Given an array arr[] of non-negative integers. We have to delete elements from this array such that arr[i + 1] > arr[j] for each valid i and this will be counted as one step. We have to apply the same operations until the array has become strictly decreasing. Now the task is to count the number of steps required to get the desired array. Examples:
Input: arr[] = {6, 5, 8, 4, 7, 10, 9} Output: 2 Initially 8, 7 and 10 do not satisfy the condition so they all are deleted in the first step and the array becomes {6, 5, 4, 9} In the next step 9 gets deleted and the array becomes {6, 5, 4} which is strictly decreasing. Input: arr[] = {1, 2, 3, 4, 5} Output: 1
Approach: The idea is to keep the indices of only required elements that are to be checked against a particular element. Thus, we use a vector to store only the required indices. We insert every index at the back and then remove the indices from back if the following condition is satisfied.
arr[vect.back()] ? val[i]
We take another array in which we update the no of steps particular element takes to delete. If status[i] = -1 then element is not to be deleted, 0 denotes first step and so on…. That’s why we will add 1 to the answer. While popping the indices, we repeatedly update the status of elements. If all indices are popped i.e. vect.size() = 0 then this element is not to be deleted so change its status to -1. Below is the implementation of the above approach:
CPP
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;int status[100000];// Function to return the required// number of stepsint countSteps(int* val, int n){ int sol = 0; vector<int> vec(1, 0); status[0] = -1; // Compute the number of steps for (int i = 1; i < n; ++i) { // Current status is to // delete in first step status[i] = 0; // Pop the indices while // condition is satisfied while (vec.size() > 0 && val[vec.back()] >= val[i]) { // Inserting the correct // step no to delete status[i] = max(status[i], status[vec.back()] + 1); vec.pop_back(); } if (vec.size() == 0) { // Status changed to not delete status[i] = -1; } // Pushing a new index in the vector vec.push_back(i); // Build the solution from // smaller to larger size sol = max(sol, status[i] + 1); } return sol;}// Driver codeint main(){ int val[] = { 6, 5, 8, 4, 7, 10, 9 }; int n = sizeof(val) / sizeof(val[0]); cout << countSteps(val, n); return 0;} |
Java
// A Java implementation of the approachimport java.util.*;class GFG { static int []status = new int[100000];// Function to return the required// number of stepsstatic int countSteps(int[]val, int n){ int sol = 0; Vector<Integer> vec = new Vector<>(1); vec.add(0); status[0] = -1; // Compute the number of steps for (int i = 1; i < n; ++i) { // Current status is to // delete in first step status[i] = 0; // Pop the indices while // condition is satisfied while (vec.size() > 0 && val[vec.lastElement()] >= val[i]) { // Inserting the correct // step no to delete status[i] = Math.max(status[i], status[vec.lastElement()] + 1); vec.remove(vec.lastElement()); } if (vec.isEmpty()) { // Status changed to not delete status[i] = -1; } // Pushing a new index in the vector vec.add(i); // Build the solution from // smaller to larger size sol = Math.max(sol, status[i] + 1); } return sol;}// Driver codepublic static void main(String[] args) { int val[] = { 6, 5, 8, 4, 7, 10, 9 }; int n = val.length; System.out.println(countSteps(val, n));}}/* This code contributed by PrinciRaj1992 */ |
Python3
# Python3 implementation of the approach status = [0]*100000; # Function to return the required # number of steps def countSteps(val, n) : sol = 0; vec = [1, 0]; status[0] = -1; # Compute the number of steps for i in range(n) : # Current status is to # delete in first step status[i] = 0; # Pop the indices while # condition is satisfied while (len(vec) > 0 and val[vec[len(vec)-1]] >= val[i]) : # Inserting the correct # step no to delete status[i] = max(status[i], status[len(vec)-1] + 1); vec.pop(); if (len(vec) == 0) : # Status changed to not delete status[i] = -1; # Pushing a new index in the vector vec.append(i); # Build the solution from # smaller to larger size sol = max(sol, status[i] + 1); return sol; # Driver code if __name__ == "__main__" : val = [ 6, 5, 8, 4, 7, 10, 9 ]; n = len(val); print(countSteps(val, n)); # This code is contributed by AnkitRai01 |
C#
// A C# implementation of the approach using System;using System.Collections.Generic;class GFG { static int []status = new int[100000]; // Function to return the required // number of steps static int countSteps(int[]val, int n) { int sol = 0; List<int> vec = new List<int>(1); vec.Add(0); status[0] = -1; // Compute the number of steps for (int i = 1; i < n; ++i) { // Current status is to // delete in first step status[i] = 0; // Pop the indices while // condition is satisfied while (vec.Count > 0 && val[vec[vec.Count-1]] >= val[i]) { // Inserting the correct // step no to delete status[i] = Math.Max(status[i], status[vec[vec.Count-1]] + 1); vec.Remove(vec[vec.Count-1]); } if (vec.Count == 0) { // Status changed to not delete status[i] = -1; } // Pushing a new index in the vector vec.Add(i); // Build the solution from // smaller to larger size sol = Math.Max(sol, status[i] + 1); } return sol; } // Driver code public static void Main(String[] args) { int []val = { 6, 5, 8, 4, 7, 10, 9 }; int n = val.Length; Console.WriteLine(countSteps(val, n)); } } // This code contributed by Rajput-Ji |
Javascript
let status = [];function countSteps(val, n){ let sol = 0; let vec = [0]; vec.push(0); status[0] = -1; // Compute the number of steps for (let i = 1; i < n; ++i) { // Current status is to // delete in first step status[i] = 0; // Pop the indices while // condition is satisfied while (vec.length > 0 && val[vec.length-1] >= val[i]) { // Inserting the correct // step no to delete status[i] = Math.max(status[i],status[vec.length-1] + 1); vec.pop(); } if (vec.length==0) { // Status changed to not delete status[i] = -1; } // Pushing a new index in the vector vec.push(i); // Build the solution from // smaller to larger size sol = Math.max(sol, status[i] + 1); } return sol + 1;} let val = [ 6, 5, 8, 4, 7, 10, 9 ]; let n = val.length; console.log(countSteps(val, n));// This code is contributed by utkarshshirode02. |
Time Complexity: O(N), as we are using a loop to traverse N times. Where N is the number of elements in the array.
Auxiliary Space: O(100000), as we are using extra space for the array status
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