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 : 4Input : arr[] = {1 2 3 4}
Output : 1Input : 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 programint 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;} |
Java
// Java program to count the Minimum number of// increasing subsequencesimport java.util.*;public class Main { static int MinimumNumIncreasingSubsequences(int[] arr, int n) { TreeSet<Integer> last = new TreeSet<>(); // last element in each increasing subsequence // found so far for (int i = 0; i < n; i++) { Integer it = last.ceiling(arr[i]); if (it == null) // if all the elements in last larger // than or to arr[i] then insert it into // last last.add(arr[i]); else { last.remove(it); last.add(arr[i]); } } return last.size(); } // Driver program public static void main(String[] args) { int[] arr = { 8, 4, 1, 2, 9 }; int n = arr.length; System.out.println( "Minimum number of increasing subsequences are: " + MinimumNumIncreasingSubsequences(arr, n)); }}//contributed by Aditya Sharma |
Python3
# Python code implementationdef MinimumNumIncreasingSubsequences(arr, n): last = set() # last element in each increasing subsequence found # so far for i in range(n): it = -1 for value in last: if value >= arr[i]: it = value break if it == -1: # if all the elements in last larger than # or to arr[i] then insert it into last last.add(arr[i]) else: last.remove(it) last.add(arr[i]) return len(last)# Codearr = [8, 4, 1, 2, 9]n = len(arr)print("Minimum number of increasing subsequences are:", MinimumNumIncreasingSubsequences(arr, n))# This code is contributed by sankar. |
C#
// C# program to count the minimum no. of increasing// subsequencesusing System;using System.Collections.Generic;public class GFG { static int MinimumNumIncreasingSubsequences(int[] arr, int n) { HashSet<int> last = new HashSet<int>(); // last element in each increasing subsequence found // so far for (int i = 0; i < n; i++) { int it = -1; foreach(int value in last) { if (value >= arr[i]) { it = value; break; } } if (it == -1) // if all the elements in last larger than // or to arr[i] then insert it into last last.Add(arr[i]); else { last.Remove(it); last.Add(arr[i]); } } return last.Count; } static public void Main() { // Code int[] arr = { 8, 4, 1, 2, 9 }; int n = arr.Length; Console.WriteLine( "Minimum number of increasing subsequences are: " + MinimumNumIncreasingSubsequences(arr, n)); }}// This code is contributed by karthik. |
Javascript
// Javascript program to count the Minimum number of// increasing subsequencesfunction MinimumNumIncreasingSubsequences(arr, n) { let last = new Set(); // last element in each increasing subsequence // found so far for (let i = 0; i < n; i++) { let it = null; for (let value of last) { if (value >= arr[i]) { it = value; break; } } if (it === null) // if all the elements in last larger // than or to arr[i] then insert it into // last last.add(arr[i]); else { last.delete(it); last.add(arr[i]); }}return last.size;}// Driver programlet arr = [8, 4, 1, 2, 9];let n = arr.length;console.log("Minimum number of increasing subsequences are: " +MinimumNumIncreasingSubsequences(arr, n)); |
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 :
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// To search for correct position of num in array dpint search(vector<int>dp, int num){ // Initialise low,high and ans int low = 0,high = dp.size() - 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;}int longestDecrasingSubsequence(vector<int>A,int N){ // Initialise Dp array vector<int>dp(N+1,INT_MAX); // 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] = INT_MIN; // 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 the position int index = search(dp, A[i]); // update the dp array if (index != -1) dp[index] = min(dp[index], A[i]); } int Len = 0; for(int i = 1; i < N; i++){ if (dp[i] != INT_MAX) Len = max(i, Len); } return Len;}// Driver codeint main(){ int n = 4; vector<int> a = { 1, 2, 3, 4 }; for(int i=0;i<n;i++) a[i] = -a[i]; cout << longestDecrasingSubsequence(a, n) << endl; return 0;}// This code is contributed by shinjanpatra |
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 the 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)); }} |
Python3
# Python program for the above approachimport sysdef longestDecrasingSubsequence(A,N): # Initialise Dp array dp = [sys.maxsize for i in range(N+1)] # 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] = -sys.maxsize -1 # For each number search for the correct position # of number and insert the number in array for i in range(N): # search for the position index = search(dp, A[i]) # update the dp array if (index != -1): dp[index] = min(dp[index], A[i]) Len = 0 for i in range(1,N): if (dp[i] != sys.maxsize): Len = max(i, Len) return Len# to search for correct position of num in array dpdef search(dp, num): # initialise low,high and ans low,high = 0, len(dp) - 1 ans = -1 while (low <= high): # get mid 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 Coden = 4a = [ 1, 2, 3, 4 ]for i in range(n): a[i] = -a[i]print(longestDecrasingSubsequence(a, n))# This code is contributed by shinjanpatra |
C#
// C# code for the above approachusing System;class GFG { static int longestDecrasingSubsequence(int[] A, int N) { // Initialise Dp array int[] dp = new int[N + 1]; for (int i = 0; i < dp.Length; i++) dp[i] = int.MaxValue; // 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] = int.MinValue; // 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 the 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] != int.MaxValue) len = Math.Max(i, len); } return len/4; } // 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]; Console.WriteLine( longestDecrasingSubsequence(a, n)); }}// This code is contributed by pradeepkumarppk2003 |
Javascript
// JS program for the above approach// To search for correct position of num in array dpfunction search(dp, num) { // Initialise low,high and ans let low = 0, high = dp.length - 1; let ans = -1; while (low <= high) { // Get mid let mid = low + Math.floor((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;}function longestDecrasingSubsequence(A, N) { // Initialise Dp array let dp = new Array(N + 1).fill(Number.MAX_SAFE_INTEGER); // 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] = Number.MIN_SAFE_INTEGER; // For each number search for the correct position // of number and insert the number in array for (let i = 0; i < N; i++) { // search for the position let index = search(dp, A[i]); // update the dp array if (index !== -1) dp[index] = Math.min(dp[index], A[i]); } let Len = 0; for (let i = 1; i < N; i++) { if (dp[i] !== Number.MAX_SAFE_INTEGER) Len = Math.max(i, Len); } return Len;}// Driver codelet n = 4;let a = [1, 2, 3, 4];for (let i = 0; i < n; i++) a[i] = -a[i];console.log(longestDecrasingSubsequence(a, n));// This code is contributed by lokeshpotta20. |
Output
1
Time complexity: O(N * log(N))
Auxiliary Space: O(N)
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