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 the stock’s price for all N days. 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 its price on the given day.
Examples:
Input: N = 7, price[] = [100 80 60 70 60 75 85]
Output: 1 1 1 2 1 4 6
Explanation: Traversing the given input span for 100 will be 1, 80 is smaller than 100 so the span is 1, 60 is smaller than 80 so the span is 1, 70 is greater than 60 so the span is 2 and so on. Hence the output will be 1 1 1 2 1 4 6.Input: N = 6, price[] = [10 4 5 90 120 80]
Output:1 1 2 4 5 1
Explanation: Traversing the given input span for 10 will be 1, 4 is smaller than 10 so the span will be 1, 5 is greater than 4 so the span will be 2 and so on. Hence, the output will be 1 1 2 4 5 1.
Naive Approach: To solve the problem follow the below idea:
Traverse the input price array. For every element being visited, traverse elements on the left of it and increment the span value of it while elements on the left side are smaller
Below is the implementation of the above approach:
C++
// C++ program for brute force method// to calculate stock span values#include <bits/stdc++.h>using namespace std;// Fills array S[] with span valuesvoid calculateSpan(int price[], int n, int S[]){ // Span value of first day is always 1 S[0] = 1; // Calculate span value of remaining days // by linearly checking previous days for (int i = 1; i < n; i++) { S[i] = 1; // Initialize span value // Traverse left while the next element // on left is smaller than price[i] for (int j = i - 1; (j >= 0) && (price[i] >= price[j]); j--) S[i]++; }}// A utility function to print elements of arrayvoid printArray(int arr[], int n){ for (int i = 0; i < n; i++) cout << arr[i] << " ";}// Driver codeint main(){ int price[] = { 10, 4, 5, 90, 120, 80 }; int n = sizeof(price) / sizeof(price[0]); int S[n]; // Fill the span values in array S[] calculateSpan(price, n, S); // print the calculated span values printArray(S, n); return 0;}// This is code is contributed by rathbhupendra |
C
// C program for brute force method to calculate stock span// values#include <stdio.h>// Fills array S[] with span valuesvoid calculateSpan(int price[], int n, int S[]){ // Span value of first day is always 1 S[0] = 1; // Calculate span value of remaining days by linearly // checking previous days for (int i = 1; i < n; i++) { S[i] = 1; // Initialize span value // Traverse left while the next element on left is // smaller than price[i] for (int j = i - 1; (j >= 0) && (price[i] >= price[j]); j--) S[i]++; }}// A utility function to print elements of arrayvoid printArray(int arr[], int n){ for (int i = 0; i < n; i++) printf("%d ", arr[i]);}// Driver program to test above functionint main(){ int price[] = { 10, 4, 5, 90, 120, 80 }; int n = sizeof(price) / sizeof(price[0]); int S[n]; // Fill the span values in array S[] calculateSpan(price, n, S); // print the calculated span values printArray(S, n); return 0;} |
Java
// Java implementation for brute force method to calculate// stock span valuesimport java.util.Arrays;class GFG { // method to calculate stock span values static void calculateSpan(int price[], int n, int S[]) { // Span value of first day is always 1 S[0] = 1; // Calculate span value of remaining days by // linearly checking previous days for (int i = 1; i < n; i++) { S[i] = 1; // Initialize span value // Traverse left while the next element on left // is smaller than price[i] for (int j = i - 1; (j >= 0) && (price[i] >= price[j]); j--) S[i]++; } } // A utility function to print elements of array static void printArray(int arr[]) { System.out.print(Arrays.toString(arr)); } // Driver code public static void main(String[] args) { int price[] = { 10, 4, 5, 90, 120, 80 }; int n = price.length; int S[] = new int[n]; // Fill the span values in array S[] calculateSpan(price, n, S); // print the calculated span values printArray(S); }}// This code is contributed by Sumit Ghosh |
Python3
# Python3 program for brute force method to calculate stock span values# Fills list S[] with span valuesdef calculateSpan(price, n, S): # Span value of first day is always 1 S[0] = 1 # Calculate span value of remaining days by linearly # checking previous days for i in range(1, n, 1): S[i] = 1 # Initialize span value # Traverse left while the next element on left is # smaller than price[i] j = i - 1 while (j >= 0) and (price[i] >= price[j]): S[i] += 1 j -= 1# A utility function to print elements of arraydef printArray(arr, n): for i in range(n): print(arr[i], end=" ")# Driver program to test above functionprice = [10, 4, 5, 90, 120, 80]n = len(price)S = [None] * n# Fill the span values in list S[]calculateSpan(price, n, S)# print the calculated span valuesprintArray(S, n)# This code is contributed by Sunny Karira |
C#
// C# implementation for brute force method// to calculate stock span valuesusing System;class GFG { // method to calculate stock span values static void calculateSpan(int[] price, int n, int[] S) { // Span value of first day is always 1 S[0] = 1; // Calculate span value of remaining // days by linearly checking previous // days for (int i = 1; i < n; i++) { S[i] = 1; // Initialize span value // Traverse left while the next // element on left is smaller // than price[i] for (int j = i - 1; (j >= 0) && (price[i] >= price[j]); j--) S[i]++; } } // A utility function to print elements // of array static void printArray(int[] arr) { string result = string.Join(" ", arr); Console.WriteLine(result); } // Driver code public static void Main() { int[] price = { 10, 4, 5, 90, 120, 80 }; int n = price.Length; int[] S = new int[n]; // Fill the span values in array S[] calculateSpan(price, n, S); // print the calculated span values printArray(S); }}// This code is contributed by Sam007. |
PHP
<?php// PHP program for brute force method// to calculate stock span values// Fills array S[] with span valuesfunction calculateSpan($price, $n, $S){ // Span value of first // day is always 1 $S[0] = 1; // Calculate span value of // remaining days by linearly // checking previous days for ($i = 1; $i < $n; $i++) { // Initialize span value $S[$i] = 1; // Traverse left while the next // element on left is smaller // than price[i] for ($j = $i - 1; ($j >= 0) && ($price[$i] >= $price[$j]); $j--) $S[$i]++; } // print the calculated // span values for ($i = 0; $i < $n; $i++) echo $S[$i] . " ";; } // Driver Code $price = array(10, 4, 5, 90, 120, 80); $n = count($price); $S = array($n); // Fill the span values in array S[] calculateSpan($price, $n, $S);// This code is contributed by Sam007?> |
Javascript
<script> // Javascript implementation for brute force method // to calculate stock span values // method to calculate stock span values function calculateSpan(price, n, S) { // Span value of first day is always 1 S[0] = 1; // Calculate span value of remaining // days by linearly checking previous // days for (let i = 1; i < n; i++) { S[i] = 1; // Initialize span value // Traverse left while the next // element on left is smaller // than price[i] for (let j = i - 1; (j >= 0) && (price[i] >= price[j]); j--) S[i]++; } } // A utility function to print elements // of array function printArray(arr) { let result = arr.join(" "); document.write(result); } let price = [ 10, 4, 5, 90, 120, 80 ]; let n = price.length; let S = new Array(n); S.fill(0); // Fill the span values in array S[] calculateSpan(price, n, S); // print the calculated span values printArray(S);</script> |
Output
1 1 2 4 5 1
Time Complexity: O(N2)
Auxiliary Space: O(N)
The stock span problem using stack:
To solve the problem follow the below idea:
We see that S[i] on the day i can be easily computed if we know the closest day preceding i, such that the price is greater than on that day than the price on the day i. If such a day exists, let’s call it h(i), otherwise, we define h(i) = -1
The span is now computed as S[i] = i – h(i). See the following diagram
To implement this logic, we use a stack as an abstract data type to store the days i, h(i), h(h(i)), and so on. When we go from day i-1 to i, we pop the days when the price of the stock was less than or equal to price[i] and then push the value of day i back into the stack.
Follow the below steps to solve the problem:
- Create a stack of type int and push 0 in it
- Set the answer of day 1 as 1 and run a for loop to traverse the days
- While the stack is not empty and the price of st.top is less than or equal to the price of the current day, pop out the top value
- Set the answer of the current day as i+1 if the stack is empty else equal to i – st.top
- Push the current day into the stack
- Print the answer using the answer array
Below is the implementation of the above approach:
Note: We have to check also for a case when all the stock prices should be the same so therefore we have to just check whether the current stock price is bigger than the previous one or not. We will not pop from the stack when the current and previous stock prices are the same.
C++
// C++ linear time solution for stock span problem#include <bits/stdc++.h>using namespace std;// A stack based efficient method to calculate// stock span valuesvoid calculateSpan(int price[], int n, int S[]){ // Create a stack and push index of first // element to it stack<int> st; st.push(0); // Span value of first element is always 1 S[0] = 1; // Calculate span values for rest of the elements for (int i = 1; i < n; i++) { // Pop elements from stack while stack is not // empty and top of stack is smaller than // price[i] while (!st.empty() && price[st.top()] <= price[i]) st.pop(); // If stack becomes empty, then price[i] is // greater than all elements on left of it, // i.e., price[0], price[1], ..price[i-1]. Else // price[i] is greater than elements after // top of stack S[i] = (st.empty()) ? (i + 1) : (i - st.top()); // Push this element to stack st.push(i); }}// A utility function to print elements of arrayvoid printArray(int arr[], int n){ for (int i = 0; i < n; i++) cout << arr[i] << " ";}// Driver codeint main(){ int price[] = { 10, 4, 5, 90, 120, 80 }; int n = sizeof(price) / sizeof(price[0]); int S[n]; // Fill the span values in array S[] calculateSpan(price, n, S); // print the calculated span values printArray(S, n); return 0;} |
Java
// Java linear time solution for stock span problemimport java.util.ArrayDeque;import java.util.Arrays;import java.util.Deque;public class GFG { // A stack based efficient method to calculate // stock span values static void calculateSpan(int price[], int n, int S[]) { // Create a stack and push index of first element // to it Deque<Integer> st = new ArrayDeque<Integer>(); // Stack<Integer> st = new Stack<>(); st.push(0); // Span value of first element is always 1 S[0] = 1; // Calculate span values for rest of the elements for (int i = 1; i < n; i++) { // Pop elements from stack while stack is not // empty and top of stack is smaller than // price[i] while (!st.isEmpty() && price[st.peek()] <= price[i]) st.pop(); // If stack becomes empty, then price[i] is // greater than all elements on left of it, // i.e., price[0], price[1], ..price[i-1]. Else // price[i] is greater than elements after top // of stack S[i] = (st.isEmpty()) ? (i + 1) : (i - st.peek()); // Push this element to stack st.push(i); } } // A utility function to print elements of array static void printArray(int arr[]) { System.out.print(Arrays.toString(arr)); } // Driver code public static void main(String[] args) { int price[] = { 10, 4, 5, 90, 120, 80 }; int n = price.length; int S[] = new int[n]; // Fill the span values in array S[] calculateSpan(price, n, S); // print the calculated span values printArray(S); }}// This code is contributed by Sumit Ghosh |
Python3
# Python linear time solution for stock span problem# A stack based efficient method to calculate sdef calculateSpan(price, S): n = len(price) # Create a stack and push index of first element to it st = [] st.append(0) # Span value of first element is always 1 S[0] = 1 # Calculate span values for rest of the elements for i in range(1, n): # Pop elements from stack while stack is not # empty and top of stack is smaller than price[i] while(len(st) > 0 and price[st[-1]] <= price[i]): st.pop() # If stack becomes empty, then price[i] is greater # than all elements on left of it, i.e. price[0], # price[1], ..price[i-1]. Else the price[i] is # greater than elements after top of stack S[i] = i + 1 if len(st) == 0 else (i - st[-1]) # Push this element to stack st.append(i)# A utility function to print elements of arraydef printArray(arr, n): for i in range(0, n): print(arr[i], end=" ")# Driver program to test above functionprice = [10, 4, 5, 90, 120, 80]S = [0 for i in range(len(price)+1)]# Fill the span values in array S[]calculateSpan(price, S)# Print the calculated span valuesprintArray(S, len(price))# This code is contributed by Nikhil Kumar Singh (nickzuck_007) |
C#
// C# linear time solution for// stock span problemusing System;using System.Collections;class GFG { // a linear time solution for // stock span problem A stack // based efficient method to calculate // stock span values static void calculateSpan(int[] price, int n, int[] S) { // Create a stack and Push // index of first element to it Stack st = new Stack(); st.Push(0); // Span value of first // element is always 1 S[0] = 1; // Calculate span values // for rest of the elements for (int i = 1; i < n; i++) { // Pop elements from stack // while stack is not empty // and top of stack is smaller // than price[i] while (st.Count > 0 && price[(int)st.Peek()] <= price[i]) st.Pop(); // If stack becomes empty, then price[i] is // greater than all elements on left of it, // i.e., price[0], price[1], ..price[i-1]. Else // price[i] is greater than elements after top // of stack S[i] = (st.Count == 0) ? (i + 1) : (i - (int)st.Peek()); // Push this element to stack st.Push(i); } } // A utility function to print elements of array static void printArray(int[] arr) { for (int i = 0; i < arr.Length; i++) Console.Write(arr[i] + " "); } // Driver method public static void Main(String[] args) { int[] price = { 10, 4, 5, 90, 120, 80 }; int n = price.Length; int[] S = new int[n]; // Fill the span values in array S[] calculateSpan(price, n, S); // print the calculated span values printArray(S); }}// This code is contributed by Arnab Kundu |
Javascript
<script>// javascript linear time solution for stock span problem // A stack based efficient method to calculate // stock span values function calculateSpan(price , n , S) { // Create a stack and push index of first element // to it var st = []; st.push(0); // Span value of first element is always 1 S[0] = 1; // Calculate span values for rest of the elements for (var i = 1; i < n; i++) { // Pop elements from stack while stack is not // empty and top of stack is smaller than // price[i] while (st.length!==0 && price[st[st.length - 1]] <= price[i]) st.pop(); // If stack becomes empty, then price[i] is // greater than all elements on left of it, i.e., // price[0], price[1], ..price[i-1]. Else price[i] // is greater than elements after top of stack S[i] = (st.length===0) ? (i + 1) : (i - st[st.length - 1]); // Push this element to stack st.push(i); } } // A utility function to print elements of array function printArray(arr) { document.write(arr); } // Driver method var price = [ 10, 4, 5, 90, 120, 80 ]; var n = price.length; var S = Array(n).fill(0); // Fill the span values in array S calculateSpan(price, n, S); // print the calculated span values printArray(S);// This code contributed by Rajput-Ji</script> |
Output
1 1 2 4 5 1
Time Complexity: O(N). It seems more than O(N) at first look. If we take a closer look, we can observe that every element of the array is added and removed from the stack at most once.
Auxiliary Space: O(N) in the worst case when all elements are sorted in decreasing order.
The stock span problem using dynamic programming:
To solve the problem follow the below idea:
Since the same subproblems are called again, this problem has the Overlapping Subproblems property. S0, the recomputations of the same subproblems can be avoided by constructing a temporary array to store already calculated answers
- Store the answer for every index in an array and calculate the value of the next index using previous values
- i.e check the answer for previous element and if the value of the current element is greater than the previous element
- Add the answer to the previous index into current answer and then check the value of (previous index – answer of previous index)
- Check this condition repeatedly
Below is the implementation of the above approach:
C++
// C++ program for a linear time solution for stock// span problem without using stack#include <bits/stdc++.h>using namespace std;// An efficient method to calculate stock span values// implementing the same idea without using stackvoid calculateSpan(int A[], int n, int ans[]){ // Span value of first element is always 1 ans[0] = 1; // Calculate span values for rest of the elements for (int i = 1; i < n; i++) { int counter = 1; while ((i - counter) >= 0 && A[i] >= A[i - counter]) { counter += ans[i - counter]; } ans[i] = counter; }}// A utility function to print elements of arrayvoid printArray(int arr[], int n){ for (int i = 0; i < n; i++) cout << arr[i] << " ";}// Driver codeint main(){ int price[] = { 10, 4, 5, 90, 120, 80 }; int n = sizeof(price) / sizeof(price[0]); int S[n]; // Fill the span values in array S[] calculateSpan(price, n, S); // print the calculated span values printArray(S, n); return 0;} |
Java
// Java program for a linear time// solution for stock span problem// without using stackimport java.io.*;class GFG { // An efficient method to calculate // stock span values implementing the // same idea without using stack static void calculateSpan(int A[], int n, int ans[]) { // Span value of first element is always 1 ans[0] = 1; // Calculate span values for rest of the elements for (int i = 1; i < n; i++) { int counter = 1; while ((i - counter) >= 0 && A[i] >= A[i - counter]) { counter += ans[i - counter]; } ans[i] = counter; } } // A utility function to print elements of array static void printArray(int arr[], int n) { for (int i = 0; i < n; i++) System.out.print(arr[i] + " "); } // Driver code public static void main(String[] args) { int price[] = { 10, 4, 5, 90, 120, 80 }; int n = price.length; int S[] = new int[n]; // Fill the span values in array S[] calculateSpan(price, n, S); // print the calculated span values printArray(S, n); }}/* This code contributed by PrinciRaj1992 */ |
Python3
# Python3 program for a linear time# solution for stock span problem# without using stack# An efficient method to calculate# stock span values implementing# the same idea without using stackdef calculateSpan(A, n, ans): # Span value of first element # is always 1 ans[0] = 1 # Calculate span values for rest # of the elements for i in range(1, n): counter = 1 while ((i - counter) >= 0 and A[i] >= A[i - counter]): counter += ans[i - counter] ans[i] = counter# A utility function to print elements# of arraydef printArray(arr, n): for i in range(n): print(arr[i], end=' ') print()# Driver codeprice = [10, 4, 5, 90, 120, 80]n = len(price)S = [0] * (n)# Fill the span values in array S[]calculateSpan(price, n, S)# Print the calculated span valuesprintArray(S, n)# This code is contributed by Prateek Gupta |
C#
// C# program for a linear time// solution for stock span problem// without using stackusing System;public class GFG { // An efficient method to calculate // stock span values implementing the // same idea without using stack static void calculateSpan(int[] A, int n, int[] ans) { // Span value of first element is always 1 ans[0] = 1; // Calculate span values for rest of the elements for (int i = 1; i < n; i++) { int counter = 1; while ((i - counter) >= 0 && A[i] >= A[i - counter]) { counter += ans[i - counter]; } ans[i] = counter; } } // A utility function to print elements of array static void printArray(int[] arr, int n) { for (int i = 0; i < n; i++) Console.Write(arr[i] + " "); } // Driver code public static void Main(String[] args) { int[] price = { 10, 4, 5, 90, 120, 80 }; int n = price.Length; int[] S = new int[n]; // Fill the span values in array S[] calculateSpan(price, n, S); // print the calculated span values printArray(S, n); }}// This code has been contributed by 29AjayKumar |
Javascript
<script> // JavaScript program for the above approach; // An efficient method to calculate stock span values // implementing the same idea without using stack function calculateSpan(A, n, ans) { // Span value of first element is always 1 ans[0] = 1; // Calculate span values for rest of the elements for (let i = 1; i < n; i++) { let counter = 1; while ((i - counter) >= 0 && A[i] >= A[i - counter]) { counter += ans[i - counter]; } ans[i] = counter; } } // A utility function to print elements of array function printArray(arr, n) { for (let i = 0; i < n; i++) document.write(arr[i] + " "); } // Driver program to test above function let price = [10, 4, 5, 90, 120, 80]; let n = price.length; let S = new Array(n); // Fill the span values in array S[] calculateSpan(price, n, S); // print the calculated span values printArray(S, n); // This code is contributed by Potta Lokesh </script> |
Output
1 1 2 4 5 1
Time Complexity: O(N)
Auxiliary Space: O(N)
The stock span problem using two stacks:
Follow the below steps to solve the problem:
- In this approach, I have used the data structure stack to implement this task.
- Here, two stacks are used. One stack stores the actual stock prices whereas, the other stack is a temporary stack.
- The stock span problem is solved using only the Push and Pop functions of the Stack.
- Just to take input values, I have taken array ‘price’ and to store output, used array ‘span’.
Below is the implementation of the above approach:
C++
// C++ program for brute force method// to calculate stock span values#include <bits/stdc++.h>using namespace std;vector<int> calculateSpan(int arr[], int n){ // Your code here stack<int> s; vector<int> ans; for (int i = 0; i < n; i++) { while (!s.empty() and arr[s.top()] <= arr[i]) s.pop(); if (s.empty()) ans.push_back(i + 1); else { int top = s.top(); ans.push_back(i - top); } s.push(i); } return ans;}// A utility function to print elements of arrayvoid printArray(vector<int> arr){ for (int i = 0; i < arr.size(); i++) cout << arr[i] << " ";}// Driver codeint main(){ int price[] = { 10, 4, 5, 90, 120, 80 }; int n = sizeof(price) / sizeof(price[0]); int S[n]; // Function call vector<int> arr = calculateSpan(price, n); printArray(arr); return 0;}// This is code is contributed by Arpit Jain |
C
// C program for the above approach#include <limits.h>#include <stdio.h>#include <stdlib.h>#define SIZE 6// change size of stack from here// change this char to int if// you want to create stack of// int. rest all program will work finetypedef int stackentry;typedef struct stack { stackentry entry[SIZE]; int top;} STACK;// stack is initialized by setting top pointer = -1.void initialiseStack(STACK* s) { s->top = -1; }// to check if stack is full.int IsStackfull(STACK s){ if (s.top == SIZE - 1) { return (1); } return (0);}// to check if stack is empty.int IsStackempty(STACK s){ if (s.top == -1) { return (1); } else { return (0); }}// to push elements into the stack.void push(stackentry d, STACK* s){ if (!IsStackfull(*s)) { s->entry[(s->top) + 1] = d; s->top = s->top + 1; }}// to pop element from stack.stackentry pop(STACK* s){ stackentry ans; if (!IsStackempty(*s)) { ans = s->entry[s->top]; s->top = s->top - 1; } else { // '\0' will be returned if // stack is empty and of // char type. if (sizeof(stackentry) == 1) ans = '\0'; else // INT_MIN will be returned // if stack is empty // and of int type. ans = INT_MIN; } return (ans);}// The code for implementing stock// span problem is written// here in main function.int main(){ // Just to store prices on 7 adjacent days int price[6] = { 10, 4, 5, 90, 120, 80 }; // in span array , span of each day will be stored. int span[6] = { 0 }; int i; // stack 's' will store stock values of each // day. stack 'temp' is temporary stack STACK s, temp; // setting top pointer to -1. initialiseStack(&s); initialiseStack(&temp); // count basically signifies span of // particular day. int count = 1; // since first day span is 1 only. span[0] = 1; push(price[0], &s); // calculate span of remaining days. for (i = 1; i < 6; i++) { // count will be span of that particular day. count = 1; // if current day stock is larger than previous day // span, then it will be popped out into temp stack. // popping will be carried out till span gets over // and count will be incremented . while (!IsStackempty(s) && s.entry[s.top] <= price[i]) { push(pop(&s), &temp); count++; } // now, one by one all stocks from temp will be // popped and pushed back to s. while (!IsStackempty(temp)) { push(pop(&temp), &s); } // pushing current stock push(price[i], &s); // appending span of that particular // day into output array. span[i] = count; } // printing the output. for (i = 0; i < 6; i++) printf("%d ", span[i]);} |
Java
// Java program for brute force method// to calculate stock span valuesimport java.util.ArrayDeque;import java.util.ArrayList;import java.util.Deque;class GFG { static ArrayList<Integer> calculateSpan(int arr[], int n) { // Your code here Deque<Integer> s = new ArrayDeque<Integer>(); ArrayList<Integer> ans = new ArrayList<Integer>(); for (int i = 0; i < n; i++) { while (!s.isEmpty() && arr[s.peek()] <= arr[i]) s.pop(); if (s.isEmpty()) ans.add(i + 1); else { int top = s.peek(); ans.add(i - top); } s.push(i); } return ans; } // A utility function to print elements of array static void printArray(ArrayList<Integer> arr) { for (int i = 0; i < arr.size(); i++) System.out.print(arr.get(i) + " "); } // Driver code public static void main(String args[]) { int price[] = { 10, 4, 5, 90, 120, 80 }; int n = price.length; ArrayList<Integer> arr = calculateSpan(price, n); printArray(arr); }}// This is code is contributed by Lovely Jain |
Python3
# Python3 program for a linear time# solution for stock span problem# using stackdef calculateSpan(a, n): s = [] ans = [] for i in range(0, n): while(s != [] and a[s[-1]] <= a[i]): s.pop() if(s == []): ans.append(i+1) else: top = s[-1] ans.append(i - top) s.append(i) return ans# A utility function to print elements# of arraydef printArray(arr, n): for i in range(n): print(arr[i], end=' ') print()# Driver codeprice = [10, 4, 5, 90, 120, 80]n = len(price)ans = calculateSpan(price, n)# Print the calculated span valuesprintArray(ans, n)# This code is contributed by Arpit Jain |
C#
// C# program for brute force method// to calculate stock span valuesusing System;using System.Collections;using System.Collections.Generic;class GFG { public static List<int> calculateSpan(int[] arr, int n) { // Your code here Stack s = new Stack(); List<int> ans = new List<int>(); for (int i = 0; i < n; i++) { while (s.Count != 0 && arr[(int)s.Peek()] <= arr[i]) s.Pop(); if (s.Count == 0) ans.Add(i + 1); else { int top = (int)s.Peek(); ans.Add(i - top); } s.Push(i); } return ans; } // A utility function to print elements of array static void printArray(List<int> arr) { for (int i = 0; i < arr.Count; i++) Console.Write(arr[i] + " "); } // Driver code public static void Main(string[] args) { int[] price = { 10, 4, 5, 90, 120, 80 }; int n = price.Length; List<int> arr = calculateSpan(price, n); printArray(arr); }}// This code is contributed by karandeep1234 |
Javascript
// javascript program for brute force method // to calculate stock span values function calculateSpan(arr, n) { // Your code here let s =[]; let ans=[]; for(let i=0;i<n;i++) { while(s.length!=0 && arr[s[s.length-1]] <= arr[i]) s.pop(); if(s.length==0) ans.push(i+1); else { let top = s[s.length-1]; ans.push(i-top); } s.push(i); } return ans; } // Driver code let price = [ 10, 4, 5, 90, 120, 80 ]; let n = price.length; let arr = calculateSpan(price, n); document.write(arr); |
Output
1 1 2 4 5 1
Time Complexity: O(N), where N is the size of the array.
Auxiliary Space: O(N), where N is the size of the array.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
