Given an array of integers of size N, you have to divide it into the minimum number of “strictly increasing subsequences”

For example: let the sequence be {1, 3, 2, 4}, then the answer would be 2. In this case, the first increasing sequence would be {1, 3, 4} and the second would be {2}.

Examples:

Input : arr[] = {1 3 2 4}

Output: 2

There are two increasing subsequences {1, 3, 4} and {2}

Input : arr[] = {4 3 2 1}

Output : 4

Input : arr[] = {1 2 3 4}

Output : 1

Input : arr[] = {1 6 2 4 3}

Output : 3

If we focus on the example we can see that the Minimum number of increasing subsequences equals to the length of longest decreasing subsequence where each element from the longest decreasing subsequence represents an increasing subsequence, so it can be found in N*Log(N) time complexity in the same way as longest increasing subsequence by multiplying all the elements with -1.

We iterator over all elements and store in a sorted array (multiset) S the last element in each one of the increasing subsequences found so far and for every element X, we pick the largest element smaller than X -using binary search- in the S and replace it with X which means that we added the current element to increasing subsequence ending with X, otherwise, if there is no element smaller than X in S we insert it in S which forms a new increasing subsequence and so on until the last element and our answer in the last will be the size of S.

## CPP

`// C++ program to count the Minimum number of` `// increasing subsequences` `#include <bits/stdc++.h>` `using` `namespace` `std;` `int` `MinimumNumIncreasingSubsequences(` `int` `arr[], ` `int` `n)` `{` ` ` `multiset<` `int` `> last;` ` ` `// last element in each increasing subsequence` ` ` `// found so far` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` `// here our current element is arr[i]` ` ` `multiset<` `int` `>::iterator it = last.lower_bound(arr[i]);` ` ` `// iterator to the first element larger` ` ` `// than or equal to arr[i]` ` ` `if` `(it == last.begin())` ` ` `// if all the elements in last larger` ` ` `// than or to arr[i] then insert it into last` ` ` `last.insert(arr[i]);` ` ` `else` `{` ` ` `it--;` ` ` `// the largest element smaller than arr[i] is the number` ` ` `// before *it which is it--` ` ` `last.erase(it); ` `// erase the largest element smaller than arr[i]` ` ` `last.insert(arr[i]); ` `// and replace it with arr[i]` ` ` `}` ` ` `}` ` ` `return` `last.size(); ` `// our answer is the size of last` `}` `// Driver program` `int` `main()` `{` ` ` `int` `arr[] = { 8, 4, 1, 2, 9 };` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(` `int` `);` ` ` `cout << ` `"Minimum number of increasing subsequences are : "` ` ` `<< MinimumNumIncreasingSubsequences(arr, n);` ` ` `return` `0;` `}` |

**Output**

Minimum number of increasing subsequences are : 3

Time complexity: O(N log(N))

Auxiliary Space : O(N) **Approach 2**: The idea is to find the longest decreasing subsequence

- Initialize a dp array of length n.
- Inverting all the elements of the array.
- for each element in the array.
- find the index if the current element in the dp array.
- find the maximum index which is valid.
- dp[i] indicate that minimum element ending at the length i subsequence.

**Below is the implementation of above approach** :

## Java

`// Java program for the above approach` `import` `java.util.*;` `public` `class` `Main {` ` ` ` ` `static` `int` `longestDecrasingSubsequence(` `int` `A[], ` `int` `N)` ` ` `{` ` ` `// Initialise Dp array` ` ` `int` `dp[] = ` `new` `int` `[N + ` `1` `];` ` ` `Arrays.fill(dp, Integer.MAX_VALUE);` ` ` ` ` `// All the elements are in range` ` ` `// of Integer minvalue` ` ` `// to maxvalue` ` ` `// dp[i] indicate the min element` ` ` `// of subsequence of` ` ` `// length i is dp[i]` ` ` `dp[` `0` `] = Integer.MIN_VALUE;` ` ` ` ` `// For each number search for the correct position` ` ` `// of number and insert the number in array` ` ` `for` `(` `int` `i = ` `0` `; i < N; i++) {` ` ` ` ` `// search for th position` ` ` `int` `index = search(dp, A[i]);` ` ` ` ` `// update the dp array` ` ` `if` `(index != -` `1` `)` ` ` `dp[index] = Math.min(dp[index], A[i]);` ` ` `}` ` ` `int` `len = ` `0` `;` ` ` `for` `(` `int` `i = ` `1` `; i <= N; i++) {` ` ` `if` `(dp[i] != Integer.MAX_VALUE)` ` ` `len = Math.max(i, len);` ` ` `}` ` ` `return` `len;` ` ` `}` ` ` ` ` `// to search for correct position of num in array dp` ` ` `static` `int` `search(` `int` `dp[], ` `int` `num)` ` ` `{` ` ` ` ` `// initialise low,high and ans` ` ` `int` `low = ` `0` `, high = dp.length - ` `1` `;` ` ` `int` `ans = -` `1` `;` ` ` `while` `(low <= high) {` ` ` `// get mid` ` ` `int` `mid = low + (high - low) / ` `2` `;` ` ` ` ` `// if mid element is >=num search for left half` ` ` `if` `(dp[mid] >= num) {` ` ` `ans = mid;` ` ` `high = mid - ` `1` `;` ` ` `}` ` ` `else` ` ` `low = mid + ` `1` `;` ` ` `}` ` ` `return` `ans;` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `main(String args[])` ` ` `{` ` ` `int` `n = ` `4` `;` ` ` `int` `a[] = { ` `1` `, ` `2` `, ` `3` `, ` `4` `};` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++)` ` ` `a[i] = -a[i];` ` ` `System.out.print(longestDecrasingSubsequence(a, n));` ` ` `}` `}` |

**Output**

1

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