        ``// If all elements are in decreasing order` `        ``if` `(lis.size() < 2)` `            ``break``;`   `        ``// Remove lis elements from current array. Note` `        ``// that both lis[] and arr[] are sorted in` `        ``// increasing order.` `        ``for` `(``int` `i=0; i0; i++)` `        ``{` `            ``// If first element of lis[] is found` `            ``if` `(arr[i] == lis[0])` `            ``{` `                ``// Remove lis element from arr[]` `                ``arr.erase(arr.begin()+i) ;` `                ``i--;`   `                ``// Erase first element of lis[]` `                ``lis.erase(lis.begin()) ;` `            ``}` `        ``}` `    ``}`   `    ``// print remaining element of array` `    ``int` `i;` `    ``for` `(i=0; i < arr.size(); i++)` `        ``cout  << arr[i] << ``" "``;`   `    ``// print -1 for empty array` `    ``if` `(i == 0)` `        ``cout << ``"-1"``;` `}`   `// Driver function` `int` `main()` `{` `    ``int` `input[] = { 3, 2, 6, 4, 5, 1 };` `    ``int` `n = ``sizeof``(input) / ``sizeof``(input[0]);`   `    ``// minimize array after deleting LIS` `    ``minimize(input, n);`   `    ``return` `0;` `}`

Java

 `// Java program to find size` `//  of array after repeated` `// deletion of LIS` `import` `java.util.*;` `class` `GFG{`   `// Function to conMaximum Sum LIS` `static` `Vector findLIS(Vector arr, ` `                               ``int` `n)` `{` `  ``// L[i] - The Maximum Sum Increasing` `  ``// Subsequence that ends with arr[i]` `  ``Vector []L = ``new` `Vector[n];` `  `  `  ``for` `(``int` `i = ``0``; i < L.length; i++)` `    ``L[i] = ``new` `Vector();` `  `  `  ``// L[0] is equal to arr[0]` `  ``L[``0``].add(arr.elementAt(``0``));`   `  ``// Start from index 1` `  ``for` `(``int` `i = ``1``; i < n; i++)` `  ``{` `    ``// For every j less than i` `    ``for` `(``int` `j = ``0``; j < i; j++)` `    ``{` `      ``// L[i] = {MaxSum(L[j])} + arr[i]` `      ``// where j < i and arr[j] < arr[i]` `      ``if` `(arr.elementAt(i) > arr.elementAt(j) && ` `          ``(L[i].size() < L[j].size()))` `        ``L[i] = L[j];` `    ``}`   `    ``// L[i] ends with arr[i]` `    ``L[i].add(arr.elementAt(i));` `  ``}`   `  ``// Set lis =  LIS` `  ``// whose size is max among all` `  ``int` `maxSize = ``1``;` `  ``Vector lis = ``new` `Vector<>();` `  `  `  ``for` `(Vector x : L)` `  ``{` `    ``// The > sign makes sure that the LIS` `    ``// ending first is chose.    ` `    ``if` `(x.size() > maxSize)` `    ``{` `      ``lis = x;` `      ``maxSize = x.size();` `    ``}` `  ``}` `  ``return` `lis;` `}`   `// Function to minimize array` `static` `void` `minimize(``int` `input[], ` `                     ``int` `n)` `{` `  ``Vector arr = ``new` `Vector<>();` `  ``for``(``int` `i = ``0``; i < n; i++)` `    ``arr.add(input[i]);`   `  ``while` `(arr.size() != ``0``)` `  ``{` `    ``// Find LIS of current array` `    ``Vector lis = findLIS(arr, ` `                                  ``arr.size());`   `    ``// If all elements are ` `    ``// in decreasing order` `    ``if` `(lis.size() < ``2``)` `      ``break``;`   `    ``// Remove lis elements from ` `    ``// current array. Note that both ` `    ``// lis[] and arr[] are sorted in` `    ``// increasing order.` `    ``for` `(``int` `i = ``0``; i < arr.size() && ` `             ``lis.size() > ``0``; i++)` `    ``{` `      ``// If first element of lis[] is found` `      ``if` `(arr.elementAt(i) == lis.elementAt(``0``))` `      ``{` `        ``// Remove lis element from arr[]` `        ``arr.removeAll(lis);` `        ``i--;`   `        ``// Erase first element of lis[]` `        ``lis.remove(``0``);` `      ``}` `    ``}` `  ``}`   `  ``// print remaining element of array` `  ``int` `i;` `  ``for` `(i = ``1``; i < arr.size(); i++)` `    ``System.out.print(arr.elementAt(i) + ``" "``);`   `  ``// print -1 for empty array` `  ``if` `(i == ``0``)` `    ``System.out.print(``"-1"``);` `}`   `// Driver function` `public` `static` `void` `main(String[] args)` `{` `  ``int` `input[] = {``3``, ``2``, ``6``, ``4``, ``5``, ``1``};` `  ``int` `n = input.length;`   `  ``// minimize array after deleting LIS` `  ``minimize(input, n);` `}` `}`   `// This code is contributed by gauravrajput1`

Python3

 `''' Python program to find size of array after repeated` `  ``deletion of LIS '''`   `# Function to construct Maximum Sum LIS` `from` `typing ``import` `List`   `def` `findLIS(arr: ``List``[``int``], n: ``int``) ``-``> ``List``[``int``]:`   `    ``# L[i] - The Maximum Sum Increasing` `    ``# Subsequence that ends with arr[i]` `    ``L ``=` `[``0``] ``*` `n` `    ``for` `i ``in` `range``(n):` `        ``L[i] ``=` `[]`   `    ``# L[0] is equal to arr[0]` `    ``L[``0``].append(arr[``0``])`   `    ``# start from index 1` `    ``for` `i ``in` `range``(``1``, n):`   `        ``# for every j less than i` `        ``for` `j ``in` `range``(i):` `          `  `            ``''' L[i] = MaxSum(L[j]) + arr[i]` `            ``where j < i and arr[j] < arr[i] '''` `            ``if` `(arr[i] > arr[j] ``and` `                ``(``len``(L[i]) < ``len``(L[j]))):` `                ``L[i] ``=` `L[j].copy()`   `        ``# L[i] ends with arr[i]` `        ``L[i].append(arr[i])`   `    ``# set lis =  LIS` `    ``# whose size is max among all` `    ``maxSize ``=` `1` `    ``lis: ``List``[``int``] ``=` `[]` `    ``for` `x ``in` `L:`   `        ``# The > sign makes sure that the LIS` `        ``# ending first is chose.` `        ``if` `(``len``(x) > maxSize):`   `            ``lis ``=` `x.copy()` `            ``maxSize ``=` `len``(x)`   `    ``return` `lis`   `# Function to minimize array` `def` `minimize(``input``: ``List``[``int``], n: ``int``) ``-``> ``None``:`   `    ``arr ``=` `input``.copy()`   `    ``while` `len``(arr):`   `        ``# Find LIS of current array` `        ``lis ``=` `findLIS(arr, ``len``(arr))`   `        ``# If all elements are in decreasing order` `        ``if` `(``len``(lis) < ``2``):` `            ``break`   `        ``# Remove lis elements from current array. Note` `        ``# that both lis[] and arr[] are sorted in` `        ``# increasing order.` `        ``i ``=` `0` `        ``while` `i < ``len``(arr) ``and` `len``(lis) > ``0``:`   `            ``# If first element of lis[] is found` `            ``if` `(arr[i] ``=``=` `lis[``0``]):`   `                ``# Remove lis element from arr[]` `                ``arr.remove(arr[i])` `                ``i ``-``=` `1`   `                ``# Erase first element of lis[]` `                ``lis.remove(lis[``0``])` `            ``i ``+``=` `1`   `    ``# print remaining element of array` `    ``i ``=` `0` `    ``while` `i < ``len``(arr):` `        ``print``(arr[i], end``=``" "``)` `        ``i ``+``=` `1`   `    ``# print -1 for empty array` `    ``if` `(i ``=``=` `0``):` `        ``print``(``"-1"``)`   `# Driver function` `if` `__name__ ``=``=` `"__main__"``:`   `    ``input` `=` `[``3``, ``2``, ``6``, ``4``, ``5``, ``1``]` `    ``n ``=` `len``(``input``)`   `    ``# minimize array after deleting LIS` `    ``minimize(``input``, n)`   `# This code is contributed by sanjeev2552`

Output:

`1`

This article is contributed by Shivam Pradhan (anuj_charm). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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