给定非负整数的数组arr [] 。我们必须从此数组中删除元素，以使每个有效i的arr [i + 1]> arr [j]都将被计为一个步骤。我们必须应用相同的操作，直到数组严格减小为止。现在的任务是计算获得所需阵列所需的步骤数。
例子：
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
方法：想法是仅保留要针对特定元素进行检查的必需元素的索引。因此，我们使用向量仅存储所需的索引。我们将每个索引插入后面，如果满足以下条件，则从后面删除索引。
arr[vect.back()] ≥ val[i]
我们采用另一个数组，在该数组中我们更新特定元素要删除的步骤。
如果status [i] = -1，则不删除元素，0表示第一步，依此类推。这就是为什么我们将答案加1。
弹出索引时，我们反复更新元素的状态。如果弹出所有索引，即vect.size()= 0，则不删除此元素，因此将其状态更改为-1 。
下面是上述方法的实现：
CPP
// C++ implementation of the approach
#include
using namespace std;
int status[100000];
// Function to return the required
// number of steps
int countSteps(int* val, int n)
{
int sol = 0;
vector 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 code
int 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 approach
import java.util.*;
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;
Vector 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 code
public 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 AnkitRai01C#
// 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 vec = new List(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 输出：
2
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