Given a permutation from 1 to N of size N. Find the length of the longest subsequence having the property that the first and last element is greater than all the other subsequence elements.
Examples:
Input : N = 6
4 2 6 5 3 1
Output : 3
The subsequence that has the longest size is [4, 2, 3] or [4, 2, 6] or [4, 2, 5]. No other subsequence holds the above property with first and last element greater than all remaining subsequence elements.
Input : N = 5
5 4 1 2 3
Output : 4
The longest subsequence is [5, 1, 2, 3] or [4, 1, 2, 3]. With 5 and 3 greater than all other subsequence elements i.e 1 and 2.
Approach:
Idea is to use Fenwick tree data structure.Start iterating from the highest value in the permutation to lowest one. For each iteration, move left or a right pointer to the current element only if the left pointer can be moved to more left and right pointer to more right.
If left index is L and right index is R then there are elements from [L+1, R-1] in between. For this particular iteration i, count the number of elements that are lesser than min(arr[L], arr[R]) using the fenwick tree.
Store each query of type {left+1, right-1, val, index}, where left and right are current left and right pointers and val is the min of arr[L] and arr[R] and idx is the index of that particular query.
- Sort the array in ascending order.
- Sort the queries according to val i.e min(arr[L], arr[R] ) in ascending order, initialize Fenwick array as 0.
- Start from the first query and traverse the array until the value in the array is less than equal to val. For each such element update the Fenwick tree with value equal to 1.
- Query the Fenwick array in the range l to r.
- Print maximum of all the query results .
Below is the implementation of the above approach:
CPP
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;struct Query { int l, r, x, index;};struct Arrays { int val, index;};// Comparison functionsbool cmp1(Query q1, Query q2){ return q1.x < q2.x;}bool cmp2(Arrays x, Arrays y){ return x.val < y.val;}// Function to update the value in Fenwick treevoid update(int* Fenwick, int index, int val, int n){ while (index <= n) { Fenwick[index] += val; index += index & (-index); }}// Function to return the query resultint query(int* Fenwick, int index, int n){ int sum = 0; while (index > 0) { sum = sum + Fenwick[index]; index -= index & (-index); } return sum;}// Function to return the length of subsequenceint maxLength(int n, vector<int>& v){ int where[n + 2]; memset(where, 0, sizeof where); Arrays arr[n]; // Store the value in struct Array for (int i = 1; i <= n; ++i) { v[i - 1] = v[i - 1] - 1; int x = v[i - 1]; where[x] = i - 1; arr[i - 1].val = x; arr[i - 1].index = i - 1; } // If less than 2 elements are // present return that element. if (n <= 2) { cout << n << endl; return 0; } // Set the left and right pointers to extreme int left = n, right = 0, mx = 0; Query queries[4 * n]; int j = 0; for (int i = n - 1; i >= 0; --i) { // Calculate left and right pointer index. left = min(left, where[i]); right = max(right, where[i]); int diff = right - left; if (diff == 0 || diff == 1) { continue; } int val1 = v[left]; int val2 = v[right]; int minn = min(val1, val2); // Store the queries from [L+1, R-1]. queries[j].l = left + 1; queries[j].r = right - 1; queries[j].x = minn; queries[j].index = j; ++j; } int Fenwick[n + 1]; memset(Fenwick, 0, sizeof Fenwick); int q = j - 1; // Sort array and queries for fenwick updates sort(arr, arr + n + 1, cmp2); sort(queries, queries + q + 1, cmp1); int curr = 0; int ans[q]; memset(ans, 0, sizeof ans); // For each query calculate maxx for // the answer and store it in ans array. for (int i = 0; i <= q; ++i) { while (arr[curr].val <= queries[i].x and curr < n) { update(Fenwick, arr[curr].index + 1, 1, n); curr++; } ans[queries[i].index] = query(Fenwick, queries[i].r + 1, n) - query(Fenwick, queries[i].l, n); } for (int i = 0; i <= q; ++i) { mx = max(mx, ans[i]); } // Mx will be mx + 2 as while calculating // mx, we excluded element // at index left and right mx = mx + 2; return mx;}// Driver Codeint main(){ int n = 6; vector<int> v = { 4, 2, 6, 5, 3, 1 }; cout << maxLength(n, v) << endl; return 0;} |
Java
import java.util.*;public class GFG { static class Query { int l, r, x, index; } static class ArrayElement { int val, index; } // Comparison functions static boolean cmp1(Query q1, Query q2) { return q1.x < q2.x; } static boolean cmp2(ArrayElement x, ArrayElement y) { return x.val < y.val; } // Function to update the value in Fenwick tree static void update(int[] Fenwick, int index, int val, int n) { while (index <= n) { Fenwick[index] += val; index += index & (-index); } } // Function to return the query result static int query(int[] Fenwick, int index, int n) { int sum = 0; while (index > 0) { sum = sum + Fenwick[index]; index -= index & (-index); } return sum; } // Function to return the length of subsequence static int maxLength(int n, ArrayList<Integer> v) { int[] where = new int[n + 2]; ArrayElement[] arr = new ArrayElement[n]; Query[] queries = new Query[4 * n]; Query[] ans = new Query[n]; for (int i = 1; i <= n; ++i) { v.set(i - 1, v.get(i - 1) - 1); int x = v.get(i - 1); where[x] = i - 1; arr[i - 1] = new ArrayElement(); arr[i - 1].val = x; arr[i - 1].index = i - 1; } if (n <= 2) { System.out.println(n); return 0; } int left = n, right = 0, mx = 0; int j = 0; for (int i = n - 1; i >= 0; --i) { left = Math.min(left, where[i]); right = Math.max(right, where[i]); int diff = right - left; if (diff == 0 || diff == 1) { continue; } int val1 = v.get(left); int val2 = v.get(right); int minn = Math.min(val1, val2); queries[j] = new Query(); queries[j].x = minn; queries[j].index = j; queries[j].l = left + 1; queries[j].r = right - 1; ++j; } int[] Fenwick = new int[n + 1]; int q = j - 1; ArrayElement[] sortedArr = Arrays.copyOfRange(arr, 0, n + 1); Query[] sortedQueries = Arrays.copyOfRange(queries, 0, q + 1); Query[] sortedAns = new Query[n]; int curr = 0; Query[] sortedAnsTemp = Arrays.copyOfRange(sortedAns, 0, n); // For each query calculate maxx for // the answer and store it in ans array. for (int i = 0; i <= q; ++i) { while (sortedArr[curr].val <= sortedQueries[i].x && curr < n) { update(Fenwick, sortedArr[curr].index + 1, 1, n); curr++; } sortedAnsTemp[sortedQueries[i].index] = new Query(); sortedAnsTemp[sortedQueries[i].index].x = query(Fenwick, sortedQueries[i].r + 1, n) - query(Fenwick, sortedQueries[i].l, n); sortedAnsTemp[sortedQueries[i].index].index = sortedQueries[i].index; } for (int i = 0; i <= q; ++i) { mx = Math.max(mx, sortedAnsTemp[i].x); } // Mx will be mx + 2 as while calculating // mx, we excluded element // at index left and right mx = mx + 2; return mx; } public static void main(String[] args) { int n = 6; ArrayList<Integer> v = new ArrayList<>(Arrays.asList(4, 2, 6, 5, 3, 1)); System.out.println(maxLength(n, v)); }} |
Python3
# Python implementation of the above approachfrom typing import Listclass Query: def __init__(self) -> None: self.l = 0 self.r = 0 self.x = 0 self.index = 0class Arrays: def __init__(self) -> None: self.val = 0 self.index = 0# Function to update the value in Fenwick treedef update(Fenwick: List[int], index: int, val: int, n: int) -> None: while (index <= n): Fenwick[index] += val index += index & (-index)# Function to return the query resultdef query(Fenwick: List[int], index: int, n: int) -> int: summ = 0 while (index > 0): summ = summ + Fenwick[index] index -= index & (-index) return summ# Function to return the length of subsequencedef maxLength(n: int, v: List[int]) -> int: where = [0 for _ in range(n + 2)] arr = [Arrays() for _ in range(n)] # Store the value in struct Array for i in range(1, n + 1): v[i - 1] = v[i - 1] - 1 x = v[i - 1] where[x] = i - 1 arr[i - 1].val = x arr[i - 1].index = i - 1 # If less than 2 elements are # present return that element. if (n <= 2): print(n) return 0 # Set the left and right pointers to extreme left = n right = 0 mx = 0 queries = [Query() for _ in range(4 * n)] j = 0 for i in range(n - 1, -1, -1): # Calculate left and right pointer index. left = min(left, where[i]) right = max(right, where[i]) diff = right - left if (diff == 0 or diff == 1): continue val1 = v[left] val2 = v[right] minn = min(val1, val2) # Store the queries from [L+1, R-1]. queries[j].l = left + 1 queries[j].r = right - 1 queries[j].x = minn queries[j].index = j j += 1 Fenwick = [0 for _ in range(n + 1)] q = j - 1 # Sort array and queries for fenwick updates arr[:n + 1].sort(key=lambda x: x.val) queries[:q + 1].sort(key=lambda val: val.x) curr = 0 ans = [0 for _ in range(q + 1)] # For each query calculate maxx for # the answer and store it in ans array. for i in range(q + 1): while (arr[curr].val <= queries[i].x and curr < n): update(Fenwick, arr[curr].index + 1, 1, n) curr += 1 # if queries[i].index < q: ans[queries[i].index] = query(Fenwick, queries[i].r + 1, n) - query( Fenwick, queries[i].l, n) for i in range(q + 1): mx = max(mx, ans[i]) # Mx will be mx + 2 as while calculating # mx, we excluded element # at index left and right mx = mx + 2 return mx# Driver Codeif __name__ == "__main__": n = 6 v = [4, 2, 6, 5, 3, 1] print(maxLength(n, v))# This code is contributed by sanjeev2552 |
Javascript
<script> //JavaScript code for the above approach function Query() { this.l = 0; this.r = 0; this.x = 0; this.index = 0; } function Arrays() { this.val = 0; this.index = 0; } // Function to update the value in Fenwick tree function update(Fenwick, index, val, n) { while (index <= n) { Fenwick[index] += val; index += index & (-index); } } // Function to return the query result function query(Fenwick, index, n) { let summ = 0; while (index > 0) { summ += Fenwick[index]; index -= index & (-index); } return summ; } // Function to return the length of subsequence function maxLength(n, v) { const where = Array(n + 2).fill(0); const arr = Array(n).fill(0).map(() => new Arrays()); // Store the value in struct Array for (let i = 1; i <= n; i++) { v[i - 1] = v[i - 1] - 1; const x = v[i - 1]; where[x] = i - 1; arr[i - 1].val = x; arr[i - 1].index = i - 1; } // If less than 2 elements are // present return that element. if (n <= 2) { console.log(n); return 0; } // Set the left and right pointers to extreme let left = n; let right = 0; let mx = 0; const queries = Array(4 * n).fill(0).map(() => new Query()); let j = 0; for (let i = n - 1; i >= 0; i--) { // Calculate left and right pointer index. left = Math.min(left, where[i]); right = Math.max(right, where[i]); const diff = right - left; if (diff === 0 || diff === 1) { continue; } const val1 = v[left]; const val2 = v[right]; const minn = Math.min(val1, val2); // Store the queries from [L+1, R-1]. queries[j].l = left + 1; queries[j].r = right - 1; queries[j].x = minn; queries[j].index = j; j++; } const Fenwick = Array(n + 1).fill(0); const q = j - 1; // Sort array and queries for fenwick updates arr.slice(0, n + 1).sort((a, b) => a.val - b.val); queries.slice(0, q + 1).sort((a, b) => a.x - b.x); let curr = 0; const ans = Array(q + 1).fill(0); // For each query calculate maxx for // the answer and store it in ans array. for (let i = 0; i <= q; i++) { while (arr[curr].val <= queries[i].x && curr < n) { update(Fenwick, arr[curr].index + 1, 1, n); curr++; } ans[queries[i].index] = query(Fenwick, queries[i].r + 1, n) - query(Fenwick, queries[i].l, n); } for (let i = 0; i <= q; i++) { mx = Math.max(mx, ans[i]); } // Mx will be mx + 2 as while calculating // mx, we excluded element // at index left and right mx = mx + 2; return mx; } // Driver Code const n = 6; const v = [4, 2, 6, 5, 3, 1]; document.write(maxLength(n, v)); // This code is contributed by Potta Lokesh </script> |
Output
3
Time Complexity: O(nlogn), where N is number of elements in the given range of the vector.
Auxiliary Space: O(n)
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