laydate 跨度_股票跨度问题

谁借莪１个温暖的怀抱￠ 2023-03-05 09:35 13阅读 0赞

laydate 跨度

**Problem statement:**

**问题陈述：**

The **stock span problem** is a financial problem where we have a series of ***n*** daily price quotes for a stock and we need to calculate the span of stock's price for all n days.

**该股跨度的问题**是，我们有一系列***n个***每日报价的股票的财务问题，我们需要计算股票的价格的跨度为所有n天。

The span ***Si*** of the stock's price on a given day ***i*** is defined as the maximum number of consecutive days just before the given day, for which the price of the stock on the current day is less than or equal to its price on the given day.

给定日***i***的股票价格跨度***Si***被定义为给定日之前连续的最大天数，为此，当日股票价格小于或等于给定日的价格。 。

For example, if an array of 7 days prices is given as \{100, 80, 60, 70, 60, 75, 85\}, then the span values for corresponding 7 days are \{1, 1, 1, 2, 1, 4, 6\}.

例如，如果将7天价格的数组指定为\{100，80，60，70，60，75，85\}，则对应的7天的跨度值为\{1、1、1、2、1、4 ，6\}。

**Input:**

**输入：**

The first line of input contains an integer ***T*** denoting the number of test cases. The first line of each test case is ***N***, ***N*** is the size of the array. The second line of each test case contains ***N*** input ***arr\[i\]***.

输入的第一行包含一个整数***T，***表示测试用例的数量。 每个测试用例的第一行是***N*** ， ***N***是数组的大小。 每个测试用例的第二行包含***N个***输入***arr \[i\]*** 。

**Output:**

**输出：**

For each testcase, print the span values for all days.

对于每个测试用例，请打印整天的跨度值。

**Examples:**

**例子：**

Input:	
        T = 1
        N = 6
        [10, 4, 5, 90, 120, 80]
        
        Output: 
        1 1 2 4 5 1
    
        Input:
        T = 1
        N = 7
        [2,4,5,6,7,8,9]
        
        Output: 
        1 2 3 4 5 6 7

**Solution Approach**

**解决方法**

**1) Brute Force Approach**

**1)蛮力法**

In this approach, we will use the *count* variable for each element and check all the elements before it, if the elements are smaller than or equal to the current element then increase the count variable.

在这种方法中，我们将对每个元素使用*count*变量，并检查它之前的所有元素，如果元素小于或等于当前元素，则增加count变量。

We will store each element's count in a vector and then print the elements.

我们将每个元素的计数存储在向量中，然后打印这些元素。

This approach will traverse the elements in the entire array hence time complexity will **be O(n\*n)**.

这种方法将遍历整个数组中的元素，因此时间复杂度将为**O(n \* n)** 。

**Pseudo Code:**

**伪代码：**

function(arr[],n):
    	//vector to store the elements which are <= to current element.
    	vector<int>v1
    
    	//1st element is always 1 as there is no other element 
    	v1.push_back(1)
    	for(int i=1;i<n;i++)
    		cnt=1;
    	// Traverse left while the next element   
    	// on left is smaller than price[i]  
    	for(int j = i - 1; (j >= 0) &&  (arr[i] >=arr[j]); j--)
    		cnt++;
    		v1.push_back(cnt);
    
    	for(int i=0;i<n;i++) //print the results.
    		cout<<v1[i]<
     Time Complexity: O(n*n) Space Complexity: (n)
     C++ Implementation:
     Output
     Enter number of test cases: 3
    Enter size of stock's price array: 5
    Enter stock prices: 1 2 5 9 2  
    Span Values: 1 2 3 4 1 
    Enter size of stock's price array: 9
    Enter stock prices: 1 2 3 4 5 6 7 8 9
    Span Values: 1 2 3 4 5 6 7 8 9 
    Enter size of stock's price array: 7       
    Enter stock prices: 100 99 98 97 96 95 94
    Span Values: 1 1 1 1 1 1 1 
    
     2) Stack based solution
     In this approach we will use stack with pair, we will store array element and its index, each time we iterate through array element. We check the previous next greater element, it the previous greater element is found at some location i, then we will take the difference between the current index element and the previous greater element.
     We will use a vector for each index to push the number of elements that are smaller than or equal to the current element.
     We will use a boolean variable flag, if the element is pushed according to condition then make variable flag true otherwise push 1 as only that element is valid and all element before is greater than it.
     Pseudo Code:
     Time Complexity for above approach: O(n) space Complexity for above approach: O(n)
     C++ Implementation:
     #include <bits/stdc++.h>
    using namespace std;
    
    typedef long long ll;
    
    int main()
    {
         
        ll t;
        cout << "Enter number of test cases: ";
        cin >> t;
    
        while (t--) {
         
            ll n; //size of stock array variable.
            cout << "Enter size of stock's price array: ";
            cin >> n;
    
            ll arr[n];
    
            cout << "Enter stock prices: ";
            for (ll i = 0; i < n; i++)
                cin >> arr[i];
            stack<pair<ll, ll> > st; //stack declaration.
            st.push({
          INT_MAX, -1 });
            vector<ll> v1;
            for (ll i = 0; i < n; i++) {
         
                bool flag = false;
                while (!st.empty()) {
         
                    // is stack top is greater than current element then
                    // put the difference value in the vector.
                    if (st.top().first > arr[i]) 
                    {
         
                        v1.push_back(i - st.top().second);
                        flag = true;
                        break;
                    }
                    else
                        st.pop(); //if top value is not greater than pop.
                }
                if (flag == false)
                    v1.push_back(1);
                st.push({
          arr[i], i });
            }
            cout << "Span Values: "; //print then span values.
            for (ll i = 0; i < n; i++)
                cout << v1[i] << " ";
            cout << "\n";
        }
        return 0;
    }
    
     Output
     Problem reference: https://leetcode.com/problems/online-stock-span/
    
    
    
    
    
    
       
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function(arr[],n):
    	//vector to store the elements which are <= to current element.
    	vector<int>v1
    
    	//1st element is always 1 as there is no other element 
    	v1.push_back(1)
    	for(int i=1;i<n;i++)
    		cnt=1;
    	// Traverse left while the next element   
    	// on left is smaller than price[i]  
    	for(int j = i - 1; (j >= 0) &&  (arr[i] >=arr[j]); j--)
    		cnt++;
    		v1.push_back(cnt);
    
    	for(int i=0;i<n;i++) //print the results.
    		cout<<v1[i]<
     Time Complexity: O(n*n) Space Complexity: (n)
     C++ Implementation: 
     Output 
     Enter number of test cases: 3
    Enter size of stock's price array: 5
    Enter stock prices: 1 2 5 9 2  
    Span Values: 1 2 3 4 1 
    Enter size of stock's price array: 9
    Enter stock prices: 1 2 3 4 5 6 7 8 9
    Span Values: 1 2 3 4 5 6 7 8 9 
    Enter size of stock's price array: 7       
    Enter stock prices: 100 99 98 97 96 95 94
    Span Values: 1 1 1 1 1 1 1 
    
     2) Stack based solution 
     In this approach we will use stack with pair, we will store array element and its index, each time we iterate through array element. We check the previous next greater element, it the previous greater element is found at some location i , then we will take the difference between the current index element and the previous greater element.
     We will use a vector for each index to push the number of elements that are smaller than or equal to the current element.
     We will use a boolean variable flag , if the element is pushed according to condition then make variable flag true otherwise push 1 as only that element is valid and all element before is greater than it.
     Pseudo Code: 
     Time Complexity for above approach: O(n) space Complexity for above approach: O(n)
     C++ Implementation: 
      # include  < bits/stdc++.h >
    using namespace std ;
    
    typedef long long ll ;
    
    int main ( )
    {
        ll t ;
        cout < < " Enter number of test cases:  " ;
        cin > > t ;
    
        while ( t - - ) {
            ll n ; //size of stock array variable.
            cout < < " Enter size of stock's price array:  " ;
            cin > > n ;
    
            ll arr [ n ] ;
    
            cout < < " Enter stock prices:  " ;
            for ( ll i = 0 ; i < n ; i + + )
                cin > > arr [ i ] ;
            stack < pair < ll , ll > > st ; //stack declaration.
            st . push ({ INT_MAX , - 1 }) ;
            vector < ll > v1 ;
            for ( ll i = 0 ; i < n ; i + + ) {
                bool flag = false ;
                while ( ! st . empty ( ) ) {
                    // is stack top is greater than current element then
                    // put the difference value in the vector.
                    if ( st . top ( ) . first > arr [ i ] ) 
                    {
                        v1 . push_back ( i - st . top ( ) . second ) ;
                        flag = true ;
                        break ;
                    }
                    else
                        st . pop ( ) ; //if top value is not greater than pop.
                }
                if ( flag = = false )
                    v1 . push_back ( 1 ) ;
                st . push ({ arr [ i ] , i }) ;
            }
            cout < < " Span Values:  " ; //print then span values.
            for ( ll i = 0 ; i < n ; i + + )
                cout < < v1 [ i ] < < "   " ;
            cout < < " \n " ;
        }
        return 0 ;
    }
    
     Output 
     Problem reference: https://leetcode.com/problems/online-stock-span/ 
    
    
    
    
    
    
       
        TOP Interview Coding Problems/Challenges Run-length encoding (find/print frequency of letters in a string) Sort an array of 0's, 1's and 2's in linear time complexity Checking Anagrams (check whether two string is anagrams or not) Relative sorting algorithm Finding subarray with given sum Find the level in a binary tree with given sum K Check whether a Binary Tree is BST (Binary Search Tree) or not 1[0]1 Pattern Count Capitalize first and last letter of each word in a line Print vertical sum of a binary tree Print Boundary Sum of a Binary Tree Reverse a single linked list Greedy Strategy to solve major algorithm problems Job sequencing problem Root to leaf Path Sum Exit Point in a Matrix Find length of loop in a linked list Toppers of Class Print All Nodes that don't have Sibling Transform to Sum Tree Shortest Source to Destination Path 评论和讨论 广告：您是博主吗？ 加入我们的Blogging论坛 。 请启用JavaScript以查看 由Disqus提供的 评论。

> 翻译自: [https://www.includehelp.com/icp/stock-span-problem.aspx][https_www.includehelp.com_icp_stock-span-problem.aspx]

laydate 跨度

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